Lakhmir Singh & Manjit Kaur 2020 Solutions for Class 10 Science Chapter 5 Refraction Of Light are provided here with simple stepbystep explanations. These solutions for Refraction Of Light are extremely popular among class 10 students for Science Refraction Of Light Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Lakhmir Singh & Manjit Kaur 2020 Book of class 10 Science Chapter 5 are provided here for you for free. You will also love the adfree experience on Meritnationâ€™s Lakhmir Singh & Manjit Kaur 2020 Solutions. All Lakhmir Singh & Manjit Kaur 2020 Solutions for class 10 Science are prepared by experts and are 100% accurate.
Page No 219:
Question 1:
If a ray of light goes from a rarer medium to a denser medium, will it bend towards the normal or away from it?
Answer:
If a ray of light goes from a rarer medium to a denser medium, it will bend towards the normal.
Page No 219:
Question 2:
If a ray of light goes from a denser medium to a rarer medium, will it bend towards the normal or away from the normal?
Answer:
If a ray of light goes form a denser medium to a rarer medium, it will bend away from the normal.
Page No 219:
Question 3:
A beam of light travelling in a rectangular glass slab emerges into air. Draw a raydiagram indicating the change in its path.
Answer:
When a beam of light travelling in a rectangular glass slab emerges into air, it bends away from the normal.
Page No 219:
Question 4:
A beam of light travelling in air is incident on water. Draw a raydiagram indicating the change in its path in water.
Answer:
When a beam of light travelling in air enters water, it bends towards the normal.
Page No 219:
Question 5:
A ray of light travelling in water emerges into air. Draw a raydiagram indicating the change in its path.
Answer:
When a ray of light travelling in water enters air, it bends away from the normal.
Page No 219:
Question 6:
A ray of light travelling in air is incident on a parallelsided glass slab (or rectangular glass slab). Draw a raydiagram indicating the change in its path in glass.
Answer:
When a ray of light travelling in air is incident on a parallel side glass slab, it bends towards the normal.
Page No 219:
Question 7:
A ray of light travelling in glass emerges into air. State whether it will bend towards the normal or away from the normal.
Answer:
We know that glass is a denser medium and air is a rarer medium. When a ray of light travels from a denser medium to a rarer medium, it bends away from the normal. So, the ray will bend away from the normal in the given case.
Page No 219:
Question 8:
A ray of light travelling in air enters obliquely into water. Does the ray light bend towards the normal or away from the normal? Why?
Answer:
We know that air is a rarer medium and water is a denser medium. When a ray of light goes from a rarer medium to a denser medium, it bends towards the normal. So, the light ray will bend towards the normal in the given case.
Page No 219:
Question 9:
A ray of light goes from water into air. Will it bend towards the normal or away from the normal?
Answer:
When a ray of light travels from a denser medium (water) to a rarer medium (air), it bends away from the normal. Therefore, the ray of light will bend away from the normal in the given case.
Page No 219:
Question 10:
State two effects caused by the refraction of light.
Answer:
Refraction of light can cause following two effects:
1. An object placed under water appears to be raised.
2. A stick held obliquely and partly immersed in water appears to be bent at the water surface.
Page No 219:
Question 11:
Name the phenomenon due to which a swimming pool appears less deep than it really is.
Answer:
This is due to the refraction of light.
Page No 219:
Question 12:
When a ray of light passes from air into glass, is the angle of refraction greater than or less than the angle of incidence?
Answer:
When a ray of light passes from air into glass, it bends towards the normal. So, the angle of refraction is smaller than the angle of incidence.
Page No 220:
Question 13:
A ray of light passes from air into a block of glass. Does it bend towards the normal or away from it?
Answer:
We know that air is a rarer medium and glass is a denser medium. When a ray of light goes from a rarer medium to a denser medium, it bends towards the normal. So, the light ray will bend towards the normal in the given case.
Page No 220:
Question 14:
As light rays pass from water into glass, are they refracted towards the normal or away from the normal?
Answer:
Glass is denser than water; therefore, if light rays pass from water into glass, the rays will be refracted towards the normal.
Page No 220:
Question 15:
In which material do you think light rays travel fasterglass or air?
Answer:
The speed of light in an optically rarer medium is more than the speed of light in a denser medium. Now, air is an optically rarer medium and glass is a denser medium. Therefore, light rays will travel fast in air.
Page No 220:
Question 16:
Which phenomenon of light makes the water to appear shallower than it really is?
Answer:
Refraction of light causes the water to appear shallower than it really is.
Page No 220:
Question 17:
State whether the following statement is true or false:
Refraction occurs because light slows down in denser materials.
Answer:
True.
The refraction of light is due to the change in the speed of light on going from one medium to other. The speed of light ray slows down in denser materials.
Page No 220:
Question 18:
Why does a ray of light bend when it travels from one medium to another?
Answer:
The bending of light, when it passes from one medium to another, is known as refraction. It occurs due to the change in the speed of light on going from one medium to another.
Page No 220:
Question 19:
Fill in the following blanks with suitable words:
(a) Light travelling along a normal is ...............refracted.
(b) Light bends when is passes from water into air. We say that it is ............
Answer:
(a) Light travelling along a normal is not refracted.
(b) Light bends when is passes from water into air. We say that it is refracted.
Page No 220:
Question 20:
What is meant by 'refraction of light'? Draw a labelled ray diagram to show the refraction of light.
Answer:
Refraction of light:
The change in direction of light when it passes from one medium to another obliquely is called refraction of light.
Here, the light ray changes its direction or refracts at point A when it travels from air to glass. The ray changes its direction again at point B when it travels from glass to air.
Page No 220:
Question 21:
A ray of light travelling in air is incident on a rectangular glass block and emerges out into the air from the opposite face. Draw a labelled ray diagram to show the completer path of this ray of light. Mark the two points where the refraction of light takes place. What can you say about the final direction of ray of light?
Answer:
Here, light ray changes its direction or refracts at point A when it travels from air to glass. The ray changes its direction again at point B when it travels from glass to air.
The final direction of light ray, i.e., BX will be parallel to the direction OAY in which it enters the glass block.
Page No 220:
Question 22:
Draw a labelled ray diagram to show how a ray of light is refracted when it passes:
(a) from air into an optically denser medium.
(b) from an optically denser medium into air.
Answer:
(a) Refraction of light ray when it passes from air to an optically denser medium:
(b) Refraction of light ray when it passes from an optically denser medium to air:
Page No 220:
Question 23:
The diagram given alongside shows a ray of light entering a rectangular block of glass.
(a) Copy the diagram and draw the normal at the point of entry.
(b) Draw the approximate path of the ray of light through the glass block and out of the other side.
Figure
Answer:
(a)
NN' is the normal at the point of entry.
(b) BC shows the path of light ray through the glass block and CD shows the path on the other side of the glass block.
Page No 220:
Question 24:
What is meant by the 'angle of incidence' and the 'angle of refraction' for a ray of light? Draw a labelled ray diagram to show the angle of incidence and the angle of refraction for a refracted ray of light.
Answer:
Angle of incidence:
The angle between the incident ray and the normal is called angle of incidence.
Angle of refraction:
The angle between the refracted ray and the normal is called angle of refraction.
Diagram for a refracted ray of light:
Here,
i = Angle of incidence
r = Angle of refraction
i' = Angle of incident at the emergent point
r' = Angle of refraction at the emergent point
Page No 220:
Question 25:
Light travels more quickly through water than through glass.
(a) Which is optically denser : water or glass?
(b) If a ray of light passes from glass into water, which way will it bend : towards the normal or away from the normal?
Answer:
(a) Glass is optically denser as compared to water.
(Speed of light in an optically rarer medium is more than the speed of light in a denser medium.)
(b) If a ray of light passes from glass to water, it will bend away from the normal because glass is a denser medium as compared to water.
Page No 220:
Question 26:
Draw a labelled ray diagram to show how a ray of light passes through a parallel sided glass block:
(a) if it hits the glass block at 90° (that is, perpendicular to the glass block)
(b) if it hits the glass block at an angle other than 90° (that is, obliquely to the glass block).
Answer:
(a) If the light ray hits the glass block at 90° (that is, perpendicular to the glass block):
(b) If it hits the glass block at an angle other than 90° (that is, obliquely to the glass block):
Page No 220:
Question 27:
When a light ray passes from air into glass, what happens to its speed? Draw a diagram to show which way the ray of light bends.
Answer:
When a ray of light travels from a rarer medium to a denser medium (air into glass), its speed decreases and it bends towards the normal.
Page No 220:
Question 28:
(a) Explain why, a stick half immersed in water appears to be bent at the surface. Draw a labelled diagram to illustrate your answer.
(b) A coin in a glass tumbler appears to rise as the glass tumbler is slowly filled with water. Name the phenomenon responsible for this effect.
Answer:
(a) When a stick is half immersed in water, it appears to be bent at the surface due to the refraction of light.
In the above figure, the half portion BO of stick AO is immersed in water and it appears to be bent at point B. The light ray OC coming from the lower end O of the stick passes from water to air and gets refracted away from the normal in the direction CX. Another ray OD is refracted in the direction DY. These two refracted rays, i.e, CX and DY meet at point I, when produced backwards. The point I is nearer to the water surface than point O. Therefore, a virtual image of end O of the stick is formed as I.
So, human eye at position E sees the end O at I; i.e., stick appears to be bent.
(b) A coin in a glass tumbler appears to rise as the glass tumbler is slowly filled with water. It occurs because of refraction.
Page No 220:
Question 29:
(a) With the help of a labelled diagram, explain why a tank full of water appears less deep than it actually is.
(b) Name the phenomenon due to which a pencil partly immersed in water and held obliquely appears to be bent at the water surface.
Answer:
(a) Refraction of light causes a tank full of water to appear less deep than it actually is.
Suppose, we have a tank of water as shown in the following figure.
Here, we take a point O at the bottom of the tank. We see this point because of the light rays coming from it. Now, a light ray OA from point O passes through water and enters air at point A. It gets refracted away from the normal in the direction AX. Similarly, another ray OB gets refracted at point B and bends away from the normal in the direction BY. These two refracted rays, i.e., AX and BY, meet at point I on producing them backwards. This point I (nearer to the water surface than O) is the image of point O.
Therefore, point O appears to be nearer at position I. Similarly, this can be applied to all points on the bottom of the tank. This causes the tank to appear less deep than it actually is.
(b) It is the phenomenon of refraction due to which a pencil partly immersed in water and held obliquely appears to be bent at the water surface.
Page No 220:
Question 30:
(a) With the help of a diagram, show how when light falls obliquely on the side of a rectangular glass slab, the emergent ray is parallel to the incident ray.
(b) Show the lateral displacement of the ray on the diagram.
(c) State two factors on which the lateral displacement of the emergent ray depends.
Answer:
(a) In the following figure, a light ray travelling in air is incident on the rectangular glass slab. It gets refracted and bends towards the normal. Again, a change in the direction takes place when the refracted ray travelling in glass emerges into air. Here, the light ray bends away from the normal. We see that the incident and emergent rays are parallel to each other. These rays are parallel because the extent of bending on the opposite and parallel faces of slab is equal and opposite.
(b) The perpendicular distance between the original path of the incident and emergent rays coming out of the glass slab is called lateral displacement of the emergent ray. It is shown in the above figure.
(c) Two factors on which the lateral displacement of the emergent ray depends:
1. Angle of incidence
2. Thickness of glass slab
Page No 220:
Question 31:
Explain with the help of a labelled ray diagram, why a pencil partly immersed in water appears to be bent at the water surface. State whether the bending of pencil will increase or decrease if water is replaced by another liquid which is optically more dense than water. Give reason for your answer.
Answer:
In the above figure, the portion BO of the pencil AO is immersed in water and it appears to be bent at point B. The light ray OC coming from the lower end O of the pencil passes from water to air and gets refracted away from the normal in the direction CX. Another ray OD is refracted in the direction DY. These two refracted rays, CX and DY, meet at point I, when they are produced backwards. Point I is nearer to the water surface than point O. Therefore, a virtual image of end O of the pencil is formed as point I.
So, human eye at position E sees the end O at point I; i.e., pencil appears to be bent.
If water is replaced by another liquid, which is optically denser than water, the bending of pencil will increase because an optically denser medium causes more refraction or more bending of light rays.
Page No 221:
Question 32:
Light travelling from a denser medium to a rarer medium along a normal to the boundary:
(a) is refracted towards the normal
(b) is refracted away from the normal
(c) goes along the boundary
(d) is not refracted
Answer:
(d) is not refracted
When a ray of light travels along the normal incident on the surface, it is not refracted.
Page No 221:
Question 33:
A ray of light passes from glass into air. The angle of refraction will be:
(a) equal to the angle of incidence
(b) greater than the angle of incidence
(c) smaller than the angle of incidence
(d) 45°
Answer:
(b) greater than the angle of incidence
Because when a light ray passes from denser medium to a rarer medium, it bends away from the normal.
Page No 221:
Question 34:
A ray of light travelling in air goes into water. The angle of refraction will be:
(a) 90°
(b) smaller than the angle of incidence
(c) equal to the angle of incidence
(d) greater than the angle of incidence
Answer:
(b) smaller than the angle of incidence
Because when a light ray passes from a rarer medium to a denser medium, it bends towards the normal.
Page No 221:
Question 35:
The speed of light in air is:
(a) 3 × 10^{8} cm/s
(b) 3 × 10^{8} mm/s
(c) 3 × 10^{8} km/s
(d) 3 × 10^{8} m/s
Answer:
The speed of light in air is:
(d) 3 × 10^{8} m/s
Page No 221:
Question 36:
When a ray of light travelling in glass enters into water obliquely:
(a) it is refracted towards the normal
(b) it is not refracted at all
(c) it goes along the normal
(d) it is refracted away from the normal
Answer:
(d) it is refracted away from the normal
Because if a ray of light goes form a denser medium (glass) to a rarer medium (water), it bends away from the normal.
Page No 221:
Question 37:
A ray of light travelling in water falls at right angles to the boundary of a parallelsided glass block. The ray of light:
(a) is refracted towards the normal
(b) is refracted away from the normal
(c) does not get refracted
(d) is reflected along the same path
Answer:
(c) does not get refracted
When a ray of light travels along the normal incident on the boundary separating two mediums, it is not refracted.
Page No 221:
Question 38:
A ray of light passes from a medium X to another medium Y. No refraction of light occurs if the ray of light hits the boundary of medium Y at an angle of:
(a) 0°
(b) 45°
(c) 90°
(d) 120°
Answer:
(c) 90°
If a ray of light travels along the normal, it is not refracted.
Page No 221:
Question 39:
Which of the following diagrams shows the ray of light refracted correctly?
Figure
Answer:
Diagram E shows that the ray of light is refracted correctly.
When a light ray goes from a rarer medium to a denser medium, it bends towards the normal. This principle is replicated in this diagram. .
Page No 221:
Question 40:
A vertical ray of light strikes the horizontal surface of some water:
(a) What is the angle of incidence?
(b) What is the angle of refraction?
Answer:
(a) Angle of incidence is ${0}^{0}$ because light ray is travelling along the normal and angle of incidence is the angle between the light ray and the normal.
(b) If the incident ray falls normally on the surface, no bending of light ray takes place. So, the angle of refraction is also ${0}^{0}$.
Page No 221:
Question 41:
How is the reflection of light ray from a plane mirror different from the refraction of light ray as it enters a block of glass?
Answer:
In case of reflection, the angle of reflection is equal to the angle of incidence. On the other hand, in case of refraction, the angle of refraction is not equal to the angle of incidence.
Page No 221:
Question 42:
How does the light have to enter the glass:
(a) to produce a large amount of bending?
(b) for no refraction to happen?
Answer:
(a) To produce a large amount of bending, the light ray has to enter the glass with a large angle of incidence.
(b) For no refraction, the light ray has to enter the glass perpendicularly.
Page No 221:
Question 43:
(a) How can you bend light away from the normal?
(b) How must light travel out of a substance if it is not going to be refracted?
Answer:
(a) We can bend light away from the normal by making the light to enter from a denser medium to a rarer medium.
(b) Light will not refract if it travels at the right angles to the surface of the substance.
Page No 221:
Question 44:
Draw and complete the following diagrams to show what happens to the beams of light as they enter the glass block and then leave it:
Figure
Answer:
When a beam of light rays enters the glass block, it gets refracted. It bends towards the normal. Also, when these light rays leave the block, they bend away from the normal.
When light rays fall normally on the surface of the glass block, there is no bending of rays; the rays travel straight.
Page No 221:
Question 45:
Why does a beam of light when it enters glass at an angle? Why does it not bend if it inters the glass at right angles?
Answer:
When a beam of light enters glass at an angle, the speed changes and therefore the direction of light changes; i.e., bending of light occurs.
When a beam of light falls at right angles to the surface of glass, all parts of light waves reach the glass at the same time, enter the glass at the same time and hence slow down at the same time. Due to this no change in direction of light takes place; i.e., bending of light does not take place.
Page No 227:
Question 1:
What name is given to the ratio of sine of angle of incidence to the sine of angle of refraction?
Answer:
The ratio of sine of angle of incidence to the sine of angle of refraction is a constant value. This value is called the refractive index of a medium.
Page No 227:
Question 2:
Write the relation between the angle of incidence and the angle of refraction for a medium.
Answer:
The relationship between the angles of incidence and refraction is given by Snell's law.
According to this law, the ratio of the sines of the angles of incidence and refraction is constant for a given pair of media.
$\frac{\mathrm{Sine}\mathrm{of}\mathrm{angle}\mathrm{of}\mathrm{incidence}\left(\mathrm{i}\right)}{\mathrm{Sine}\mathrm{of}\mathrm{angle}\mathrm{of}\mathrm{refraction}\left(\mathrm{r}\right)}=\mathrm{constant}(\mathrm{n})\phantom{\rule{0ex}{0ex}}\mathrm{This}\mathrm{constant}\mathrm{is}\mathrm{known}\mathrm{as}\mathrm{refractive}\mathrm{index}.$
Page No 228:
Question 3:
What is the unit of refractive index?
Answer:
Refractive index is a ratio of two similar quantities (the sines of angles); therefore, it has no units.
Page No 228:
Question 4:
Which has higher refraction index : water of glass?
Answer:
The refraction index of glass is higher than that of water. We know that a denser medium has high refractive index. Glass is denser than water; therefore, it has high refraction index.
Page No 228:
Question 5:
Refractive indices of carbon disulphide and ethyl alcohol are 1.63 and 1.36 respectively. Which is optically denser?
Answer:
Carbon disulphide is optically denser than ethyl alcohol. It is because a substance with high refractive index is optically denser than the other with low refractive index.
Page No 228:
Question 6:
The refractive index of diamond is 2.42. What is the meaning of this statement in relation to the speed of light?
Answer:
The refractive index of diamond is 2.42. This means that the ratio of the speed of light in air (or vacuum) to that in diamond is equal to 2.42.
Page No 228:
Question 7:
If the refractive index for light going from air to diamond be 2.42, what will be the refractive index for light going from diamond to air?
Answer:
Here,
${}^{air}n_{diamond}=2.42\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{}^{diamond}n_{air}=\frac{1}{{}^{air}{n}_{diamond}}=\frac{1}{2.42}=0.41\phantom{\rule{0ex}{0ex}}$
Page No 228:
Question 8:
How is the refractive index of a material related to the speed of light in it?
Answer:
The refractive index is the ratio of speeds of light in the two mediums.
The refractive index of medium 2 with respect to medium 1 is equal to the ratio of speeds of light in medium 1 and in medium 2.
${}_{1}{}^{2}\mathrm{n}=\frac{{\mathrm{v}}_{1}}{{\mathrm{v}}_{2}}\phantom{\rule{0ex}{0ex}}{v}_{1}=\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}1\phantom{\rule{0ex}{0ex}}{v}_{2}=\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}2$
Page No 228:
Question 9:
Fill in the following blank with a suitable word:
When a ray of light goes from air into a clear material, you see the ray bend. How much the ray bends is determined by the ............... of the material.
Answer:
When a ray of light goes from air into a clear material, you see the ray bend. How much the ray bends is determined by the refractive index of the material.
Page No 228:
Question 10:
Give three examples of materials that refract light rays. What happens to the speed of light rays when they enter these materials?
Answer:
Three examples of materials that refract light are glass, water and air.
The speed of light rays changes when they enter these materials.
Page No 228:
Question 11:
Define Snell's law of refraction. A ray of light is incident on a glass slab at an angle of incidence of 60°. If the angle of refraction be 32.7°, calculate the refractive index of glass. (Given : sin 60° = 0.866, and sin 32.7° = 0.540).
Answer:
According to Snell's law, the ratio of sines of the angles of incidence and refraction is constant for a given pair of mediums.
We get:
sin i /sin r = n (constant)
This constant is called refractive index.
According to the question:
Angle of incidence, i = 60°
Angle of refraction, r = 32.7°
Refractive index, n = ?
Applying the above formula, we get:
sin i / sin r = n
or, n = sin 60°/ sin 32.7°
= 0.866/0.540 = 1.60
Thus, the refractive index of glass is 1.60.
Page No 228:
Question 12:
The speed of light in vacuum and in two different glasses is given in the table below:
Medium  Speed of light 
Vacuum Flint glass Crown glass 
3.00 × 10^{8} m/s 1.86 × 10^{8} m/s 1.97 × 10^{8} m/s 
(a) Calculate the absolute refractive indexes of flint glass and crown glass.
(b) Calculate the relative refractive index for light going from crown glass to flint glass.
Answer:
(a) Absolute refractive index of flint glass = Speed of light in vacuum / Speed of light in flint glass
= (3.00 × 10^{8}) / (1.86 × 10^{8})
= 1.61
Absolute refractive index of crown glass = Speed of light in vacuum / Speed of light in crown glass
= (3.00 × 10^{8}) / (1.97 × 10^{8})
= 1.52
(b) Relative refractive index for light going from crown glass to flint glass is given by:
(Speed of light in crown glass) / (Speed of light in flint glass)
= (1.97 × 10^{8}) / (1.86 × 10^{8})
= 1.059
Page No 228:
Question 13:
The speed of light in air is 3 × 10^{8} m/s. In medium X its speed is 2 × 10^{8} m/s and in medium Y the speed of light is 2.5 × 10^{8} m/s Calculate:
(a) _{air}n_{x}
(b) _{air}n_{Y}
(c) _{x}n_{Y}
Answer:
(a) _{air}n_{x} = (Speed of light in air) / (Speed of light in medium X)
= (3 × 10^{8}) / (2 × 10^{8})
= 1.5
(b) _{air}n _{Y = }(Speed of light in air) / (Speed of light in medium â€‹Y)
= (3 × 10^{8}) / (2.5 × 10^{8})
= 1.2
(c) _{x}n_{Y}_{ = }(Speed of light in medium X) / (Speed of light in medium â€‹Y)
= (2 × 10^{8}) / (2.5 × 10^{8})
= 0.8
Page No 228:
Question 14:
What is the speed of light in a medium of refractive index $\frac{6}{5}$ if its speed in air is 3,00,000 km/s?
Answer:
Given:
Speed of light in air = 3,00,000 km/s
Refractive index of the medium = 6/5
Speed of light in the given medium = ?
Applying formula for refractive index, we get:
Refractive index of a medium = (Speed of light in air) / (Speed of light in medium)
or
(Speed of light in medium) = (Speed of light in air) / (Refractive index of a medium)
= 3,00,000 / (6/5)
= 2,50,000 km/s
Page No 228:
Question 15:
The refractive index of glass is 1.5. Calculate the speed of light in glass. The speed of light in air is 3.0 × 10^{8} ms^{−1}.
Answer:
Given:
Refractive index of glass = 1.5
Speed of light in air = 3.0 × 10^{8} m/s
Speed of light in glass = ?
Applying the formula for refractive index, we get:
Refractive index of a medium = (Speed of light in air)/(Speed of light in medium)
For glass:
Refractive index of glass = (Speed of light in air)/(Speed of light in glass)
(Speed of light in glass) = (Speed of light in air)/(Refractive index of glass)
= (3.0 × 10^{8})/1.5
= 2.0 × 10^{8} m/s
Page No 228:
Question 16:
The speed of light in water is 2.25 × 10^{8} m/s. If the speed of light in vacuum be 3 × 10^{8} m/s, calculate the refractive index of water.
Answer:
Given:
Speed of light in water = 2.25 × 10^{8} m/s
Speed of light in vacuum = 3 × 10^{8} m/s
Refractive index of water = ?
Applying the formula for refractive index, we get:â€‹
Refractive index of a medium = (Speed of light in vacuum) / (Speed of light in the medium)
For water:
Refractive index of water = (Speed of light in vacuum) / (Speed of light in water)
= (3 × 10^{8}) / (2.25 × 10^{8})
= 1.33
Thus, the refractive index of water is 1.33.
Page No 228:
Question 17:
Light enters from air into diamond which has a refractive index of 2.42. Calculate the speed of light in diamond. The speed of light in air is 3.0 × 10^{8} ms^{−1}.
Answer:
We know:
Refractive index = Speed of light in air/Speed of light in diamond
Let X be the speed of light in diamond. Now, we have:
2.42 = 3.0 × 10^{8} ms^{1 }/ X
X = 1.24 × 10^{8} ms^{1}
Page No 228:
Question 18:
(a) State and explain the laws of refraction of light with the help of a labelled diagram.
(b) What is meant by the refractive index of a substance?
(c) Light travels through air at 300 million ms^{−1}. On entering water it slows down to 225 million ms^{−1}. Calculate the refractive index of water.
Answer:
(a)
Laws of refraction of light:
(i) The incident ray, the refracted ray and the normal to the interface of two transparent mediums at the point of incidence, all lie in the same plane.
(ii) The ratio of sines of angles of incidence and refraction is a constant, for the light of a given colour and for a given pair of mediums. This law is also known as Snell’s law of refraction.
(b)
If i is the angle of incidence and r is the angle of refraction, we get:
$\frac{\mathrm{sin}\mathrm{i}}{\mathrm{sin}\mathrm{r}}$ = constant
This constant value is called the refractive index of the medium in which light enters through air. Refractive index of a medium can also be expressed in terms of speed of light as follows:
_{1}n_{2} = $\frac{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}1}{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}2}$
Here, _{1}n_{2} is the refractive index of medium 2 with respect to medium 1.
(c)
Here, mediums 1 and 2 are air and water, respectively.
Now,
Refractive index of water = Speed of light in air / Speed of light in water
= 300 million ms^{−1}^{ / }225 million ms^{−1}^{
= 1.33
Therefore, refractive index of water is 1.33.}
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Question 19:
The refractive indices of four substance P, Q, R and S are 1.50, 1.36, 1.77 and 1.31 respectively. The speed of light is the maximum in the substance:
(a) P
(b) Q
(c) R
(d) S
Answer:
(d) S
Explanation:
We know that:
Refractive index (n) = Speed of light in air / Speed of light in a medium
According to this formula, speed of light will be maximum in a substance whose refractive index is minimum.
Therefore, speed of light will be maximum in substance S whose refractive index is 1.31.
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Question 20:
The refractive indices of four materials A, B, C and D are 1.33, 1.43, 1.71 and 1.52 respectively. When the light rays pass from air into these materials, they refract the maximum in:
(a) material A
(b) material B
(c) material C
(d) material D
Answer:
(c) material C
Explanation:
The refraction in a material depends on its refractive index. Refractive index is calculated by $\frac{\mathrm{sin}\mathrm{i}}{\mathrm{sin}\mathrm{r}}$.
This ratio is maximum for material C; therefore, it produces maximum refraction.
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Question 21:
The refractive index of glass for light going from air to glass is $\frac{3}{2}$. The refractive index for light going from glass to air will be:
(a) $\frac{1}{3}$
(b) $\frac{4}{5}$
(c) $\frac{4}{6}$
(d) $\frac{5}{2}$
Answer:
(c) 4/6
Explanation:
Refractive index of material 2 with respect to material 1 is given by:
_{1}n_{2} = $\frac{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}1}{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}2}$
By the same argument, refractive index of medium 1 with respect to medium 2 is given by:
_{2}n_{1} = $\frac{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}2}{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}1}$ = 1/_{1}n_{2}
_{1}n_{2}_{ is $\frac{3}{2}$;} therefore, _{2}n_{1}_{ will be} 1/$\frac{3}{2}$.
Therefore, refractive index of light going from glass to air will be 4/6.
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Question 22:
The refractive indices of four media A, B, C and D are 1.44, 1.52, 1.65 and 1.36 respectively. When light travelling in air is incident in these media at equal angles, the angle of refraction will be the minimum:
(a) in medium A
(b) in medium B
(c) in medium C
(d) in medium D
Answer:
(c) in medium C
â€‹Refractive index = $\frac{\mathrm{sin}\mathrm{i}}{\mathrm{sin}\mathrm{r}}$
The angle of incidence is equal in all the cases; therefore, the refractive index will be maximum in the case of minimum angle of refraction. According to this argument, medium C will have minimum angle of refraction because it has maximum refractive index.
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Question 23:
The speed of light in substance X is 1.25 × 10^{8} m/s and that in air is 3 × 10^{8} m/s. The refractive index of this substance will be:
(a) 2.4
(b) 0.4
(c) 4.2
(d) 3.75
Answer:
(a) 2.4
Refractive index = $\frac{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{air}}{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{the}\mathrm{substance}}$
=$\frac{3\times {10}^{8}\mathrm{m}/\mathrm{s}}{1.25\times {10}^{8}\mathrm{m}/\mathrm{s}}$ = 2.4
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Question 24:
The refractive indexes of four substances P, Q, R and S are 1.77, 1.50, 2.42 and 1.31 respectively. When light travelling in air is incident on these substances at equal angles, the angle of refraction will be the maximum in:
(a) substance P
(b) substance Q
(c) substance R
(d) substance S
Answer:
(d) substance S
Refractive index = $\frac{\mathrm{sin}\mathrm{i}}{\mathrm{sin}\mathrm{r}}$
The value of (Sin i) is same in all the cases; therefore, the value of (Sin r) will be maximum for minimum refractive index. This means that the angle of refraction will be maximum for minimum refractive index and substance S has minimum refractive index.
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Question 25:
The refractive index of water is:
(a) 1.33
(b) 1.50
(c) 2.42
(d) 1.36
Answer:
(a) 1.33
Explanation:
Velocity of light in water = 225,563,010 m/s
Velocity of light in air = 300,000,000 m/s
Refractive index = $\frac{\mathrm{Velocity}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{air}}{\mathrm{Velocity}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{water}}$
= $\frac{300,000,000}{225,563,010}\phantom{\rule{0ex}{0ex}}=1.33$
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Question 26:
The refractive index of water with respect to air is $\frac{4}{3}$. The refractive index of air with respect to water will be:
(a) 1.75
(b) 0.50
(c) 0.75
(d) 0.25
Answer:
(c) 0.75
Explanation:
Refractive index of air with respect to water = $\frac{1}{\mathrm{Refractive}\mathrm{index}\mathrm{of}\mathrm{water}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{air}}$
Refractive index of air with respect to water = 3/4 = 0.75
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Question 27:
Refractive indices of water, sulphuric acid, glass and carbon disulphide are 1.33, 1.43, 1.53 and 1.63 respectively. the light travels slowest in:
(a) sulphuric acid
(b) glass
(c) water
(d) carbon disulphide
Answer:
(d) carbon disulphide
Explanation:
Refractive index = $\frac{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{air}}{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}}$
Speed of light in the medium is slowest; therefore refractive index will be maximum as the speed of light in air is constant. Thus, light will travel slowest in the substance with refractive index 1.63.
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Question 28:
The refractive index of glass with respect to air is $\frac{3}{2}$ and the refractive index of water with respect to air is $\frac{4}{3}$. The refractive index of glass with respect to water will be:
(a) 1.525
(b) 1.225
(c) 1.425
(d) 1.125
Answer:
(d) 1.125
Explanation:
Refractive index of glass with respect to water = $\frac{\mathrm{Refractive}\mathrm{index}\mathrm{of}\mathrm{glass}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{air}}{\mathrm{Refractive}\mathrm{index}\mathrm{of}\mathrm{water}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{air}}$
= $\frac{{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{4}{3}}}$
=1.125
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Question 29:
The following table gives the refractive indices of a few media:
1  2  3  4  5  
Medium :  Water  Crown glass  Rock salt  Ruby  Diamond 
Refractive index :  1.33  1.52  1.54  1.71  2.42 
(i) a medium pair so that light speeds up when it goes from one of these medium to another.
(ii) a medium pair so that light slows down when it goes from one of these medium to another.
Answer:
(1) Light speeds up as it travels from a denser medium to a rarer one, ie., from a medium with a higher refractive index to one with a lower refractive index. An example would be the medium pair of diamond and water, where light travels from diamond to water.
(2) Using the same argument, light slows down as it travels from a medium with a lower refractive index to one with a higher refractive index. Therefore, the medium pair of crown glass and ruby can be taken as an example, where light moves from the crown glass to the ruby.
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Question 30:
Refractive indices of four media A, B, C and D are given below:

Medium Refractive index A 1.33 B 1.44 C 1.52 D 1.65
Answer:
The refractive index of a medium is related to the speed of light as follows:
Refractive index =$\frac{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{vacuum}}{\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{a}\mathrm{medium}}$
Since the speed of light in vacuum is a constant, the refractive index becomes inversely proportional to the speed of light in a medium.
(1)The speed of light is maximum in a medium that has the lowest refractive index, ie., medium A .
(2) The speed of light is minimum in a medium that has the highest refractive index, ie., medium D.
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Question 1:
Name the lens which can concentrate sun's rays to a point and burn a hole in a piece of paper.
Answer:
A convex lens can concentrate the sun's rays to a point and burn a hole in a piece of paper.
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Question 2:
Give the usual name for the following:
A point inside a lens through which the light passes undeviated.
Answer:
The usual name for a point inside a lens through which light passes undeviated is the optical center .
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Question 3:
A 1 cm high object is placed at a distance of 2f from a convex lens. What is the height of the image formed?
Answer:
The height of the image formed is 1 cm. The reason being, when an object is placed at a distance of 2f from a convex lens, the size of the image formed is equal to the size of the object .
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Question 4:
If the image formed by a convex lens is of the same size as that of the object, what is the position of the image with respect to the lens?
Answer:
The image will be formed at a distance of 2F from the lens and behind it.
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Question 5:
If an object is placed at the focus of a convex lens, where is the image formed?
Answer:
For an object placed at the focus of a convex lens, the image is formed at infinity.
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Question 6:
Where should an object be placed in order to use a convex lens as a magnifying glass?
Answer:
The object should be placed at a distance that is less than F(focus) of the lens. The resultant image is virtual, erect and enlarged and formed on the same side as the object. From this theory, a convex lens can therefore, be used as a magnifying glass.
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Question 7:
Where should an object be placed in front of a convex lens so as to obtain its virtual, erect and magnified image?
Answer:
The object should be placed between the optical centre and the focus of a convex lens to obtain a virtual, erect and magnified image.
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Question 8:
Where should an object be placed in front of a convex lens so as to obtain its real, inverted and magnified image?
Answer:
The object should be placed between f and 2f of a convex lens to obtain a real, inverted and magnified image.
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Question 9:
For what position of an object a real, diminished image is formed by a convex lens?
Answer:
When an object is placed beyond 2F, the image formed is real and diminished.
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Question 10:
If an object is at a considerable distance (or infinity) in front of a convex lens, where is the image formed?
Answer:
The image will be formed at the focus.
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Question 11:
Draw the given diagram in your answer book and complete it for the path of a ray of light after passing through the lens.
Figure
Answer:
The ray will pass through the focus after refraction since it is parallel to the principle axis. See the diagram given below.
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Question 12:
What type of lens would you use as a magnifying glass? How close must the object be to the lens?
Answer:
For a magnifying glass, we use a convex lens. The object should be placed between the optical centre and the focus.
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Question 13:
Name two factors on which the focal length of a lens depends.
Answer:
Focal length depends on :
(1) Curvature of the lens
(2) Material medium of the lens
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Question 14:
State any two uses of convex lenses.
Answer:
Two uses of convex lenses are:
(1) They are used in telescopes.
(2) The convex lenses are used in magnifying glasses.They are used in the glasses of those affected with longsightedness.
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Question 15:
Fill in the following blanks with suitable words:
(a) Parallel rays of light are refracted by a convex lens to a point called the ........
(b) The image in a convex lens depends upon the distance of the ........... from the lens.
Answer:
(a) Parallel rays of light are refracted by a convex lens to a point called the focus.
(b) The image in a convex lens depends upon the distance of the object from the lens.
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Question 16:
What is a lens? Distinguish between a convex lens and concave lens. Which of the two is a converging lens : convex lens of concave lens?
Answer:
A lens is an optical device that transmits and refracts a ray of light, and converges or diverges a beam of light.
Convex Lens  Concave Lens 
1.A convex lens converges or concentrates the light rays to a point. 2.A convex lens has a real focus. 3.A convex lens has a positive focal length. 
1. A concave lens diverges the rays passing through it. 2.A concave lens has a virtual focus. 3.A concave lens has a negative focal length. 
A convex lens is a converging lens .
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Question 17:
(a) Explain with the help of a diagram, why the convex lens is also called a converging lens.
(b) Define principal axis, principal focus and focal length of a convex lens.
Answer:
A convex lense is outwardly curved and causes the light to pass through it and converge or concentrate to a point. Think of a magnifying glass that is used to burn something. The light that passes through it concentrates to a point, and this convergence is used to burn things. See the diagram given below.
(b) An imaginary line that passes through the optical center and the center of curvature of both faces of the lens, and is perpendicular to the faces is known as the principal axis.
Focus is a point on the principal axis where all the rays that are parallel to the principal axis meet after refraction from the lens.
The focal length may be defined as the distance between the optical centre and principal focus of the lens.
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Question 18:
(a) Explain with the help of a diagram, why the concave lens is also called a diverging lens.
(b) Define the principal focus of a concave lens.
Answer:
(a) A concave lense is curved inwards. Light passing through it diverges or spreads out. Think of a flashlight. The light from the small bulb passes through the lens and spreads out into a bigger beam on the other side. See the diagram given below.
(b)â€‹ The principle focus of a concave lens is a point on the principle axis where all the reflected rays parallel to the principle axis appear to diverge.
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Question 19:
Draw a ray diagram to show the formation of a real magnified image by a convex lens. (In your sketch the position of object and image with respect to the principal focus of lens should be shown clearly).
Answer:
A convex lens forms a real and magnified image when an object is placed between F and 2F.
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Question 20:
Describe with the help of a raydiagram, the formation of image of a finite object placed in front of a convex lens between f and 2f. Give two characteristics of the image so formed.
Answer:
The diagram shows the formation of the image of a finite object that is placed in front of a convex lens and between f and 2f:
Two characteristics of the image so formed are:
(1) It is Real
(2) It is Inverted
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Question 21:
Describe with the help of a ray diagram the nature, size and position of the image formed when an object is placed in front of a convex lens between focus and optical centre. State three characteristics of the image formed.
Answer:
Image formation diagram:
Characteristics of the image so formed:
1) On the same side as that of the object or behind the lens
2) Nature of image – virtual and erect
3) Size of Image – enlarged
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Question 22:
An object is placed at a distance equal to 2f in front of a convex lens. Draw a labelled ray diagram to show the formation of image. State two characteristics of the image formed.
Answer:
Image formation diagram:
Two characteristics of the image formed:
(1)Nature of image – real and inverted.
(2)Size of Image – same size as that of objectâ€‹.
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Question 23:
Describe with the help of a raydiagram, the size, nature and position of the image formed by a convex lens when an object is placed beyond 2f in front of the lens.
Answer:
Image formation diagram:
Image properties:
Image – between F and 2F
Nature of image – real and inverted
Size of Image – diminished
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Question 24:
Describe with the help of a ray diagram the nature, size and position of the image formed when an object is placed at infinity (considerable distance) in front of a convex lens. State three characteristics of the image so formed.
Answer:
Image formation diagram:
Image characteristics:
(1)Image – at F
(2)Nature of image – real and inverted
(3)Size of Image – point sized
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Question 25:
(a) What type of lens is shown in the diagram on the right? What will happen to the parallel rays of light? Show by completing the ray diagram.
(b) Your eye contains a convex lens. Why is it unwise to look at the sun?
Figure
Answer:
(a) It's a convex lens. The light rays will converge at a point called the focus of the lens.
(b) A convex lens converges a beam of light rays at a point called the focus. Therefore, it is unwise to look at the sun as it may damage the eye.
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Question 26:
Where must the object be placed for the image formed by a converging lens to be:
(a) real, inverted and smaller than the object?
(b) real, inverted and same size as the object?
(c) real, inverted and larger than the object?
(d) virtual, upright and larger than the object?
Answer:
For a convex lens:
(a) To form a real, inverted and smaller image than the object, the object should be beyond 2F .
(b) To form aâ€‹ real and inverted image, which is the same size as that of the object, the object should be at 2F.
(c) To form a real, inverted and larger image than the object, the object should be between F and 2F.
(d) To form a virtual, upright and larger image than the object, the object should be between F(focus) and the optical centre.
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Question 27:
Draw a diagram to show how a converging lens held close to the eye acts as a magnifying glass. Why is it usual to choose a lens of short focal length for this purpose rather than one of long focal length?
Answer:
Magnifying glass:
A lens with a short focal length is used so as to obtain greater magnification.
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Question 28:
How could you find the focal length of a convex lens rapidly but approximately?
Answer:
To determine the focal length of a convex lens, place the convex lens in a holder and keep it in front of a distant object like a tree. A cardboard screen is put behind the lens. Now, change the distance of the screen from the convex lens until a clear, inverted image of the tree is formed on the screen. Measure the distance of the screen from the lens. This distance will be the focal length of the convex lens. Here, we have used the fact that the image of an object at a far distance is formed at the focus of a convex lens.
Page No 240:
Question 29:
(a) With the help of a labelled diagram explain how a convex lens converges a beam of parallel light rays. Mark the principal axis, optical centre, principal focus and focal length of the convex lens on the diagram.
(b) State whether convex lens has a real focus or a virtual focus.
(c) List some things that convex lens and concave mirror have in common.
Answer:
(a) Suppose that a parallel beam of light rays falls on a convex lens as shown in the figure. These light rays are parallel to one another and also to the axis of the lens. The incident rays pass through the convex lens and get refracted according to the laws of refraction. All the rays, after passing through the convex lens, converge at the same point F, on the other side of the lens. The point F is called the principal focus of the convex lens. Thus, the point of convergence of the parallel beam of light rays to a single point is called the focus of the lens.
(b) A convex lens has a real focus because all the light rays actually pass through the focus.
(c) Some things that a convex lens and a concave lens have in common are as follows:
 Both converge a parallel beam of light rays to a single point called the focus.
 Both form real and inverted images of an object, except in one case.
 Both form a virtual, erect and magnified image when the object is placed between the pole and focus of a convex mirror, and between the centre of the lens and the focus, for a concave lens.
Page No 241:
Question 30:
(a) With the help of a labelled diagram, explain how a concave lens diverges a beam of parallel light rays. Mark the principal axis, optical centre, principal focus and focal length of the concave lens on the diagram.
(b) State whether concave lens has a real focus or a virtual focus.
(c) List some things that concave lens and concave mirror have in common.
Answer:
(a) When a parallel beam of light rays falls on a concave lens, the rays spread out (or diverge) after passing through the lens (figure). Since the refracted rays are diverging away from one another, they do not actually meet at a point. The diverging rays, when produced backwards appear to meet at a point F called the principal focus, on the left side of the lens.
(b) A concave lens has a virtual focus because the light rays do not actually pass through the focus.
(c) Few things which are common to both a concave lens and a convex mirror are as follows:
 Both diverge a parallel beam of light.
 Both form a virtual image at all times.
 Both have a virtual focus.
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Question 31:
Draw ray diagrams to represent the nature, position and relative size of the image formed by a convex lens for the object placed:
(a) at 2F_{1},
(b) between F_{1} and the optical centre O of the lens.
Which of the above two cases shows the use of convex lens as a magnifying glass? Give reasons for your choice.
Answer:
(a) When an object is placed at 2F_{1}_{}, the image so formed by the convex lens is at 2F. This image is real, inverted and of same size as the object as shown in the figure.
(b) When the object is placed between F_{1} and the optical centre O of the lens, the image formed by the convex lens is behind the object (left of the lens). This image is virtual, magnified and erect as shown in the figure.
Case (b) is used as a magnifying glass. The reason being, the image of an object that is placed between F_{1} and the optical centre is virtual and magnified, which is the requirement of a magnifying glass.
Page No 241:
Question 32:
(a) An object is placed well outside the principal focus of a convex lens. Draw a ray diagram to show how the image is formed, and say whether the image is real or virtual.
(b) What is the effect on the size and position of the image of moving the object (i) towards the lens, and (ii) away from the lens?
Answer:
(a) When an object is placed well outside the principal focus of a convex lens, the image formed is real and inverted as shown in the figure.
(b)
(i) When an object is moved towards the lens, the size of the image starts increasing. The position of the image starts moving away from the lens and towards infinity (on the right side of the lens) till the object is placed at the focus. When the object is moved closer to the lens, the image is formed behind the object (left side of the lens). This image is virtual, erect and magnified in size.
(ii) When the object is moved away from the lens, the image so formed moves closer to the lens and gets diminished in size.
Page No 241:
Question 33:
(a) Explain what is meant by a virtual, magnified image.
(b) Draw a ray diagram to show the formation of a virtual magnified image of an object by a convex lens. In your diagram, the position of object and image with respect to the principal focus should be shown clearly.
(c) Three convex lenses are available having focal lengths of 4 cm, 40 cm and 4 m respectively. Which one would you choose as a magnifying glass and why?
Answer:
(a) A virtual image is an image formed by a lens/mirror that cannot be taken on a screen. A magnified image implies that the size of the image formed is larger than the size of the object.
(b) When an object is placed between the focus and the optical centre of a convex lens, the image formed is virtual and magnified as shown in the figure.
(c) We will choose the lens with a focal length of 4 cm as the image formed will be more magnified (the smaller the focal length of a lens, the higher will be the magnification).
Page No 241:
Question 34:
(a) Explain why, a real image can be projected on a screen but a virtual image cannot.
(b) Draw a ray diagram to show the formation of a real diminished image of an object by a convex lens. In your diagram, the position of object and image with respect to the principal focus should be shown clearly.
(c) Name one simple optical instrument in which the above arrangement of convex lens is used.
Answer:
(a) A real image can be projected on a screen unlike a virtual image because a real image is formed by the actual meeting of light rays. A virtual image, on the other hand, is not formed by the actual meeting of light rays.
(b) When an object placed is beyond 2F', the image formed by the convex lens is real and diminished, as shown in the figure.
(c) A camera works on the above arrangement because the camera lens produces a small, real and inverted image of an object on the film.
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Question 35:
A convex lens has a focal length of 10 cm. At which of the following position should an object be placed so that this convex lens may act as a magnifying glass?
(a) 15 cm
(b) 7 cm
(c) 20 cm
(d) 25 cm
Answer:
(b) 7 cm,
since the image of an object placed between the focus and the optical centre of a convex lens is enlarged and virtual.
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Question 36:
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay,
since it is opaque and does not let light to pass through it.
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Question 37:
A small bulb is placed at the focal point of a converging lens. When the bulb is switched on, the lens produces:
(a) a convergent beam of light
(b) a divergent beam of light
(c) a parallel beam of light
(d) a patch of coloured light
Answer:
(c) A parallel beam of light
The reason being, a beam of light coming from the focus of a converging lens becomes parallel, after refraction from the lens.
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Question 38:
An illuminated object is placed at a distance of 20 cm from a converging lens of focal length 15 cm. The image obtained on the screen is:
(a) upright and magnified
(b) inverted and magnified
(c) inverted and diminished
(d) upright and diminished
Answer:
(b) Inverted and magnified
For an object placed between the focal length and twice the focal length of a converging lens, the image formed is real, inverted and magnified.
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Question 39:
An object is placed f and 2f of a convex lens. Which of the following statements correctly describes its image?
(a) real, larger than the object
(b) erect, smaller than the object
(c) inverted, same size as object
(d) virtual, larger than the object
Answer:
(a) Real, larger than the object
The reason being, the image of an object placed between f and 2f of a convex lens is real, inverted and magnified.
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Question 40:
Which of the following can make a parallel beam of light when light from a bulb falls on it?
(a) concave mirror as well as concave lens
(b) convex mirror as well as convex lens
(c) concave mirror as well as convex lens
(d) convex mirror as well as concave lens
Answer:
(c) Concave mirror as well as Convex lens
When an object is placed at the focus of a concave mirror (convex lens), the reflected (refracted) light rays are always parallel to each other and to the principal axis.
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Question 41:
In order to obtain a real image twice the size of the object with a convex lens of focal length 15 cm, the object distance should be:
(a) more than 5 cm but less than 10 cm
(b) more than 10 cm but less than 15 cm
(c) more than 15 cm but less than 30 cm
(d) more than 30 cm but less than 60 cm
Answer:
(c) More than 15 cm but less than 30 cm
For an object placed between F and 2F of a convex lens, the image formed is real and enlarged.
Page No 242:
Question 42:
A converging lens is used to produce an image of an object on a screen,object on a screen. What change is needed for the image to be formed nearer to the lens?
(a) increase the focal length of the lens
(b) insert a diverging lens between the lens and the screen
(c) increase the distance of the object from the lens
(d) move the object closer to the lens
Answer:
(c) Increase the distance of the object from the lens
As the object moves away from the lens, the image gets closer to the lens.
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Question 43:
A convex lens of focal length 8 cm forms a real image of the same size as the object. The distance between object and its image will be:
(a) 8 cm
(b) 16 cm
(c) 24 cm
(d) 32 cm
Answer:
(d) 32 cm
A convex lens forms a real image at 2f on the right side of the lens. The size of the image is equal to that of the object if the object is placed at 2f.
∴ Distance between the image and the object = 2f + 2f = 4f = 4 $\times $ 8 = 32 cm
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Question 44:
A virtual, erect and magnified image of an object is to be obtained with a convex lens. For this purpose,the object should be placed:
(a) between 2F and infinity
(b) between F and optical centre
(c) between F and 2F
(d) at F
Answer:
(b) between F and optical centre
When an object is placed between F and the optical centre, the image of the object formed by convex lens is virtual, erect and magnified.
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Question 45:
A burning candle whose flame is 1.5 cm tall is placed at a certain distance in front of a convex lens. An image of candle flame is received on a white screen kept behind the lens. The image of flame also measures 1.5 cm. If f is the focal length of convex lens, the candle is placed:
(a) at f
(b) between f and 2f
(c) at 2f
(d) beyond 2f
Answer:
(c) at 2f
This is because a convex lens produces an image of the same size as the object when the object is placed at 2f.
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Question 46:
A lens of focal length 12 cm forms an erect image three times the size of the object. The distance between the object and image is:
(a) 8 cm
(b) 16 cm
(c) 24 cm
(d) 36 cm
Answer:
(b) 16 cm
Given:
Magnification, m = 3
Focal length f = 12 cm
Image distance v = ?
Object distance u = ?
We know that:
m = $\frac{v}{u}$
Therefore
3 = $\frac{\mathrm{v}}{\mathrm{u}}$
3u = v
Putting these values in lens formula, we get:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3u}\frac{1}{u}=\frac{1}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{13}{3u}=\frac{1}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{3u}=\frac{1}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow 3u=24$
$\Rightarrow u=\frac{24}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow u=6\mathrm{cm}$
v = 3u = $$8 $\times $ 3 = $$24 cm
Here, minus sign show that image is formed on the left side of the lens.
Distance between image and object = 24 $$ 8 = 16 cm
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Question 47:
If an object is placed 21 cm from a converging lens, the image formed is slightly smaller than the object. If the object is placed 19 cm from the lens, the image formed is slightly larger than object. The approximate focal length of the lens is:
(a) 5 cm
(b) 10 cm
(c) 18 cm
(d) 20 cm
Answer:
(b) 10 cm
We know that a converging lens forms an image of same size as object when object is placed at a distance of 2f from the lens. It is given that the image is smaller than the object if object is kept at a distance of 21 cm. Similarly, the image is bigger than the object if object is kept at a distance of 19 cm. Therefore, at 20 cm, the distance should be 2f. This means that the focal length is approximately 10 cm.
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Question 48:
An object is placed at the following distance from a convex lens of focal length 15 cm:
(a) 35 cm
(b) 30 cm
(c) 20 cm
(d) 10 cm
Which position of the object will produce:
(i) a magnified real image?
(ii) a magnified virtual image?
(iii) a diminished real image?
(iv) an image of same size as the object?
Answer:
(a) The object should be placed at 20 cm because a convex mirror forms a real, magnified image when an object is placed between f and 2f.
(b) The object should be placed at 10 cm because a convex mirror forms a virtual, magnified image when an object is placed placed between f (focus) and the optic centre.
(c) The object should be placed at 35 cm because a convex mirror forms a real, diminished image when an object is placed beyond 2f.
(d) The object should be placed at 30 cm because a convex mirror forms a real image of the same size when an object is placed at 2f.
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Question 49:
When an object is placed at a distance of 36 cm from a convex lens, an image of the same size as the object is formed. What will be the nature of image formed when the object is placed at a distance of:
(a) 10 cm from the lens?
(b) 20 cm from the lens?
Answer:
(a) The image will be virtual and magnified. We know that a convex lens forms the image of the same size as that of the object when the object is placed at 2f. Thus, the focal length of the lens is 13 cm. In this case, the object is placed between the focusâ€‹ (f) and the optic centre. This position of the object results in the formation of a virtual and magnified image.
(b) The image will be real and magnified because the object is placed between 2f and f.
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Question 50:
(a) Draw a diagram to show how a converging lens focusses parallel rays of light?
(b) How would you alter the above diagram to show how a converging lens can produce a beam of parallel rays of light.
Answer:
(a) When rays of light from a distant object pass through a converging lens, the light rays converge at the focus of lens (fig.).
(b) When rays of light come from the focus of lens, the emergent rays of light get parallel to the principal axis.
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Question 1:
Write the formula for a lens connecting image distance (v), object distance (u) and the focal length (f). How does the lens formula differ from the mirror formula?
Answer:
Lens formula is:
$\frac{1}{\mathrm{v}}\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
Now, the mirror formula is given by:
$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
In both the formulas,
u = Object distance
v = Image distance
f = Focal length
In mirror formula, positive sign is present between the reciprocals of image distance and object distance .
In lens formula, negative sign is present between the reciprocals of image distance and object distance.
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Question 2:
Write down the magnification formula for a lens in terms of object distance and image distance. How does this magnification formula for a lens differ from the corresponding formula for a mirror?
Answer:
Magnification formula for a lens is given by:
$\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\mathrm{Magnification}\mathrm{formula}\mathrm{for}\mathrm{a}\mathrm{mirror}\mathrm{is}\mathrm{given}\mathrm{by}:\phantom{\rule{0ex}{0ex}}\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{both}\mathrm{the}\mathrm{formulas},\phantom{\rule{0ex}{0ex}}\mathrm{m}=\mathrm{Magnification}\phantom{\rule{0ex}{0ex}}\mathrm{v}=\mathrm{Image}\mathrm{distance}\phantom{\rule{0ex}{0ex}}\mathrm{u}=\mathrm{Object}\mathrm{distance}$
There is a difference of negative sign between the lens formula and the mirror formula. In mirror magnification formula, negative sign is present, whereas in lens magnification formula, this negative sign is not present.
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Question 3:
What is the nature of the image formed by a convex lens if the magnification produced by the lens is +3?
Answer:
If the magnification produced by the lens is +3, the image will be virtual and erect. It is because if the magnification has a positive value, the image is always virtual and erect.
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Question 4:
What is the nature of the image formed by a convex lens if the magnification produced by the lens is, – 0.5?
Answer:
If the magnification produced by the lens is 0.5, the image will be real and inverted. It is because if the magnification produced by a lens has a negative value, the image is always real and inverted.
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Question 5:
What is the position of image when an object is placed at a distance of 10 cm from a convex lens of focal length 10 cm?
Answer:
Given:
Object distance, u = $$10 cm
(Since object is always placed on the left side of lens)
Focal length, f = +10 cm
According to lens formula:
$\frac{1}{\mathrm{v}}\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{v}}=\frac{1}{10}+(\frac{1}{10})=0\phantom{\rule{0ex}{0ex}}\mathrm{v}=\frac{1}{0}=\infty \phantom{\rule{0ex}{0ex}}$
Therefore, the position of image will be at infinity.
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Question 6:
Describe the nature of image formed when an object is placed at a distance of 30 cm from a convex lens of focal length 15 cm.
Answer:
Given:
u = $$30 cm
(Since the object is placed on the left side of lens)
Focal length f = +15 cm
According to lens formula:
$\frac{1}{\mathrm{v}}\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}=\frac{1}{15}+(\frac{1}{30})\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{v}=30\mathrm{cm}\phantom{\rule{0ex}{0ex}}$
Thus, the image is formed at a distance of 30 cm from the convex lens. The plus sign for image distance shows that the image is formed on the right side of the convex lens. Only real and inverted image is formed on the right side of a convex lens; therefore, the image is real and inverted.
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Question 7:
At what distance from a converging lens of focal length 12 cm must an object be placed in order that an image of magnification 1 will be produced?
Answer:
Magnification 1 means that the image distance is the same as the object distance and the size of image is the same as the object.
This happens only if the position of the object is at 2f, i.e, 2 $\times $ focal length.
Thus, the object must be placed at a distance of 24 cm (2 $\times $ 12) from the converging lens.
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Question 8:
State and explain the New Cartesian Sign Convention for spherical lenses.
Answer:
New Cartesian Sign Convention for spherical lenses:
New Cartesian Sign Convention is used for measuring various distances in the ray diagrams of lenses.
According to this convention:
I. Object is always placed on the left of the lens; i.e., light must fall on the lens from left to right.
II. All distances parallel to the principal axis are measured from the optical centre of the lens.
III. The distances along the direction of incident rays (along positive xaxis) are taken as positive, whereas the distances opposite to the direction of incident rays (along negative xaxis) are taken as negative.
IV. Distances measured above the principal axis (along positive yaxis) are taken as positive.
V. Distances measured below the principal axis (along negative yaxis) are taken as negative.
Page No 246:
Question 9:
An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.
Answer:
Given:
Object distance, u = $$10 cm (It is to the left of the lens.)
Focal length, f = + 20 cm (It is a convex lens.)
Putting these values in the lens formula, we get:
1/v $$ 1/u = 1/f (v = Image distance)
1/v $$ 1/($$10) = 1/20
or, v = $$20 cm
Thus, the image is formed at a distance of 20 cm from the convex lens (on its left side).
Only a virtual and erect image is formed on the left side of a convex lens. So, the image formed is virtual and erect.
Now,
Magnification, m = v/u
m = $$20 / ($$10) = 2
Because the value of magnification is more than 1, the image will be larger than the object.
The positive sign for magnification suggests that the image is formed above principal axis.
Height of the object, h = +4 cm
magnification m=h'/h (h=height of object)
Putting these values in the above formula, we get:
2 = h'/4 (h' = Height of the image)
h' = 8 cm
Thus, the height or size of the image is 8 cm.
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Question 10:
A small object is so placed in front of a convex lens of 5 cm focal length that a virtual image is formed at a distance of 25 cm. Find the magnification.
Answer:
Given:
Focal length, f = 5 cm
Image distance, v = $$25 cm (Image is virtual.)
Applying the lens formula, we get:
1/v $$ 1/u = 1/f (u = Object distance)
1/($$25) $$ 1/u = 1/5
or, 1/u =1/(25)1/5
or, 1/u =6/25
or,
u = $$25/6 cm
Now,
Magnification, m = v/u
= ($$25)/($$25/6)
= +6
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Question 11:
Find the position and nature of the image of an object 5 cm high and 10 cm in front of a convex lens of focal length 6 cm.
Answer:
Given:
Height of the object, h = 5 cm
Object distance, u = $$10 cm
Focal length, f = 6 cm
Applying lens formula, we get:
1/v $$ 1/u = 1/f (v = Image distance)
1/v $$ 1/($$10) = 1/6
or, v = 15 cm
Thus, the image is formed at a distance of 15 cm behind the convex lens (on the right side).
Now
Magnification, m = v/u = h'/h
m = 15/($$10)
m = $$1.5
The value of magnification is negative; therefore, the image will be real and inverted.
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Question 12:
Calculate the focal length of a convex lens which produces a virtual image at a distance of 50 cm of an object placed 20 cm in front of it.
Answer:
Given:
Object distance, u = $$20 cm
Image distance, v = $$50 cm (Image is virtual.)
Applying lens formula, we get:
1/v $$ 1/u = 1/f (f = focal distance)
1/($$50) $$ 1/ ($$20) =1/f
or, 1/f = (2+5)/100
or, 1/f =3/100
or, f = 100/3 = + 33.3 cm
Thus, the focal length of convex lens is 33.3 cm.
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Question 13:
An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm.
(i) What is the nature of image?
(ii) What is the position of image?
Answer:
Given:
Object distance, u = $$100 cm
Focal length, f = 40 cm
Applying lens formula, we get:
1/v $$1/u = 1/f
1/v $$ 1/($$100) = 1/40
or, 1/v = 1/40  1/100
or, 1/v = 6/400
or, 1/v = 3/200
or, v = 200/3 = +66.6 cm
Now
Magnification, m = v/u
m= (200/3)/($$100)
m= $$2/3
(i) The value of magnification is negative, therefore, the image will be real and inverted.
(ii) The value of v is (+66.6 cm); therefore, the image is formed 66.6 cm behind the convex lens.
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Question 14:
A convex lens produces an inverted image magnified three times of an object placed at a distance of 15 cm from it. Calculate focal length of the lens.
Answer:
Given:
Object distance, u = $$15 cm
Magnification, m = $$3 (Image is inverted.)
Applying magnification formula, we get:
m = v/u = $$3
or, v/($$15) = $$3
or, v = 45 cm
Applying lens formula, we get:
1/v $$ 1/u = 1/f
1/45 $$ 1/($$15) = 1/f
1/f = ( 1+3 )/45
1/f = 4/45
or f = 11.2 cm
Hence, focal length of the lens is 11.2 cm.
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Question 15:
A converging lens of focal length 5 cm is placed at a distance of 20 cm from a screen. How far from the lens should an object be placed so as to form its real image on the screen?
Answer:
Given:
Focal length, f = 5 cm
Image distance, v = Distance of the lens from the screen = +20 cm (for real image)
Object distance, u = ?
Applying lens formula, we get:
1/v $$ 1/u = 1/f
1/20 $$ 1/u = 1/5
Or,
1/u = 1/20 $$ 1/5 = $$3/20
Or,
u = $$20/3 = $$6.6 cm
Thus, to form a real image, the object should be placed at a distance of 6.6 cm from the lens.
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Question 16:
An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.
Answer:
Given:
Object distance, u = $$25 cm
Focal length, f = 10 cm
Height of the object, h = 5 cm
Applying the lens formula, we get:
$\frac{1}{f}=\frac{1}{v}\frac{1}{u}$
$\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{10}+\frac{1}{25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{52}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{3}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{50}{3}=+16.6\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}16.6\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{lens}.\phantom{\rule{0ex}{0ex}}m=\frac{v}{u}=\frac{50}{3}\phantom{\rule{0ex}{0ex}}m=\frac{{h}^{,}}{h}=\frac{{h}^{,}}{5}\phantom{\rule{0ex}{0ex}}\frac{{h}^{,}}{5}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}{h}^{,}=\frac{10}{3}=3.3\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}3.3\mathrm{cm}\mathrm{high}.\mathrm{It}\mathrm{will}\mathrm{be}\mathrm{real}\mathrm{and}\mathrm{inverted}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Page No 247:
Question 17:
At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side? What will be the magnification produced in this case?
Answer:
Given:
Focal length, f = 18 cm
Image distance,v = 24 cm
Putting these values in lens formula, we get:
1/v $$ 1/u = 1/f
or, 1/u = 1/v $$1/f
or, 1/u = 1/24 $$1/18 = $$(1/72)
or, u = $$72 cm
Thus, the object should be placed at a distance of 72 cm from the lens.
Now
Magnification, m = v /u = 24/ ($$72) = $$(1/3)
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Question 18:
An object 2 cm tall is placed on the axis of a convex lens of focal length 5 cm at a distance of 10 m from the optical centre of the lens. Find the nature, position and size of the image formed. Which case of image formation by convex lenses is illustrated by this example?
Answer:
Given:
Size of object, h = 2 cm
Focal length, f = 5 cm
Object distance, u = $$10 m = $$1000 cm
Applying lens formula, we get:
1/v $$ 1/u = 1/f
1/v = 1/f + 1/u
or, 1/v = 1/5 + 1/($$1000)
or, 1/v=(200+1)/1000
or, 1/v=199/1000
or, 1/v=1000/199
or, v=5 cm (approx)
Image distance, v =5 cm behind the lens;
Applying magnification formula, we get:
m = h'/h = v/u
m = h'/2 = (5)/($$1000)
Thus, the size of the image is 0.01 cm.
So, the image is inverted (h' is negative), highly diminished (h' is smaller than h) and real (v is positive).
When an object is placed beyond 2f, the image is real, inverted and diminished and is formed between f and 2f.
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Question 19:
The filament of a lamp is 80 cm from a screen and a converging lens forms an image of it on a screen, magnified three times. Find the distance of the lens from the filament and the focal length of the lens.
Answer:
Here, filament of the lamp acts as an object.
v = Image distance
and u = Object distance
According to the question:
v + u = 80 (taking magnitude only) ...(i)
and magnification m = v/u = 3
or, v = 3 u
or, v $$3u = 0 ...(ii)
Solving (i) and (ii), we get:
u = 20 cm
For the focal length of the lens, we have:
Distance of lens from the filament, u = 20 cm
v = 3 $\times $ u = 3 $\times $ 20 = 60 cm
Applying lens formula, we get:
1/v $$ 1/u = 1/f
1/60 + 1/20 = 1/f
(1+3)/60= 1/f
1/f=4/60
f=15 cm
So, focal length f = 15 cm
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Question 20:
An erect image 2.0 cm high is formed 12 cm from a lens, the object being 0.5 cm high. Find the focal length of the lens.
Answer:
Given:
Image distance, v = $$12 cm (Image is erect.)
Height of the object, h = 0.5 cm
Height of the image, h' = 2.0 cm
Applying magnification formula, we get:
m = v/u = h'/h
$$12/u = 2.0/0.5
or, object distance, u = $$3 cm
Applying lens formula, we get:
1/v $$ 1/u = 1/f
1/($$12) $$1/(3) = 1/f
or, 1/f = 3/12
or, focal length, f = 4.0 cm
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Question 21:
A convex lens of focal length 0.10 m is used to form a magnified image of an object of height 5 mm placed at a distance of 0.08 m from the lens. Calculate the position, nature and size of the image.
Answer:
Given:
Focal length, f = 0.10 m = 10 cm
Object distance, u = $$0.08 m = $$8 cm
Height of the object, h = 5 mm = 0.5 cm
Applying lens formula, we get:
1/v $$1/u = 1/f
or, 1/v = 1/f + 1/u
or, 1/v = 1/10 + 1/($$8)
Image distance, v = $$80/2 = $$40 cm
or, v = $$0.40 m
Thus, the position of image is 0.40 m from the lens on the same side as the object (on the left of lens).
Magnification, m = h'/h = v/u
m = h'/0.5 = ($$40)/($$8) = + 5
or, size or height of image, h' = 2.5 cm or 25 mm
Further, the value of magnification is positive; therefore, the image is virtual and erect.
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Question 22:
A convex lens of focal length 6 cm is held 4 cm from a newspaper which has print 0.5 cm high. By calculation, determine the size and nature of the image produced.
Answer:
Given:
Focal length, f = 6 cm
Object distance, u = $$4 cm
Size of the object, h = 0.5 cm
Applying lens formula, we get:
1/v $$1/u = 1/f
or, 1/v = 1/f + 1/u
= 1/6 + 1/($$4) = $$1/12
or, image distance, v = $$12 cm
Applying magnification formula, we get:
m = v/u = h'/h
or, m = ($$12)/($$4) = h'/0.5
or, 3 = h'/0.5
or, h^{,} =3$\times $0.5
Size of the image, h' = 1.5 cm
Therefore, the height of the image is 1.5 cm. The value of magnification is positive; therefore, the image is virtual, erect and three times magnified.
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Question 23:
Determine how far an object must be placed in front of a converging lens of focal length 10 cm in order to produce an erect (upright) image of linear magnification 4.
Answer:
Given:
Focal length, f = 10 cm
Magnification, m = +4 (Image is erect.)
Object distance, u = ?
Applying magnification formula, we get:
m = v/u
or, 4 = v/u
or, v = 4u
Applying lens formula, we get:
1/v $$1/u = 1/f
1/4u $$ 1/u = 1/10
or, u = $$30/4
or, u = $$7.5 cm
Thus, the object must be placed at a distance of 7.5 cm in front of the lens.
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Question 24:
A lens of focal length 20 cm is used to produce a ten times magnified image of a film slide on a screen. How far must the slide be placed from the lens?
Answer:
Given:
m = 10 (negative sign implies real image)
f = 20 cm
Object distance = u
Lens formula is given by:
$\frac{1}{f}=\frac{1}{v}\frac{1}{u}$
Magnification in case of convex lens is given by:
m =$\frac{v}{u}$
v = mu
$\mathrm{v}=(10)\times (\mathrm{u})=10\mathrm{u}$
Putting these values in lens formula, we get:
$\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{1}{10u}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{1+10}{10u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{11}{10u}\phantom{\rule{0ex}{0ex}}\Rightarrow 10u=220\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{220}{10}=22\mathrm{cm}\phantom{\rule{0ex}{0ex}}$
Thus, object distance = u = 22 cm
So, the slide is placed at a distance 22 cm from the poles of the lens.
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Question 25:
An object placed 4 cm in front of a converging lens produces a real image 12 cm from the lens.
(a) What is the magnification of the image?
(b) What is the focal length of the lens?
(c) Draw a ray diagram to show the formation of image. Mark clearly F and 2F in the diagram.
Answer:
Converging lens is a convex lens.
Distance of the object from the lens (u) = $$4
Distance of the image from the lens (v) = 12
(a)
$\mathrm{Magnification}\mathrm{of}\mathrm{the}\mathrm{lens}\mathrm{is}\mathrm{given}\mathrm{by}:\phantom{\rule{0ex}{0ex}}m=\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{12}{4}=3\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{magnifiction}\mathrm{produced}\mathrm{by}\mathrm{the}\mathrm{lens}\mathrm{is}3.\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{Lens}\mathrm{formula}\mathrm{is}\mathrm{given}\mathrm{by}:\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{12}\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{12}+\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1+3}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow f=3\mathrm{cm}$
(c)
Page No 247:
Question 26:
(a) An object 2 cm tall stands on the principal axis of a converging lens of focal length 8 cm. Find the position, nature and size of the image formed if the object is:
(i) 12 cm from the lens
(ii) 6 cm from the lens
(b) State one practical application each of the use of such a lens with the object in position (i) and (ii).
Answer:
Converging lens is a convex lens
Given:
Focal length (f) = +8 cm
Height of the object (h) = +2
(a)
(i) Object distance (u) = $$12
Lens formula is given as:
$\frac{1}{f}=\frac{1}{v}\frac{1}{u}$
$\Rightarrow \frac{1}{8}=\frac{1}{v}\frac{1}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}=\frac{1}{v}+\frac{1}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{8}\frac{1}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{32}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow v=24\mathrm{cm}$
Image is at a distance of 24 cm from the convex lens; therefore, we have:
Magnification =$\frac{v}{u}$
Magnification (m) = $\frac{24}{12}$
m = $$2
Hence, the image is real and inverted.
(ii) Object distance (u) = $$6
$\mathrm{According}\mathrm{to}\mathrm{lens}\mathrm{formula}:\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}=\frac{1}{v}\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}=\frac{1}{v}+\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}\frac{1}{6}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{34}{24}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{24}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{v}=24\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Image}\mathrm{is}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}24\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{lens};\mathrm{therefore},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\mathrm{Magnification}\left(m\right)=\frac{v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{24}{6}\phantom{\rule{0ex}{0ex}}m=4\phantom{\rule{0ex}{0ex}}m=\frac{{h}_{i}}{{h}_{o}}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{{h}_{i}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{i}=2\times 4\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{i}=8cm\phantom{\rule{0ex}{0ex}}\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{is}8\mathrm{cm}.\phantom{\rule{0ex}{0ex}}\mathrm{Here},\mathrm{height}\mathrm{is}\mathrm{positive};\mathrm{therefore},\mathrm{image}\mathrm{is}\mathrm{virtual}\mathrm{and}\mathrm{erect}.$
(c) The practical application for case (1) is that it can be used as a corrective lens for a farsighted person and for case (2), it can be used as a magnifying lens for reading purposes.
Page No 247:
Question 27:
(a) An object 3 cm high is placed 24 cm away from a convex lens of focal length 8 cm. Find by calculations, the position, height and nature of the image.
(b) If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image?
(c) Which of the above two cases illustrates the working of a magnifying glass?
Answer:
(a) Given:
Object distance (u) = $$24
Focal length (f) = +8
Object height (h) = 3
$\mathrm{Lens}\mathrm{formula}\mathrm{is}\mathrm{given}\mathrm{by}:\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}=\frac{1}{v}\frac{1}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}=\frac{1}{v}+\frac{1}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}\frac{1}{24}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{31}{24}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{24}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow v=12\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Image}\mathrm{will}\mathrm{be}\mathrm{form}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}12\mathrm{cm}\mathrm{on}\mathrm{the}\mathrm{right}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{convex}\mathrm{lens}.\phantom{\rule{0ex}{0ex}}\mathrm{Magnification}m=\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{12}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{diminished}.\phantom{\rule{0ex}{0ex}}\mathrm{Negative}\mathrm{value}\mathrm{of}\mathrm{magnification}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{real}\mathrm{and}\mathrm{inverted}.$
$\mathrm{m}=\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}\phantom{\rule{0ex}{0ex}}\frac{1}{2}=\frac{{\mathrm{h}}_{\mathrm{i}}}{3}\phantom{\rule{0ex}{0ex}}{\mathrm{h}}_{\mathrm{i}}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}{\mathrm{h}}_{\mathrm{i}}=1.5\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}1.5\mathrm{cm}.\mathrm{Here},\mathrm{negative}\mathrm{sign}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{in}\mathrm{the}\mathrm{downward}\mathrm{direction}.$
(b) Object distance (u) = $$3
$\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}=\frac{1}{v}\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}=\frac{1}{v}+\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{8}\frac{1}{3}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{38}{24}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{24}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{24}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow v=4.8\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}4.8\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{lens}.\phantom{\rule{0ex}{0ex}}\mathrm{Magnification}\left(m\right)=\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{4.8}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow m=1.6\phantom{\rule{0ex}{0ex}}\mathrm{Positive}\mathrm{value}\mathrm{of}\mathrm{magnification}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{virtual}\mathrm{and}\mathrm{erect}.\phantom{\rule{0ex}{0ex}}\mathrm{m}=\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1.6=\frac{{\mathrm{h}}_{\mathrm{i}}}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{h}}_{\mathrm{i}}=3\times 1.6=4.8\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Positive}\mathrm{sign}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}\mathrm{above}\mathrm{the}\mathrm{principal}\mathrm{axis}.$
(c) Case (b) illustrates the working of magnifying lens as the object is between the focus and optical centre.
Page No 247:
Question 28:
(a) Find the nature, position and magnification of the images formed by a convex lens of focal length 0.20 m if the object is placed at a distance of:
(i) 0.50 m
(ii) 0.25 m
(iii) 0.15 m
(b) Which of the above cases represents the use of convex lens in a film projector, in a camera, and as a magnifying glass?
Answer:
(i) u = $$0.50 m = $$50 cm
f = 0.20 m = 20 cm
Lens formula:
$\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{1}{v}\frac{1}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{20}\frac{1}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{52}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{3}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{100}{3}\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}33.3\mathrm{cm}\mathrm{on}\mathrm{the}\mathrm{right}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{mirror}\phantom{\rule{0ex}{0ex}}\mathrm{magnification}m=\frac{v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{33.3}{50}\phantom{\rule{0ex}{0ex}}m=0.66\phantom{\rule{0ex}{0ex}}\mathrm{magnification}\mathrm{is}\mathrm{negative}\mathrm{therefore}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{real}\mathrm{and}\mathrm{inverted}.\phantom{\rule{0ex}{0ex}}$
(ii) u = $$25 cm
f = 20 cm
$\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{1}{v}\frac{1}{25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{20}\frac{1}{25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{54}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{v}=100\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}100\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirrror}.\phantom{\rule{0ex}{0ex}}\mathrm{Magnification},\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{m}=\frac{100}{25}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{m}=4\phantom{\rule{0ex}{0ex}}\mathrm{Magnification}\mathrm{is}\mathrm{negative},\mathrm{which}\mathrm{means}\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{real}\mathrm{and}\mathrm{inverted}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
(c) u = 15 cm
f = 20
$\mathrm{Substituting}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{lens}\mathrm{formula},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{1}{v}\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}\frac{1}{15}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{34}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{60}\phantom{\rule{0ex}{0ex}}v=60\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}60\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}m=\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{60}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow m=4\phantom{\rule{0ex}{0ex}}\mathrm{Magnification}\mathrm{is}\mathrm{positive};\mathrm{therefore},\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{virtual}\mathrm{and}\mathrm{erect}.$
(b)
We can use case (ii) for a film projector.
We can use case (i) for a camera.
We can use case (ii) for a magnifying glass.
Page No 247:
Question 29:
A spherical mirror and a spherical lens each have a focal length of, –15 cm. The mirror and the lens are likely to be:
(a) both concave.
(b) both convex.
(c) the mirror is concave but the lens is convex.
(d) the mirror is convex but the lens is concave.
Answer:
(a) Both concave
According to the sign convention, the focal length for both a concave mirror and a concave lens is negative.
Page No 247:
Question 30:
Linear magnification produced by a convex lens can be:
(a) less than 1 or more than 1
(b) less than 1 or equal to 1
(c) more than 1 or equal to 1
(d) less than 1, equal to 1 or more than 1
Answer:
(d) Less than 1, equal to 1 or more than 1
The size of the image formed by a convex lens may be less than, equal to or greater than the size of the object .
Page No 248:
Question 31:
Magnification produced by a concave lens is always:
(a) more than 1
(b) equal to 1
(c) less than 1
(d) more than 1 or less than 1
Answer:
(c) Less than 1:
$M\mathrm{agnification}=\frac{\mathrm{size}\mathrm{of}\mathrm{the}\mathrm{image}}{\mathrm{size}\mathrm{of}\mathrm{the}\mathrm{object}}$
From the formula, it is clear that the magnification varies directly with the size of the image. The image formed by a concave mirror is always smaller than the size of the object. Therefore, the magnification of a concave mirror is always less than 1.
Page No 248:
Question 32:
In order to obtain a magnification of, –3 (minus 3) with a convex lens, the object should be placed:
(a) between optical centre and F
(b) between F and 2F
(c) at 2F
(d) beyond 2F
Answer:
(b) Between F and 2F
In the case of a convex lens, for an object placed between F and 2F, the image formed will be real, inverted and enlarged.
Page No 248:
Question 33:
A convex lens produces a magnification of + 5. The object is placed:
(a) at focus
(b) between f and 2f
(c) at less than f
(d) beyond 2f
Answer:
(c) At less than f
since the magnification is positive, the image formed is virtual, erect and enlarged. This is the case when an object is placed at a distance of less than f of the lens.
Page No 248:
Question 34:
If a magnification of, –1 (minus 1) is obtained by using a converging lens, then the object has to be placed:
(a) within f
(b) at 2f
(c) beyond 2f
(d) at infinity
Answer:
(b) At 2f
When an object is placed at 2f, the size of the image formed will be equal to the size of the object. Therefore, we get a magnification of 1.
Page No 248:
Question 35:
To obtain a magnification of, –0.5 with a convex lens, the object should be placed:
(a) at F
(b) between optical centre and F
(c) between F and 2F
(d) beyond 2F
Answer:
(d) Beyond 2F: since the magnification is negative and less than 1, the image formed is real, inverted and diminished. Therefore, the object should be placed beyond 2F.
Page No 248:
Question 36:
An object is 0.09 m from a magnifying lens and the image is formed 36 cm from the lens. The magnification produced is:
(a) 0.4
(b) 1.4
(c) 4.0
(d) 4.5
Answer:
(c) 4.0
Magnification is given by:
Magnification (m) =$\frac{\mathrm{distance}\mathrm{of}\mathrm{image}\mathrm{from}\mathrm{the}\mathrm{lens}}{\mathrm{distance}\mathrm{of}\mathrm{object}\mathrm{from}\mathrm{the}\mathrm{lens}}$
$\mathrm{m}=\frac{36\mathrm{cm}}{0.09\mathrm{m}}$=$\frac{36\mathrm{cm}}{9\mathrm{cm}}=4$
Page No 248:
Question 37:
To obtain a magnification of, –2 with a convex lens of focal length 10 cm, the object should be placed:
(a) between 5 cm and 10 cm
(b) between 10 cm and 20 cm
(c) at 20 cm
(d) beyond 20 cm
Answer:
(b) Between 10 cm and 20 cm:
the magnification is 2, which means that the image is real, inverted and magnified. A convex mirror forms a real, inverted and magnified image when an object is placed between F and 2F .
Page No 248:
Question 38:
A convex lens of focal length 15 cm produces a magnification of +4. The object is placed:
(a) at a distance of 15 cm
(b) between 15 cm and 30 cm
(c) at less than 15 cm
(d) beyond 30 cm
Answer:
(c) At less than 15 cm:
A convex lens forms a virtual, erect and magnified image when an object is placed within the focus.
Page No 248:
Question 39:
If a magnification of, –1 is to be obtained by using a converging lens of focal length 12 cm, then the object must be placed:
(a) within 12 cm
(b) at 24 cm
(c) at 6 cm
(d) beyond 24 cm
Answer:
(b) At 24 cm
For an object placed at a distance of 2F from a convex lens, the size of the image so formed is equal to the size of the object.
Page No 248:
Question 40:
In order to obtain a magnification of, –0.75 with a convex lens of focal length 8 cm, the object should be placed:
(a) at less than 8 cm
(b) between 8 cm and 16 cm
(c) beyond 16 cm
(d) at 16 cma
Answer:
(c) Beyond 16 cm
â€‹For an object placed beyond 2f of a convex lens, the image formed is real, inverted and smaller than the object.
Page No 248:
Question 41:
A student did an experiment with a convex lens. He put an object at different distances 25 cm, 30 cm, 40 cm, 60 cm and 120 cm from the lens. In each case he measured the distance of the image from the lens. His results were 100 cm, 24 cm, 60 cm, 30 cm and 40 cm, respectively. Unfortunately his results are written in wrong order.
(a) Rewrite the image distances in the correct order.
(b) What would be the image distance if the object distance was 90 cm?
(c) Which of the object distances gives the biggest image?
(d) What is the focal length of this lens?
Answer:
(a) Since the focal length is a constant quantity, we have to pair the object distance(u) and the image distance (v) such that the focal length always comes out to be the same. From the above argument, we get the correct order of the image distance as 100 , 60 , 40 , 30 and 24. The reason being, as the object is carried far from a convex lens, the image is formed closer to the lens.
(b) Lens formula is given by:
$\frac{1}{f}=\frac{1}{v}\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{20}=\frac{1}{v}\frac{1}{90}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{90}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{20}\frac{1}{90}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{92}{180}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{7}{180}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{180}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow v=25.7\mathrm{cm}$
The image distance will be 25.7 cm if the object distance is 90 cm.
(c) The object distance of 25 gives the biggest image because at this position, the object is between f and 2f . We know that when an object is placed between f and 2f of a convex lens, we get a real, inverted and magnified image.
(d) The image of an object at 2f is formed at 2f. This means that the pair of u and v that is equal in value gives us the value of 2f, which is 40. Hence, the value of f (focal length) is $\frac{40}{2}$ .
∴ f=20 cm.
Page No 248:
Question 42:
A magnifying lens has a focal length of 100 mm. An object whose size is 16 mm is placed at some distance from the lens so that an image is formed at a distance of 25 cm in front of the lens.
(a) What is the distance between the object and the lens?
(b) Where should the object be placed if the image is to form at infinity?
Answer:
(a) Given, focal length (f) = 100 mm
Size of object (h) = 16 mm
Image distance (v) = 25 cm = 250 mm
$L\mathrm{ens}\mathrm{formula}\mathrm{is}\mathrm{given}\mathrm{by}:\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}\frac{1}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\frac{1}{100}=\frac{1}{250}\frac{1}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{u}}=\frac{1}{250}\frac{1}{100}\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{u}}=\frac{25}{500}\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{u}}=\frac{3}{500}\phantom{\rule{0ex}{0ex}}\mathrm{u}=166.67\mathrm{mm}\phantom{\rule{0ex}{0ex}}\mathrm{u}=16.66\mathrm{cm}$
The object is at a distance of 16.66 cm and in front of the mirror. The image is at a distance of
25 cm and in front of the mirror. Therefore, the distance between the object and the image is 25 cm16.66 cm = 7.14 cm.
(b) The object should be placed at the focus for the image to be formed at infinity.
Page No 248:
Question 43:
A lens forms a real image 3 cm high of an object 1 cm high. If the separation of object and image is 15 cm, find the focal length of the lens.
Answer:
Given,
Height of image (â€‹h') =  3 cm (negative because image is real)
Height of â€‹object = 1 cm
Let us first define the sign convention:
Since the object distance is always negative, let it be u.
Since the focal length is positive for a convex lens, let it be +f .
Magnification =$\frac{{\mathrm{h}}^{\text{'}}}{\mathrm{h}}$ =$\frac{\mathrm{v}}{\mathrm{u}}$
$\frac{\mathrm{v}}{\mathrm{u}}=3$
∴â€‹ $\mathrm{v}=3\mathrm{u}$
(i) Since object distance (u) is always negative, and the ratio $\frac{\mathrm{v}}{\mathrm{u}}$ here is negative, v must be positive, i.e., it is on the right side of the lens.
Now, we have object distance + image distance = 15
v + (u) = 15 (using sign convention)
$3\mathrm{u}\mathrm{u}=15$
$4\mathrm{u}=15\phantom{\rule{0ex}{0ex}}\mathrm{u}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}Applyingthel\mathrm{ens}\mathrm{formula}:\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}\frac{1}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{f}}=\frac{1}{3\mathrm{u}}\frac{1}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{f}}=\frac{4}{3\mathrm{u}}\phantom{\rule{0ex}{0ex}}\mathrm{f}=\frac{3\mathrm{u}}{4}=\frac{3\times {\displaystyle \frac{15}{4}}}{4}=\frac{45}{16}=2.81\mathrm{cm}\phantom{\rule{0ex}{0ex}}H\mathrm{ence},the\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{lens}\mathrm{is}+2.81\mathrm{cm}.$
Page No 249:
Question 44:
An object 50 cm tall is placed on the principal axis of a convex lens. Its 20 cm tall image is formed on the screen placed at a distance of 10 cm from the lens. Calculate the focal length of the lens.
Answer:
Height of object (h) = 50 cm
Height of image (h') = 20 cm (real and inverted)
Distance of image from the lens (v) = 10 cm
Distance of object from the lens (u) = ?
Focal length of the lens (f)= ?
We know, magnification (m) of the lens is given by:
$\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}}$
Thus, substituting the values of v, h and h', we get:
$\frac{10}{\mathrm{u}}=\frac{20}{50}\phantom{\rule{0ex}{0ex}}\frac{10}{\mathrm{u}}=\frac{20}{50}\phantom{\rule{0ex}{0ex}}\mathrm{u}=\frac{5}{2}\mathrm{x}10\phantom{\rule{0ex}{0ex}}\therefore \mathrm{u}=25\mathrm{cm}.$
Using the lens formula:
$\frac{1}{\mathrm{v}}\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
$\frac{1}{10}\frac{1}{25}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{10}+\frac{1}{25}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5+2}{50}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7}{50}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{f}=\frac{50}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{f}=7.14\mathrm{cm}.$
Page No 251:
Question 1:
If the image formed by a lens is always diminished and erect, what is the nature of the lens?
Answer:
It is a concave lens because only a concave lens forms a diminished and erect image of the object at all times.
Page No 251:
Question 2:
Copy and complete the diagram below to show what happens to the rays of light when they pass through the concave lens:
Figure
Answer:
The rays will diverge and appear to meet at the focus of the lens.
Page No 252:
Question 3:
Which type of lenses are:
(a) thinner in the middle than at the edges?
(b) thicker in the middle than at the edges?
Answer:
(a) A concave lens is thinner in the middle than at the edges.
(b) A convex lens is thicker in the middle than at the edges.
Page No 252:
Question 4:
A ray of light is going towards the focus of a concave lens. draw a ray diagram to show the path of this ray of light after refraction through the lens.
Answer:
A ray of light passing through the focus of a concave lens will become parallel to the principal axis after refraction.
Page No 252:
Question 5:
(a) What type of images can a convex lens make?
(b) What type of image is always made by a concave lens?
Answer:
(a) A convex lens can form two types of images: (i) real and inverted and (ii) virtual and erect. Real and inverted images are formed when an object is placed beyond the focus of the lens and virtual and erect images are formed when an object is placed between the focus and the optical centre.
(b) A concave lens always forms a virtual and erect image.
Page No 252:
Question 6:
Take down this figure into your answer book and complete the path of the ray.
Figure
Answer:
The complete path of the light ray is as follows:
Page No 252:
Question 7:
Fill in the following blanks with suitable words:
(a) A convex lens .................. rays of light, whereas a concave lens .................. rays of light.
(b) Lenses refract light to form images: a..................... lens can form both real and virtual images, but a diverging lens forms only ...................... images.
Answer:
(a) A convex lens converges rays of light, whereas a concave lens diverges rays of light.
(b) Lenses refract light to form images: a converging lens can form both real and virtual images, but a diverging lens forms only virtual images.
Page No 252:
Question 8:
Things always look small on viewing through a lens. What is the nature of the lens?
Answer:
It is a concave lens because only a concave lens forms a diminished image at all times.
Page No 252:
Question 9:
An object lies at a distance of 2f from a concave lens of focal length f. Draw a raydiagram to illustrate the image formation.
Answer:
For a concave lens, when the object lies anywhere between the optical center (C) and infinity, the image is formed between the optical center (C) and the focus. Therefore, when the object is placed at 2f from the concave lens of focal length f, the image is formed between the optical center (C) and the focus (f). Also, the image formed is virtual, erect and diminished. The following ray diagram illustrates the image formed by the concave lens:
Page No 252:
Question 10:
Show by drawing a raydiagram that the image of an object formed by a concave lens is virtual, erect and diminished.
Answer:
The image formed by a concave lens is always virtual because it is formed on the left side of the lens. Let us take an object AB beyond the focus (f) as shown in the figure. The image A'B' is formed on the left side of the concave lens and is therefore, virtual. The image is upright. Also the height of A'B' is less than the height of AB. Therefore, the image is diminished.
Page No 252:
Question 11:
Give the position, size and nature of image formed by a concave lens when the object is placed:
(a) anywhere between optical centre and infinity.
(b) at infinity.
Answer:
(a) In the case of a concave lens, when an object is placed anywhere between the optical centre and infinity, the image is formed between the optical centre and the focus. The image formed is virtual, erect and diminished.
(b) In the case of a concave lens, when an object is placed at infinity, the image is formed at the focus. The image formed is virtual, erect and highly diminished.
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Question 12:
Which type of lens is : (a) a converging lens, and which is (b) a diverging lens? Explain your answer with diagrams.
Answer:
(a) A lens that converges a parallel beam of light passing through it to a point is known as a converging lens. The convex lens is a converging lens. The ray diagram below shows the parallel light rays converging at the focus of a convex lens.
(b) The lens which diverges a parallel beam of light passing through it is known as a diverging lens. A concave lens is a diverging lens. The ray diagram below shows the parallel light rays getting diverged when passed through a concave lens.
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Question 13:
With the help of a diagram, explain why the image of an object viewed through a concave lens appears smaller and closer than the object.
Answer:
For an object placed anywhere in front of a concave lens, the image so formed is always virtual, erect and diminished in size. The position of the image is between the optic centre and the focus of the lens, as shown in the figure.
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Question 14:
How would a pencil look like if you saw it through (a) a concave lens, and (b) a convex lens? (Assume the pencil is close to the lens). Is the image real or virtual?
Answer:
(a) When a pencil is seen through a concave lens placed close to it, it appears smaller than it's actual size. Because a concave lens always produces a virtual image, the image formed is virtual.
(b) When a pencil is seen through a convex lens placed close to it, the image so formed appears larger than it's actual size. The reason for this is, the pencil is placed within the focus of the lens and therefore, the image formed is virtual.
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Question 15:
(a) An object is placed 10 cm from a lens of focal length 5 cm. Draw the ray diagrams to show the formation of image if the lens is (i) converging, and (ii) diverging.
(b) State one practical use each of convex mirror, concave mirror, convex lens and concave lens.
Answer:
(a)(i) When an object is placed at 2F' of a converging lens, the image formed is real, inverted and of same size as the object. The image position at 2F is shown in the figure:
(a)(ii) When an object is placed beyond F of a diverging lens, the image formed is virtual, erect and diminished. The position between the focus and the optic centre is as shown in the figure:
(b)
 A convex lens is used as a magnifying glass.
 A concave mirror is used as a shaving mirror.
 A convex mirror is used as rear view mirror in vehicles.
 A concave lens is used for correcting myopia.
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Question 16:
(a) Construct ray diagrams to illustrate the formation of a virtual image using (i) a converging lens, and (ii) a diverging lens.
(b) What is the difference between the two images formed above?
Answer:
(a)(i) When an object is placed between the optic centre and the focus of a converging lens, the image formed is virtual, erect and magnified as shown in the figure.
(a)(ii) When an object is placed anywhere between the optic centre and infinity of a converging lens, the image formed is virtual, erect, and diminished as shown in the figure.
(b)
S.No  Virtual image made by a converging lens  Virtual image made by a diverging lens 
1  Image is larger than the object  Image is smaller than the object 
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Question 17:
A diverging lens is used in:
(a) a magnifying glass
(b) a car to see objects on rear side
(c) spectacles for the correction of short sight
(d) a simple camera
Answer:
(c) Spectacles for the correction of short sight
A diverging lens is used in spectacles to correct shortsightedness.
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Question 18:
When an object is kept at any distance in front of a concave lens, the image formed is always:
(a) virtual, erect and magnified
(b) virtual, inverted and diminished.
(c) virtual, erect and diminished
(d) virtual, erect and same size as object
Answer:
(c) virtual, erect and diminished
A concave lens always forms a virtual, erect and diminished image.
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Question 19:
When sunlight is concentrated on a piece of paper by a spherical mirror or lens, then a hole can be burnt in it. For doing this, the paper must be placed at he focus of:
(a) either a convex mirror or convex lens
(b) either a concave mirror or concave lens
(c) either a concave mirror or convex lens
(d) either a convex mirror or concave lens
Answer:
(c) either a concave mirror or a convex lens
Both a concave mirror and a convex lens focus parallel light beams coming from a distant object onto the focus.
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Question 20:
A beam of parallel light rays is incident through the holes on one side of a box and emerges out through the holes on its opposite side as shown in the diagram below:
Figure
Which of the following could be inside the box?
(a) a rectangular glass block
(b) a concave lens
(c) a convex lens
(d) a glass prism
Answer:
(c) a convex lens
The beam of light is converging when it is coming out of the box; therefore, there should be a convex lens inside the box.
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Question 21:
A beam of light is incident through the holes on one side of a box and emerges out through the holes on its opposite side as shown in the following figure:
Figure
The box contains:
(a) a glass prism
(b) a concave lens
(c) a convex lens
(d) a parallelsided glass slab
Answer:
(b) a concave lens
This is because the emergent rays of light are diverging.
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Question 22:
Which of the following can form a virtual image which is always smaller than the object?
(a) a plane mirror
(b) a convex lens
(c) a concave lens
(d) a concave mirror
Answer:
(c) a concave lens
A concave lens always forms a virtual image, smaller than the size of the object.
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Question 23:
When an object is placed 10 cm in front of lens A, the image is real, inverted, magnified and formed at a great distance. When the same object is placed 10 cm in front of lens B, the image formed is real, inverted and same size as the object.
(a) What is the focal length of lens A?
(b) What is the focal length of lens B?
(c) What is the nature of lens A?
(d) What is the nature of lens B?
Answer:
(a) The focal length of lens A is 10 cm as only a converging lens forms a real, inverted, magnified image at great distance when object is placed at the focus of the lens.
(b) The focal length of lens B is 5 cm as only a converging lens forms a real, inverted and same size image of object when the object is placed at 2F position of the lens.
(c) Lens A is a converging (convex) lens as only a converging lens forms a real image.
(d) Lens B is a converging (convex) lens as only a converging lens forms a real image.
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Question 24:
When a fork is seen through lenses A and B one by one, it appears as shown in the diagrams. What is the nature of (i) lens A, and (ii) lens B? Give reason for your answer.
Figure
Answer:
When the fork is seen through lens A, it appears to be diminished. Such diminished object is observed when it is placed near a concave lens. Therefore, lens A is concave, i.e., diverging in nature.
When the fork is seen through lens B, it appears to be enlarged. Such enlarged image is formed by a convex lens when an object is placed between the lens and its focus. Therefore, lens B is a convex, i.e., converging in nature.
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Question 25:
What kind of lens can form:
(a) an inverted magnified image?
(b) an erect magnified image?
(c) an inverted diminished image?
(d) am erect diminished image?
Answer:
(a) A convex lens can form an inverted magnified image.
(b) A convex lens can form an erect magnified image.
(c) A convex lens can form an inverted diminished image.
(d) A concave lens can form an erect diminished image.
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Question 1:
The lens A produces a magnification of, − 0.6 whereas lens B produces a magnification of + 0.6.
(a) What is the nature of lens A?
(b) What is the nature of lens B?
Answer:
â€‹(a) Negative magnification shows that the image is real and inverted. Therefore, it's a convex lens.
â€‹(b) Positive magnification shows that image is virtual and erect. The lens can be concave or convex in nature. However, a convex lens always forms a magnified virtual image. According to the given magnification, the size of the image is smaller than the size of the object. Therefore, the lens must be concave in nature.
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Question 2:
A 50 cm tall object is at a very large distance from a diverging lens. A virtual, erect and diminished image of the object is formed at a distance of 20 cm in front of the lens. How much is the focal length of the lens?
Answer:
Since the object is at infinity, the image will be formed at the focus. The image distance will be equal to the focal length of the lens. Therefore, the focal length of the lens is 20 cm.
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Question 3:
An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Fine the position and nature of the image.
Answer:
Object distance u = 4 cm (left side of the lens)
Focal length f = 12 cm (left side of the lens)
Image distance v = ?
From the lens formula, we know that:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}$
Substituting the values of v and u, we get:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\frac{1}{4}=\frac{1}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{12}\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{31}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{4}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{12}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow v=3\mathrm{cm}$
Thus, the image is virtual and formed 3 cm away from the lens, on the left side.
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Question 4:
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the raydiagram.
Answer:
Focal length f = 15 cm (negative due to sign convention)
Image distance v =  10 cm (negative due to sign convention)
Object distance u = ?
Applying the lens formula:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{10}\frac{\mathit{1}}{\mathit{u}}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{1}{15}\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{23}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{1}{30}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{u}=30\mathrm{cm}.$
The object should be place at a distance of 30 cm from the concave lens, and on the left side.
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Question 5:
An object 60 cm from a lens gives a virtual image at a distance of 20 cm in front of the lens. What is the focal length of the lens? Is the lens converging or diverging? Give reasons for your answer.
Answer:
Object distance u =  60 cm (As the object is on the left side of the lens)
Image distance v =  20 cm (As the image is virtual)
Magnification $\mathit{\text{m=}}\frac{\mathit{\text{v}}}{\mathit{\text{u}}}\text{=}\frac{\text{20}}{\text{60}}\text{=}\frac{\text{1}}{\text{3}}$
Because the image is diminished and virtual, the lens is concave and diverging in nature.
Using the lens formula:
$\frac{\mathit{\text{1}}}{\mathit{\text{f}}}\mathit{=}\frac{\mathit{\text{1}}}{\mathit{\text{v}}}\mathit{}\frac{\mathit{\text{1}}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathit{\text{1}}}{\mathit{\text{f}}}\mathit{=}\frac{\text{1}}{\text{20}}\frac{\text{1}}{\text{60}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathit{\text{1}}}{\mathit{\text{f}}}=\frac{\text{1}}{\text{20}}+\frac{\text{1}}{\text{60}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathit{\text{1}}}{\mathit{\text{f}}}=\frac{\text{3+1}}{\text{60}}=\frac{\text{1}}{\text{30}}$
∴ f =  30 cm.
Therefore, the focal length of the lens is 30 cm.
The lens is diverging in nature because a negative focal length indicates a concave lens .
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Question 6:
A concave lens of 20 cm focal length forms an image 15 cm from the lens. Compute the object distance.
Answer:
Focal length of concave lens f =  20 cm (focal length is negative for a concave lens)
Image distance v =  15 cm ( negative sign is given because the image formed by a concave lens is on the left side of the lens)
Using the lens formula:
$\frac{\text{1}}{\mathit{\text{f}}}\mathit{=}\frac{\text{1}}{\mathit{\text{v}}}\mathit{}\frac{\text{1}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\text{20}}=\frac{\text{1}}{\text{15}}\frac{\text{1}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\mathit{\text{u}}}=\frac{\text{1}}{\text{20}}+\frac{\text{1}}{\text{(15)}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\mathit{\text{u}}}=\frac{\text{34}}{\text{60}}=\frac{\text{1}}{\text{60}}$
∴ u =  60 cm.
The negative sign shows that the object is on the left side of the concave lens.
Therefore, the object is at a distance of 60 cm and on the left side of the concave lens.
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Question 7:
A concave lens has focal length 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification produced by the lens.
Answer:
Focal length of concave lens f =  15 cm
Image distance v =  10 cm ( image formed by a concave lens is on the left side of the lens)
Using the lens formula:
$\frac{\text{1}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{}}\frac{\text{1}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{\text{1}}{\mathit{\text{15}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{10}}}\mathit{\text{}}\frac{\text{1}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\mathit{\text{u}}}\text{=}\frac{\text{1}}{\text{15}}\text{}\frac{\text{1}}{\text{10}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\mathit{\text{u}}}\text{=}\frac{\text{23}}{\text{30}}\text{=}\frac{\text{1}}{\text{30}}$
∴ u =  30 cm.
The object is at a distance of 30 cm from the lens and on its left.
The magnification of the concave lens is given as:
$\text{Magnification=}\frac{\text{imagedistance}}{\text{objectdistance}}\phantom{\rule{0ex}{0ex}}\text{\u21d2}\mathit{\text{m=}}\frac{\mathit{\text{v}}}{\mathit{\text{u}}}\text{=}\frac{\text{10}}{\text{30}}\text{=}\frac{\text{1}}{\text{3}}$
Since the magnification is less than 1, the image is smaller than the object.
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Question 8:
Calculate the image distance for an object of height 12 mm at a distance of 0.20 m from a concave lens of focal length 0.30 m, and state the nature and size of the image.
Answer:
Focal length of concave lens f =  0.30 m
â€‹Object distance u =  0.20 m
Height of object h_{1} = 12 mm = 0.012 m
Height of image h_{2} =?
â€‹Using the lens formula:
$\frac{\mathit{\text{1}}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\mathit{\text{1}}}{\mathit{\text{v}}}\mathit{\text{}}\frac{\mathit{\text{1}}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{\mathit{\text{1}}}{\mathit{\text{0.30}}}\mathit{\text{=}}\frac{\mathit{\text{1}}}{\mathit{\text{v}}}\text{}\frac{\text{1}}{(\text{0.20)}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathit{\text{1}}}{\mathit{\text{v}}}\text{=}\frac{\text{1}}{\text{0.30}}\text{}\frac{\text{1}}{\text{0.20}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathit{\text{1}}}{\mathit{\text{v}}}\text{=}\frac{\text{23}}{\text{0.60}}\text{=}\frac{\text{1}}{\text{0.12}}$
∴ v =  0.12 m =  12 cm.
The negative sign indicates that the image is formed on the left side of the lens. Therefore, the image is virtual.
Magnification is given as:
$\text{Magnification=}\frac{\text{imagedistance}}{\text{objectdistance}}\text{=}\frac{\text{heightofimage}}{\text{heightofobject}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathit{\text{v}}}{\mathit{\text{u}}}\mathit{\text{=}}\frac{{\mathit{\text{h}}}_{\mathit{2}}}{{\mathit{\text{h}}}_{\mathit{1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{0.12}}{\text{0.20}}\text{=}\frac{{\text{h}}_{2}}{\text{0.012}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\text{h}}_{2}\text{=}\frac{\text{0.12\xd70.012}}{\text{0.20}}\text{=0.0072m=+7.2mm}$
Here, the positive sign shows that the image is erect. Also, the size of the image is smaller than the size of the object and is diminished.
Therefore, the image formed by the concave lens is virtual, diminished and erect.
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Question 9:
A concave lens has a focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.
Answer:
Focal length of the concave lens, f =  20 cm
Image distance from the concave lens, v =  15 cm (image formed by a concave lens is on the left side of the lens)
Height of the object, h_{1} = 5 cm
Height of the image = h_{2}
Using the lens formula:
$\frac{\text{1}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{}}\frac{\text{1}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\text{20}}\text{=}\frac{\text{1}}{\text{15}}\text{}\frac{\text{1}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\mathit{\text{u}}}\text{=}\frac{\text{1}}{\text{20}}\text{}\frac{\text{1}}{\text{15}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\mathit{\text{u}}}\text{=}\frac{\text{34}}{\text{60}}\text{=}\frac{\text{1}}{\text{60}}$
∴ u =  60 cm.
Therefore, the object is at a distance of 60 cm from the concave lens and on the left side of it.
The magnification of the concave lens is given as:
$\text{Magnification=}\frac{\text{imagedistance}}{\text{objectdistance}}\text{=}\frac{\text{heightofimage}}{\text{heightofobject}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathit{\text{v}}}{\mathit{\text{u}}}\mathit{\text{=}}\frac{{\mathit{\text{h}}}_{\mathit{2}}}{{\mathit{\text{h}}}_{\mathit{1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{15}}{\text{60}}\text{=}\frac{{\text{h}}_{2}}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow {\text{h}}_{2}\text{=}\frac{5}{4}\text{=1.25cm.}$
Therefore, the size of the image is 1.25 cm.
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Question 10:
An object is placed 20 cm from (a) a converging lens, and (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case.
Answer:
(a) Focal length of the converging lens, i.e., convex lens f = + 15 cm
Object distance u =  20 cm
Using the lens formula:
$\frac{\text{1}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{}}\frac{\text{1}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\mathit{\text{15}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\text{}\frac{\text{1}}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{15}=\frac{1}{v}+\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\mathit{\text{v}}}\text{=}\frac{\text{1}}{\text{15}}\text{}\frac{\text{1}}{\text{20}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\text{=}\frac{\text{43}}{\text{60}}\text{=}\frac{\text{1}}{\text{60}}$
∴ v = +60 cm.
Therefore, the image formed is real. It is at a distance of 60 cm from the lens and to its right.
$\text{Magnification=}\frac{\text{imagedistance}}{\text{objectdistance}}\phantom{\rule{0ex}{0ex}}\therefore \mathit{\text{m=}}\frac{\mathit{\text{v}}}{\mathit{\text{u}}}\mathit{\text{=}}\frac{\mathit{\text{60}}}{\mathit{\text{20}}}\text{=3.}$
The magnification is greater than 1. Therefore, the image is magnified.
The negative sign shows that the image is inverted.
(b) â€‹Focal length of the diverging lens, i.e., concave lens f =  15 cm
Object distance u =  20 cm
Using the lens formula:
$\frac{\text{1}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{}}\frac{\text{1}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{\text{1}}{\mathit{\text{15}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{}}\frac{\text{1}}{\mathit{}\mathit{20}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{15}}\text{}\frac{\text{1}}{\text{20}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\text{=}\frac{\text{43}}{\text{60}}\text{=}\frac{\text{7}}{\text{60}}$
∴ v =  8.57 cm.
Therefore, the image formed is virtual. It is at a distance of 8.57 cm from the lens and to its left.
$\text{Magnification=}\frac{\text{imagedistance}}{\text{objectdistance}}\phantom{\rule{0ex}{0ex}}\therefore \mathit{\text{m=}}\frac{\mathit{\text{v}}}{\mathit{\text{u}}}\text{=}\frac{8.57}{\text{20}}\text{=+0.42.}$
The magnification is less than 1. Therefore, the image is diminished. The positive sign indicates that the image is erect.
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Question 11:
A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image.
Answer:
A concave lens is also known as a diverging lens.
Focal length of concave lens, f = $$15 cm
Object distance from the lens, u = $$40 cm
Height of the object, h_{1} = 2.0 cm
Height of the image, h_{2}_{ = }?
Using the lens formula, we get:
$\frac{\mathit{\text{1}}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\mathit{\text{1}}}{\mathit{\text{v}}}\mathit{\text{}}\frac{\mathit{\text{1}}}{\mathit{\text{u}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\text{15}}\text{=}\frac{\text{1}}{\text{v}}\text{}\frac{\text{1}}{\text{40}}\text{}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{15}=\frac{1}{v}+\frac{1}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{15}}\text{}\frac{\text{1}}{\text{40}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\text{=}\frac{\text{83}}{\text{120}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\text{}\frac{\text{11}}{\text{120}}$
∴ v =  10.90 cm
Therefore, the image is formed at a distance of 10.90 cm and to the left of the lens.
Magnification of the lens:
$\text{Magnification=}\frac{\text{Imagedistance}}{\text{Objectdistance}}\text{=}\frac{\text{Heightoftheimage}}{\text{Heightoftheobject}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathit{\text{v}}}{\mathit{\text{u}}}\mathit{\text{=}}\frac{{\mathit{\text{h}}}_{\mathit{2}}}{{\mathit{\text{h}}}_{\mathit{1}}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{\mathit{\text{10.90}}}{\mathit{\text{40}}}\mathit{\text{=}}\frac{{\mathit{\text{h}}}_{\mathit{2}}}{\mathit{2}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}{\mathit{\text{h}}}_{\mathit{2}}\text{=+0.54cm}$
The height of the image formed is 0.54 cm. Also, the positive sign of the height of the image shows that the image is erect.
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Question 12:
(a) Find the position and size of the virtual image formed when an object 2 cm tall is placed 20 cm from:
(i) a diverging lens of focal length 40 cm.
(ii) a converging lens of focal length 40 cm.
(b) Draw labelled ray diagrams to show the formation of images in case (i) and (ii) above (The diagrams may not be according to scale).
Answer:
(a) Object height (h) = 2 cm
Image height (h') =?
Object distance (u) = 20 cm (sign convention)
(i) Focal length (f) = 40 cm (sign convention)
Image distance (v) = ?
Applying the lens formula:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\frac{1}{20}=\frac{1}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{20}=\frac{1}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{40}\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{12}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{3}{40}\phantom{\rule{0ex}{0ex}}\therefore v=\frac{40}{3}=13.33\mathrm{cm}.$
Negative sign shows that the image formed is virtual and erect.
Now, magnification m = $\frac{v}{u}=\frac{h\text{'}}{h}\phantom{\rule{0ex}{0ex}}$
$h\text{'}=\frac{v}{u}xh\phantom{\rule{0ex}{0ex}}h\text{'}=\frac{13.33}{20}\mathrm{x}2=1.33\mathrm{cm}$
Height of the image is 1.33 cm.
(ii) Focal length (f) = 40 cm (sign convention)
Image distance (v) = ?
Applying the lens formula:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\frac{1}{20}=\frac{1}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{20}=\frac{1}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{40}\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{12}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{40}\phantom{\rule{0ex}{0ex}}\therefore v=40\mathrm{cm}.$
Negative sign shows that the image formed is virtual and erect.
Now, magnification = $\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}}\phantom{\rule{0ex}{0ex}}$
$h\text{'}=\frac{v}{u}xh\phantom{\rule{0ex}{0ex}}h\text{'}=\frac{40}{20}\mathrm{x}2=4\mathrm{cm}$.
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Question 13:
(a) A small object is placed 150 mm away from a diverging lens of focal length 100 mm.
(i) Copy the figure below and draw rays to show how an image is formed by the lens.
Figure
(ii) Calculate the distance of the image from the lens by using the lens formula.
(b) The diverging lens in part (a) is replaced by a converging lens also of focal length 100 mm. The object remains in the same position and an image is formed by the converging lens. Compare two properties of this image with those of the image formed by the diverging lens in part (a).
Answer:
(a)(i) The image formed by the lens is virtual, erect and diminished in size.
(a)(ii) Object distance (u) = 150 mm = 15 cm (sign convention)
Focal length (f) = 100 mm = 10 cm (sign convention)
Image distance (v) = ?
Applying the lens formula, we have:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\frac{1}{15}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{15}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{10}\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{32}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{5}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow v=6\mathrm{cm}.$
(b)
S.No  Image formed by a diverging lens  Image formed by a converging lens 
1  Image formed is virtual and erect  Image formed is real and inverted 
2  Image is smaller in size than the object  Image is larger in size than the object 
Page No 256:
Question 14:
A concave lens produces an image 20 cm from the lens of an object placed 30 cm from the lens. The focal length of the lens is:
(a) 50 cm
(b) 40 cm
(c) 60 cm
(d) 30 cm
Answer:
(c) 60 cm
Object distance from the lens, u = $$30 cm
Image distance from the lens, v = $$20 cm
Using the lens formula, we get:
$\frac{\text{1}}{\text{f}}=\frac{\text{1}}{\text{v}}\frac{\text{1}}{\text{u}}$
$\frac{\text{1}}{\text{f}}=\frac{\text{1}}{\text{20}}\frac{\text{1}}{\text{30}}\phantom{\rule{0ex}{0ex}}\frac{\text{1}}{\text{f}}=\frac{\text{1}}{\text{20}}+\frac{\text{1}}{\text{30}}\phantom{\rule{0ex}{0ex}}\frac{\text{1}}{\text{f}}=\frac{\text{3+2}}{\text{60}}=\frac{\text{1}}{\text{60}}$
∴ f = $$60 cm
Therefore, the focal length of the lens is 60 cm.
Page No 256:
Question 15:
Only one of the following applies to a concave lens. This is:
(a) focal length is positive
(b) image distance can be positive or negative
(c) height of image can be positive or negative
(d) image distance is always negative
Answer:
(d) image distance is always negative
Because a concave lens always forms a virtual image on the left side of the lens.
Page No 256:
Question 16:
The magnification produced by a spherical mirror and a spherical lens is + 0.8.
(a) The mirror and lens are both convex
(b) The mirror and lens are both concave
(c) The mirror is concave but the lens is convex
(d) The mirror is convex but the lens is concave
Answer:
(d) The mirror is convex but the lens is concave
Because both convex mirror and concave lens have positive magnification.
Page No 256:
Question 17:
The magnification produced by a spherical lens and a spherical mirror is + 2.0.
(a) The lens and mirror are both concave
(b) The lens and mirror are both convex
(c) The lens is convex but the mirror is concave
(d) The lens is concave but the mirror is convex
Answer:
(c) The lens is convex but the mirror is concave.
Because both concave mirror and convex lens have positive magnification.
Page No 257:
Question 18:
A camera fitted with a lens of focal length 50 mm is being used to photograph a flower that is 5 cm in diameter. The flower is placed 20 cm in front of the camera lens.
(a) At what distance from the film should the lens be adjusted to obtain a sharp image of the flower?
(b) What would be the diameter of the image of the flower on the film?
(c) What is the nature of camera lens?
Answer:
(a)
Given:
Focal length, f = 50 mm = 5 cm
Object distance, u = $$20 cm
Image distance, v = ?
Putting these values in lens formula, we get:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\frac{1}{20}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{20}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{5}\frac{1}{20}$
$\Rightarrow \frac{1}{v}=\frac{205}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{15}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{20}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow v=6.66\mathrm{cm}$
Film should be adjusted at a distance of 6.66 cm behind the lens.
(b)
Given:
Diameter of object, h = 5 cm
Diameter of image, h' = ?
Magnification, m = $\frac{h\text{'}}{h}=\frac{v}{u}$
Therefore
h' =$\frac{v}{u}xh$
h' = $\frac{6.66}{20}\mathrm{x}5=1.66\mathrm{cm}$
(c) It is a convex lens.
Page No 257:
Question 19:
An object is 2 m from a lens which forms an erect image onefourth (exactly) the size of the object. Determine the focal length of the lens. What type of lens is this?
Answer:
Given:
Object distance (u) = $$2 m
Image distance (v) = ?
Magnification (m) = 0.25 (onefourth of the size of the image)
Focal length (f) = ?
Magnification (m) = $\frac{\mathrm{v}}{\mathrm{u}}$
$0.25=\frac{\mathrm{v}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{v}=0.5\mathrm{m}$
Putting these values in lens formula, we get:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{0.5}\frac{1}{2}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{0.5}+\frac{1}{2}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{2}\frac{1}{0.5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{2}\frac{10}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow f=0.66\mathrm{m}$
Negative sign of focal length shows that lens is diverging in nature. Hence, it is a concave lens.
Page No 257:
Question 20:
An image formed on a screen is three times the size of the object. The object and screen are 80 cm apart when the image is sharply focussed.
(a) State which type of lens is used.
(b) Calculate focal length of the lens.
Answer:
(a) The image here can be taken on a screen. This means that the image is real. Further, we know that only convex lens forms a real image; therefore, a convex lens has been used here.
(b) Magnification (m) = 3
Object distance (u) = ?
Image distance (v) = ?
Focal length (f) = ?
Distance between image and object (v + u) = 80 cm
v = 80 $$ u
We know that:
m = $\frac{\mathrm{v}}{\mathrm{u}}$
or 3 = $\frac{80\mathrm{u}}{\mathrm{u}}$
or 3u = 80 $$ u
or 4u = 80
or u = 20 cm
Thus, v = 80 $$ u
= 80 $$ 20 = 60 cm
Putting these values in lens formula, we get:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{60}\frac{1}{20}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1+3}{60}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow f=\frac{60}{4}=15\mathrm{cm}$
Page No 261:
Question 1:
The lens A has a focal length of 25 cm whereas another lens B has a focal length of 60 cm. Giving reason state, which lens has more power : A or B.
Answer:
Power varies inversely with the focal length. Lens A has a small focal length and hence, more power.
Page No 261:
Question 2:
Which causes more bending (or more refraction) of light rays passing through it : a convex lens of long focal length or a convex lens of short focal length?
Answer:
The bending of light by a convex lens depends on the power of the lens. The more the power, the more the light bends. A convex lens having a short focal length will have more power and hence, a higher refraction.
Page No 261:
Question 3:
Name the physical quantity whose unit is dioptre.
Answer:
The power of a lens is a physical quantity with dioptre as it's unit.
Page No 261:
Question 4:
Define 1 dioptre power of a lens.
Answer:
1 dioptre is the power of a lens whose focal length is 1 metre. â€‹
Page No 261:
Question 5:
Which type of lens has (a) a positive power, and (b) a negative power?
Answer:
(a) Convex lens has positive power.
(b) Concave lens has negative power.
Page No 261:
Question 6:
Which of the two has a greater power : a lens of short focal length or a lens of large focal length?
Answer:
A lens with a shorter focal length will have more power because power varies inversely with focal length .
Page No 261:
Question 7:
How is the power of a lens related to its focal length?
Answer:
The power of a lens is related to the focal length as:
Power (P) =$\frac{1}{\mathrm{f}}$,
where, f is measured in metres and P in dioptres.
Page No 261:
Question 8:
Which has more power : a thick convex lens or a thin convex lens, made of the same glass? Give reason for your choice.
Answer:
A thick convex lens has more power than a thin one because it has greater curvature or lesser focal length than a thin lens.
Page No 261:
Question 9:
The focal length of a convex lens is 25 cm. What is its power?
Answer:
Given, f = 25 cm
â€‹â€‹Power of a lens is given by:
Power (P) =$\frac{1}{f}$
∴ P =$\frac{1}{25\times {10}^{2}}$ =$\frac{100}{25}$= 4 D.
Page No 261:
Question 10:
What is the power of a convex lens of focal length 0.5 m?
Answer:
â€‹â€‹Power of a lens is given by:
Power (P) = $\frac{1}{f}$
∴ P =$\frac{1}{0.5}$=2 D.
Page No 261:
Question 11:
A converging lens has focal length of 50 mm. What is the power of the lens?
Answer:
Power of a lens is given by:
Power (P) = $\frac{1}{f}$
∴ P = $\frac{1}{50\times {10}^{3}}$ = $\frac{1000}{50}$= 20 D.
Page No 261:
Question 12:
What is the power of a convex lens lens whose focal length is 80 cm?
Answer:
Power of a lens is given by:
Power (P) = $\frac{1}{f}$
∴â€‹ P = $\frac{1}{80\times {10}^{2}}$ = $\frac{100}{80}$ = 1.25 D.
Page No 261:
Question 13:
A diverging lens has focal length of 3 cm. Calculate the power.
Answer:
Power of a lens is given by:
Power (P) = $\frac{1}{f}$
∴ P = $\frac{1}{3\times {10}^{2}}$ = $\frac{100}{3}$= 33.33 D (appx.)
Page No 261:
Question 14:
The power of a lens is + 0.2 D. Calculate its focal length.
Answer:
â€‹Power of a lens is given by:
Power (P) = $\frac{1}{\mathrm{f}}$
$\mathrm{f}=\frac{1}{\mathrm{p}}$
$\mathrm{f}=\frac{1}{0.2}$ =5 m.
Page No 261:
Question 15:
The power of a lens is, −2 D. What is its focal length?
Answer:
â€‹Power of a lens is given by:
$\mathrm{power}\mathrm{p}=\frac{1}{\mathrm{focal}\mathrm{length}}$
$p=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow f=\frac{1}{p}\Rightarrow f=\frac{1}{2}\mathrm{f}=0.5\mathrm{m}$
Negative sign indicates that it's a concave lens.
Page No 261:
Question 16:
What is the nature of a lens having a power of + 0.5 D?
Answer:
Positive sign indicates it is a converging lens. Therefore, the lens is convex.
Page No 261:
Question 17:
What is the nature of a lens whose power is, −4 D?
Answer:
Negative sign indicates it is a diverging lens. Therefore, the lens is concave in nature.
Page No 261:
Question 18:
The optician's prescription for a spectacle lens is marked +0.5 D. What is the :
(a) nature of spectacle lens?
(b) focal length of spectacle lens?
Answer:
â€‹(a) Positive power indicates it is a converging lens.
(b) Power of a lens is given by:
Power (P) = $\frac{\mathit{1}}{\mathit{f}}$
0.5 = $\frac{\mathit{1}}{\mathit{f}}$
$f=\frac{1}{0.5}$
f = 2 m
Focal length of the lens is 2 m.
Page No 261:
Question 19:
A doctor has prescribed a corrective lens of power, −1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Power of a lens is given by:
Power (P) = $\frac{1}{\mathrm{f}}$
$f=\frac{1}{p}$
$\Rightarrow f=\frac{1}{1.5}$
∴â€‹ f = 66.67 cm.
Since the power is negative, the lens is of a diverging nature. Therefore, it's a concave lens.
Page No 262:
Question 20:
A lens has a focal length of, −10 cm. What is the power of the lens and what is its nature?
Answer:
â€‹Power of a lens is given by:
Power (P) = $\frac{\mathit{1}}{\mathit{f}}$
f = 10 cmâ€‹
∴â€‹ P = $\frac{1}{10\times {10}^{2}}$ = 10 D.
Negative sign indicates it is a diverging lens.
Page No 262:
Question 21:
The focal length of a lens is +150 mm. What kind of lens is it and what is its power?
Answer:
Power of a lens is given by:
Power (P) =$\frac{1}{f}$
∴ P = â€‹$\frac{\text{'}1}{150\times {10}^{3}}$ = 6.67 D (apprx.)
Since the power is positive, the lens is convex in nature.
Page No 262:
Question 22:
Fill in the following blanks with suitable words :
(a) The reciprocal of the focal length in metres gives you the..........of the lens, which is measured in..........
(b) For converging lenses, the power is..........while for diverging lenses, the power is.........
Answer:
â€‹(a) The reciprocal of the focal length in metres gives you the power of the lens, which is measured in dioptres.
(b) For converging lenses, the power is positive while for diverging lenses, the power is negative.
Page No 262:
Question 23:
An object of height 4 cm is placed at a distance of 15 cm in front of a concave lens of power, −10 dioptres. Find the size of the image.
Answer:
Object distance u = 15 cm
Height of object h = 4 cm
Power of the lens p = 10 dioptres
Height of image h' = ?
Image distance v = ?
Focal length of the lens f = ?
We know that:
p = $\frac{1}{f}$
$f=\frac{1}{p}\phantom{\rule{0ex}{0ex}}\Rightarrow f=\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow f=0.1\mathrm{m}=10\mathrm{cm}$
From the lens formula, we have:
$\frac{1}{\mathrm{v}}\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
$\frac{1}{v}\frac{1}{15}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{15}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{15}\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{23}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{5}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow v=6\mathrm{cm}$
Thus, the image will be formed at a distance of 6 cm and in front of the mirror.
Now, magnification m = $\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}}$
or $\frac{6}{15}=\frac{h\text{'}}{4}\phantom{\rule{0ex}{0ex}}h\text{'}=\frac{6x4}{15}\phantom{\rule{0ex}{0ex}}h\text{'}=\frac{24}{15}\phantom{\rule{0ex}{0ex}}\therefore h\text{'}=1.6\mathrm{cm}.$
Page No 262:
Question 24:
An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5D. Find (i) focal length of the lens, and (ii) size of the image.
Answer:
Object height, h = 4.25 mm = 0.425 cm (1 cm = 10 mm)
Object distance, u = $$10 cm
Power, P = +5 D
Focal length, f = ?
Image distance, v = ?
Image height, h' = ?
Power, P = $\frac{1}{f}$
f = $\frac{1}{\mathrm{p}}=\frac{1}{5}=0.2\mathrm{m}=20\mathrm{cm}$
Using the lens formula, we get:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}\frac{1}{10}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{10}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{20}\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}\therefore v\mathit{}=20\mathrm{cm}$
Now, magnification, m = $\frac{\mathit{h}\mathit{\text{'}}}{\mathit{h}}\mathit{=}\frac{\mathit{v}}{\mathit{u}}$
Substituting the values in the above equation, we get:
$\frac{h\mathit{\text{'}}}{0.425}=\frac{20}{10}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}h\mathit{\text{'}}=2\mathrm{x}0.425=0.85\mathrm{cm}=8.5\mathrm{mm}$
Thus, the image is 8.5 mm long; it is also erect and virtual.
Page No 262:
Question 25:
A convex lens of power 5 D and a concave lens of power 7.5 D are placed in contact with each other. What is the :
(a) power of this combination of lenses?
(b) focal length of this combination of lenses?
Answer:
(a) Power of convex lens P_{1} = + 5 D
Power of concave lens P_{2} =  7.5 D
The combined power is the algebraic sum of the individual powers of the lenses.
∴ Power of the combination P = P_{1} + P_{2}
P = 5 D  7.5 D =  2.5 D
Therefore, the power of the combination of the lenses is  2.5 D.
(b) The power and focal length are related as:
$\mathit{\text{P=}}\frac{\mathit{\text{1}}}{\mathit{\text{f}}}\phantom{\rule{0ex}{0ex}}\mathit{\text{2.5=}}\frac{\mathit{\text{1}}}{\mathit{\text{f}}}\phantom{\rule{0ex}{0ex}}\mathit{\therefore}\mathit{\text{f}}\text{=}\frac{\text{1}}{\text{2.5}}\text{=0.4m=40cm.}$
Therefore, the focal length of the combination of the lenses is 40 cm. The negative sign shows that the combination of the two lenses causes it to behave like a concave lens.
Page No 262:
Question 26:
A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with one another.
(a) What is the power of this combination?
(b) What is the focal length of this combination?
(c) Is this combination converging or diverging?
Answer:
Focal length of convex lens f_{1}= + 25 cm = + 0.25 m
Power of convex lens ${\mathit{\text{P}}}_{\mathit{1}}\mathit{\text{=}}\frac{\mathit{\text{1}}}{{\mathit{\text{f}}}_{\mathit{1}}}\text{=}\frac{\text{1}}{\text{0.25}}\text{=4D}$
Focal length of concave lens f_{2} =  10 cm =  0.10 m
Power of concave lens ${\mathit{\text{P}}}_{\mathit{2}}\mathit{\text{=}}\frac{\text{1}}{{\mathit{\text{f}}}_{\mathit{2}}}\text{=}\frac{\text{1}}{\text{0.10}}\text{=10D}$
(a) The power of the combination of lenses is the algebraic sum of the powers of the individual lenses.
∴ Power of combination P = P_{1} + P_{2}
⇒ P = 4  10 =  6 D.
(b) Suppose, the focal length of the combination of the lenses is f.
The power of a lens and the focal length are related as:
$\mathit{\text{P}}\text{=}\frac{\text{1}}{\mathit{\text{f}}}\phantom{\rule{0ex}{0ex}}\text{6=}\frac{\text{1}}{\mathit{\text{f}}}\phantom{\rule{0ex}{0ex}}\mathit{\text{f}}\text{=}\frac{\text{1}}{\text{6}}\text{=0.167m=16.7cm}$
Therefore, the focal length of the combination of the lenses is  16.7 cm.
(c) The focal length of the combination of the lenses is  16.7 cm. Here, the negative sign shows that the combination of the two lenses acts like a concave lens. Therefore, this combination of lenses is diverging.
Page No 262:
Question 27:
The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X be 15 cm :
(a) calculate the focal length of lens Y.
(b) state the nature of lens Y.
Answer:
(a) The focal length of lens X, ie., f_{1} = 15 cm = 0.15 m
Power of lens X, ${\mathit{\text{P}}}_{\mathit{1}}\text{=}\frac{\text{1}}{{\mathit{\text{f}}}_{\mathit{1}}}\text{=}\frac{\text{1}}{\text{0.15}}\text{=6.67D}$
Let the power and the focal length of the lens Y be P_{2} and f_{2}_{, }respectively.
The resultant power of the combination of two lenses is the algebraic sum of the powers of the individual lenses.
∴ Power of combination of lenses P = P_{1} + P_{2}
$\Rightarrow $ 5 = 6.67 + P_{2}
â€‹ $\Rightarrow $ P_{2} = 5  6.67 =  1.67 D
The power of the lens is related to the focal length of the lens as:
${\mathit{\text{P}}}_{\mathit{2}}\text{=}\frac{\text{1}}{{\mathit{\text{f}}}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \text{1.67=}\frac{\text{1}}{{\mathit{\text{f}}}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathit{\text{f}}}_{\mathit{2}}\text{=}\frac{\text{1}}{\text{1.67}}\text{=0.60m=60cm}$
The focal length of the lens Y is 60 cm.
(b) The focal length of the lens Y is negative. Therefore, the lens is concave and diverging in nature.
Page No 262:
Question 28:
Two lenses A and B have focal lengths of +20 cm and, −10 cm, respectively.
(a) What is the nature of lens A and lens B?
(b) What is the power of lens A and lens B?
(c) What is the power of combination if lenses A and B are held close together?
Answer:
(a) The focal length of lens A is +20 cm. The positive sign indicates that lens A is convex, i.e., a converging lens.
The focal length of lens B is 10 cm. The negative sign indicates that lens B is concave, i.e., a diverging lens.
(b) Focal length of lens A, f_{A} = +20 cm = +0.20 m
∴ Power of lens A, ${\mathit{\text{P}}}_{A}\text{=}\frac{\text{1}}{{\mathit{\text{f}}}_{A}}\text{=}\frac{\text{1}}{\text{0.20}}\text{=5D}$.
Focal length of lens B, f_{B} = 10 cm = 0.10 m
∴ Pâ€‹ower of lens B, ${\mathit{\text{P}}}_{B}\text{=}\frac{\text{1}}{{\mathit{\text{f}}}_{B}}\text{=}\frac{\text{1}}{\text{0.10}}\text{=10D}$.
(c) When lenses are combined, the power of the combination is the algebraic sum of the powers of the individual lenses.
∴ Power of combination of lens A and B, P = P_{A} + P_{B}
P = 5 D  10 D =  5 D.
Page No 262:
Question 29:
(a) What do you understand by the power of a lens? Name one factor on which the power of a lens depends.
(b) What is the unit of power of a lens? Define the unit of power of a lens.
(c) A combination of lenses for a camera contains two converging lenses of focal lengths 20 cm and 40 cm and a diverging lens of focal length 50 cm. Find the power and focal length of the combination.
Answer:
ans
Page No 262:
Question 30:
(a) Two lenses A and B have power of (i) +2D and (ii) −4D respectively. What is the nature and focal length of each lens?
(b) An object is placed at a distance of 100 cm from each of the above lenses A and B. Calculate (i) image distance, and (ii) magnification, in each of the two cases.
Answer:
(a) Power of lens A = +2D
Power of lens B = 4D
Power = $\frac{1}{\mathrm{focal}\mathrm{length}}$
or Focal length = $\frac{1}{\mathrm{power}}$
Thus, focal length (f_{A} ) of A = $\frac{1}{2}=0.5\mathrm{m}=50\mathrm{cm}$
Focal length (f_{B }) of B = $\frac{1}{4}=0.25\mathrm{m}=25\mathrm{cm}$
Therefore, lens A is a convex lens as it has a positive focal length and lens B is concave as it has a negative focal length.
(b)
Object distance (u) = 100 cm (sign convention)
For lens A:
Image distance (v) = ?
Using the lens formula, we get:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\frac{1}{100}=\frac{1}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{100}=\frac{1}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{50}\frac{1}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{21}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{100}\phantom{\rule{0ex}{0ex}}\therefore v=100\mathrm{cm}.$
∴Magnification = $\frac{v}{u}=\frac{100}{100}=1$.
For lens B:
Image distance (v_{2}) =?
Using the lens formula, we get:
$\frac{1}{v}\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}\frac{1}{100}=\frac{1}{25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{100}=\frac{1}{25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{25}\frac{1}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{41}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{5}{100}\phantom{\rule{0ex}{0ex}}\therefore v=20\mathrm{cm}.$
Negative sign shows that the image formed is virtual.
∴ Magnification = $\frac{\mathit{v}}{\mathit{u}}=\frac{20}{100}=0.2$.
Page No 262:
Question 31:
The focal lengths of four convex lenses P, Q, R and S are 20 cm, 15 cm, 5 cm and 10 cm, respectively. The lens having greatest power is :
(a) P
(b) Q
(c) R
(d) S
Answer:
(c) R
Power = $\frac{1}{\mathrm{Focal}\mathrm{length}}$
Therefore, a lens with a small focal length will have more power.
Page No 262:
Question 32:
A converging lens has a focal length of 50 cm. The power of this lens is :
(a) +0.2D
(b) −2.0D
(c) +2.0D
(d) −0.2D
Answer:
(c) +2.0 D
Power = $\frac{1}{\mathrm{Focal}\mathrm{length}}=\frac{1}{0.5\mathrm{m}}=+2\mathrm{D}$.
Page No 262:
Question 33:
A diverging lens has a focal length of 0.10 m. The power of this lens will be :
(a) +10.0D
(b) +1.0D
(c) −1.0D
(d) −10.0D
Answer:
(d) 10.0D,
since power of lens = $\frac{1}{\mathrm{focal}\mathrm{length}}=\frac{1}{0.10}=10.0\mathrm{D}$.
Page No 262:
Question 34:
The power of a lens is +2.0D. Its focal length should be :
(a) 100 cm
(b) 50 cm
(c) 25 cm
(d) 40 cm
Answer:
(b) 50 cm
Focal length = $\frac{1}{\mathrm{Power}}=\frac{1}{2.0}=0.5\mathrm{m}=50\mathrm{cm}$
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Question 35:
If a spherical lens has a power of, −0.25 D, the focal length of this lens will be :
(a) −4 cm
(b) −400 mm
(c) −4 m
(d) −40 cm
Answer:
(c) $$4 m
Focal length = $\frac{1}{\mathrm{Power}}=\frac{1}{0.25}=\frac{100}{25}=4\mathrm{m}$
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Question 36:
The power of a concave lens is 10 D and that of a convex lens is 6 D. When these two lenses are placed in contact with each other, the power of their combination will be :
(a) +16 D
(b) +4 D
(c) −16 D
(d) −4 D
Answer:
(a) 4 D
The powers of the lenses add up when brought in contact.
So, net power = +6 D + (10 D) = 4 D.
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Question 37:
The power of a converging lens is 4.5 D and that of a diverging lens is 3 D. The power of this combination of lenses placed close together is :
(a) +1.5D
(b) +7.5D
(c) −7.5D
(d) −1.5D
Answer:
(a) 1.5D
Because power of lens adds up when placed in combination.
Therefore
Net power = 4.5 D + ($$3.0 D) = 1.5 D
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Question 38:
A convex lens of focal length 10 cm is placed in contact with a concave lens of focal length 20 cm. The focal length of this combination of lenses will be:
(a) +10 cm
(b) +20 cm
(c) −10 cm
(d) −20 cm
Answer:
(b) +20 cm
We know that:
Power (P) = $\frac{1}{\mathrm{Focal}\mathrm{length}}$
Therefore
Power of convex lens P_{1}= $\frac{1}{10\mathrm{cm}}=\frac{1}{0.1\mathrm{m}}=+10\mathrm{D}$
Power of concave lens P_{2}= $\frac{1}{20\mathrm{cm}}=\frac{1}{0.2\mathrm{m}}=5\mathrm{D}$
We know that power of lens adds up when lenses comes in contact. Therefore, we have:
Net power (P) = P_{1} + P_{2}
Net power (P) = 10 D $$ 5 D = 5 D
Net focal length = $\frac{1}{\mathrm{Power}}=\frac{1}{5}=0.2\mathrm{m}=20\mathrm{cm}$
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Question 39:
The optical prescription for a pair of spectacles is :
Right eye : −3.50 D
Left eye : −4.00 D
(a) Are these lenses thinner at the middle or at the edges?
(b) Which lens has a greater focal length?
(c) Which is the weaker eye?
Answer:
(a) These lens are thinner at the middle because they are concave lens. Further, concave lens has negative power.
(b) Lens for right eye has greater focal length because power is inversely proportional to focal length.
(c) Left eye is weaker because it has a correction with a lens of lower focal length.
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Question 40:
A person got his eyes tested by an optician. The prescription for the spectacle lenses to be made reads :
Left eye : +2.50 D
Right eye : +2.00 D
(a) State whether these lenses are thicker in the middle or at the edges.
(b) Which lens bends the light rays more strongly?
(c) State whether these spectacle lenses will converge light rays or diverge light rays.
Answer:
(a) These lenses are thicker in the middle because they are convex lens.
(b) Power is inversely proportional to focal length; therefore, shorter focal length will mean strong bending. Hence, lens for left eye will bend light rays more strongly.
(c) These spectacles will converge light rays because convex lens has positive power.
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