Science Ncert Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection And Refraction are provided here with simple step-by-step explanations. These solutions for Light Reflection And Refraction are extremely popular among Class 10 students for Science Light Reflection And Refraction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Science Ncert Exemplar Book of Class 10 Science Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Science Ncert Exemplar Solutions. All Science Ncert Exemplar Solutions for class Class 10 Science are prepared by experts and are 100% accurate.

#### Page No 79:

#### Question 1:

Which of the following can make a parallel beam of light when light from a point source is incident on it?

(a) Concave mirror as well as convex lens

(b) Convex mirror as well as concave lens

(c) Two plane mirrors placed at 90° to each other

(d) Concave mirror as well as concave lens

#### Answer:

When the point source is kept at the focus of the concave mirror and convex lens, the light rays after reflection and refraction respectively, will become parallel to the principal axis.

Hence, the correct answer is option A.

#### Page No 79:

#### Question 2:

A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is

(a) –30 cm

(b) –20 cm

(c) –40 cm

(d) –60 cm

#### Answer:

Given:

${h}_{\mathrm{i}}=-5\mathrm{mm},{h}_{o}=10\mathrm{mm},v=-30\mathrm{cm}$

From the magnification formula we know,

$m=\frac{{h}_{\mathrm{i}}}{{h}_{\mathrm{o}}}=-\frac{v}{u}$

Substituting the values in the above equation,

$\frac{-5}{10}=-\frac{-30}{u}\phantom{\rule{0ex}{0ex}}u=30\times \frac{-10}{5}\phantom{\rule{0ex}{0ex}}=-60\mathrm{cm}$

Using the mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\frac{1}{-30}+\frac{1}{-60}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{-2-1}{60}\phantom{\rule{0ex}{0ex}}f=-20\mathrm{cm}$

Hence, the correct answer is option B.

#### Page No 79:

#### Question 3:

Under which of the following conditions a concave mirror can form an image larger than the actual object?

(a) When the object is kept at a distance equal to its radius of curvature

(b) When object is kept at a distance less than its focal length

(c) When object is placed between the focus and centre of curvature

(d) When object is kept at a distance greater than its radius of curvature

#### Answer:

In case of concave mirror, when the object is kept between the focus and the pole of the mirror the virtual enlarged image is formed.

Hence, the correct answer is option B.

#### Page No 79:

#### Question 4:

Figure 10.1 shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is

(a) $\sqrt{3}/\sqrt{2}$

(b) $\sqrt{2}/\sqrt{3}$

(c) $1/\sqrt{2}$

(d) $\sqrt{2}$

#### Answer:

Given:

Angle of incidence $i=60\xb0,\mathrm{Angle}\mathrm{of}\mathrm{refraction}r=45\xb0$

Using snell's law we get,

Refractive index of medium B relative to medium A ,

${n}_{\mathrm{BA}}=\frac{\mathrm{sin}i}{\mathrm{sin}r}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}60}{\mathrm{sin}45}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{\sqrt{2}}}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{\sqrt{2}}$

Hence, the correct answer is option A.

#### Page No 80:

#### Question 5:

A light ray enters from medium A to medium B as shown in Figure 10.2. The refractive index of medium B relative to A will be

(a) greater than unity

(b) less than unity

(c) equal to unity

(d) zero

#### Answer:

The ray of light travelling from medium A to medium B bends towards the normal. It shows that the medium B is denser and the medium A is rarer.

Hence, the correct answer is option A.

#### Page No 80:

#### Question 6:

Beams of light are incident through the holes A and B and emerge out of box through the holes C and D respectively as shown in the Figure10.3. Which of the following could be inside the box?

(a) A rectangular glass slab

(b) A convex lens

(c) A concave lens

(d) A prism

#### Answer:

In the given situation, the incident ray and the emergent ray are parallel to each other. It is possible only in the case of rectangular glass slab.

Hence, the correct answer is option A.

#### Page No 80:

#### Question 7:

A beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the Figure . Which of the following could be inside the box?

(a) Concave lens

(b) Rectangular glass slab

(c) Prism

(d) Convex lens

#### Answer:

In the given situation, the parallel rays are converged at a point. It is possible only in the case of convex lens.

Hence, the correct answer is option D.

#### Page No 80:

#### Question 8:

Which of the following statements is true?

(a) A convex lens has 4 dioptre power having a focal length 0.25 m

(b) A convex lens has –4 dioptre power having a focal length 0.25 m

(c) A concave lens has 4 dioptre power having a focal length 0.25 m

(d) A concave lens has –4 dioptre power having a focal length 0.25 m

#### Answer:

The focal length and the power of concave lens is always negative and of convex lens is positive. The relation between power and focal length is, $P=\frac{1}{f}$.

Using this formula to find power,

$P=\frac{1}{f}\phantom{\rule{0ex}{0ex}}=\frac{1}{0.25}\phantom{\rule{0ex}{0ex}}=4D$

Hence, the correct answer is option A.

#### Page No 81:

#### Question 9:

Magnification produced by a rear view mirror fitted in vehicles

(a) is less than one

(b) is more than one

(c) is equal to one

(d) can be more than or less than one depending upon the position of the object in front of it

#### Answer:

The mirror fitted in vehicles are convex mirror. The images formed by the convex mirror are virtual, erect and diminished or point size. It shows that the height of image will be always smaller than the height of object. The relation between the height of image and the magnification is, $m=\frac{{h}_{\mathrm{i}}}{{h}_{\mathrm{o}}}$.

Hence, the correct answer is option A.

#### Page No 81:

#### Question 10:

Rays from Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that size of its image is equal to the size of the object?

(a) 15 cm in front of the mirror

(b) 30 cm in front of the mirror

(c) between 15 cm and 30 cm in front of the mirror

(d) more than 30 cm in front of the mirror

#### Answer:

The rays from the sun are considered parallel to the principal axis. These rays will converge at the focus of the concave mirror. It shows that focal length of the mirror is 15 cm and the centre of curvature is 30 cm. In case of concave mirror, when the object is placed at the centre of curvature, the image is formed at the centre of curvature itself and of the same size.

Hence, the correct answer is option B.

#### Page No 81:

#### Question 11:

A full length image of a distant tall building can definitely be seen by using

(a) a concave mirror

(b) a convex mirror

(c) a plane mirror

(d) both concave as well as plane mirror

#### Answer:

The image formed by the convex mirror is virtual, erect and diminished. This mirror can be easily used to see full length image of the tall building.

Hence, the correct answer is option B.

#### Page No 81:

#### Question 12:

In torches, search lights and headlights of vehicles the bulb is placed

(a) between the pole and the focus of the reflector

(b) very near to the focus of the reflector

(c) between the focus and centre of curvature of the reflector

(d) at the centre of curvature of the reflector

#### Answer:

In torches, search lights and headlights of vehicles concave mirrors are used. When a light source is placed at the focus of the mirror, after reflection light rays become parallel to the principal axis. Due to this, the bulb is placed very near to the focus of the reflector.

Hence, the correct answer is option B.

#### Page No 81:

#### Question 13:

The laws of reflection hold good for

(a) plane mirror only

(b) concave mirror only

(c) convex mirror only

(d) all mirrors irrespective of their shape

#### Answer:

The laws of reflection hold good for all mirrors irrespective of their shape.

Hence, the correct answer is option D.

#### Page No 81:

#### Question 14:

The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are shown as A, B, C and D in Figure 10.5. Which one of them is correct?

(a) A

(b) B

(c) C

(d) D

#### Answer:

The light rays when pass through the rectangular glass slab, they undergo refraction such that the emergent ray is parallel to the incident ray. It is correctly shown in figure B.

Hence, the correct answer is option B.

#### Page No 82:

#### Question 15:

You are given water, mustard oil, glycerine and kerosene. In which of these media a ray of light incident obliquely at same angle would bend the most?

(a) Kerosene

(b) Water

(c) Mustard oil

(d) Glycerine

#### Answer:

The refractive index of glycerine is highest among water, mustard oil and kerosene. Due to this, a ray of light incident obliquely at same angle will bend most in case of glycerine.

Hence, the correct answer is option D.

#### Page No 82:

#### Question 16:

Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in Figure 10.6?

(a) Fig. A

(b) Fig. B

(c) Fig. C

(d) Fig. D

#### Answer:

In concave mirror, the ray of light parallel to the principal axis after reflection from the mirror will always pass through the focus.

Hence, the correct answer is option D

#### Page No 82:

#### Question 17:

Which of the following ray diagrams is correct for the ray of light incident on a lens shown in Fig. 10.7?

(a) Fig. A.

(b) Fig. B.

(c) Fig. C.

(d) Fig. D.

#### Answer:

In convex lens, the ray of light coming from the focus after refraction from the lens will become parallel to the principal axis.

Hence, the correct answer is option A.

#### Page No 83:

#### Question 18:

A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.

(a) Plane, convex and concave

(b) Convex, concave and plane

(c) Concave, plane and convex

(d) Convex, plane and concave

#### Answer:

Concave mirror can form magnified image of the object. Plane mirror always forms image of the same size as that of object. Convex mirror can form diminished or point size image of the object. The child finds image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. This shows that the order of combinations for the magic mirror from the top is concave mirror, plane mirror and convex mirror.

Hence, the correct answer is option C.

#### Page No 83:

#### Question 19:

In which of the following, the image of an object placed at infinity will be highly diminished and point sized?

(a) Concave mirror only

(b) Convex mirror only

(c) Convex lens only

(d) Concave mirror, convex mirror, concave lens and convex lens

#### Answer:

The ray of light from an object placed at infinity will always form point size image at focus, valid for concave mirror, convex mirror, concave lens and convex lens.

Hence, the correct answer is option D.

#### Page No 83:

#### Question 20:

Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.

(a) Object is placed between device and its focus, image formed is enlarged and behind it.

(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.

(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.

(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.

#### Answer:

The devices are:

(a) Concave mirror

(b) Convex lens

(c) Concave lens

(d) Convex mirror

#### Page No 83:

#### Question 21:

Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.

#### Answer:

When a ray of light enters into the glass slab, at the air and glass interface the ray will bend towards the normal and will travel in a straight line inside the glass slab. At the other end of the glass slab, the ray will emerge out from the glass slab, again at the glass and medium interface the ray will bend away from the normal. This emerged ray will be parallel to the direction of the incident ray.

#### Page No 83:

#### Question 22:

A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine. Support your answer with reason.

#### Answer:

The pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water due to the refraction of light. The refractive index of different medium is different. The refractive index of kerosene is more than the water, due to this the pencil will bend more then in case of water.

Therefore the pencil will not appear to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine.

#### Page No 83:

#### Question 23:

How is the refractive index of a medium related to the speed of light? Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media?

#### Answer:

The refractive index of a medium is the ratio of speed of light in vacuum and speed of light in medium.

$\mathrm{Refractive}\mathrm{Index}\left(n\mathit{\right)}\mathit{=}\frac{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{absolute}\mathrm{vauum}\mathit{}\mathit{\left(}c\mathit{\right)}}{\mathrm{Speed}\mathrm{of}\mathrm{light}\mathrm{in}\mathrm{medium}\mathit{}\mathit{\left(}v\mathit{\right)}}$

The expression for speed of light from air to medium *A,
${n}_{\mathrm{A}}=\frac{c}{{v}_{\mathrm{A}}}$*

The expression for speed of light from air to medium

*B,*

${n}_{\mathrm{B}}=\frac{c}{{v}_{\mathrm{B}}}$

${n}_{\mathrm{B}}=\frac{c}{{v}_{\mathrm{B}}}$

The expression for refractive index of a medium

*A*with respect to medium

*B*in terms of speed of light in these two media is,

*$\frac{{n}_{\mathrm{B}}}{{n}_{\mathrm{A}}}=\frac{{\displaystyle \frac{c}{{v}_{\mathrm{B}}}}}{{\displaystyle \frac{c}{{v}_{\mathrm{A}}}}}\phantom{\rule{0ex}{0ex}}=\frac{{v}_{\mathrm{A}}}{{v}_{\mathrm{B}}}$*

#### Page No 83:

#### Question 24:

Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.

#### Answer:

Given:

$\mathrm{Refractive}\mathrm{Index}\mathrm{of}\mathrm{glass}\left({n}_{\mathrm{g}}\mathit{\right)}\mathit{=}1.5\phantom{\rule{0ex}{0ex}}\mathrm{Refractive}\mathrm{Index}\mathrm{of}\mathrm{diamond}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{glass}\left({n}_{\mathrm{dg}}\mathit{\right)}\mathit{=}1.6$

$\mathrm{Refractive}\mathrm{index}\mathrm{of}\mathrm{diamond}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{glass}=\frac{\mathrm{Refractive}\mathrm{index}\mathrm{of}\mathrm{diamond}\left({n}_{d}\right)}{\mathrm{Refractive}\mathrm{index}\mathrm{of}\mathrm{glass}}\phantom{\rule{0ex}{0ex}}1.6=\frac{{\mathrm{n}}_{\mathrm{d}}}{1.5}\phantom{\rule{0ex}{0ex}}{\mathrm{n}}_{\mathrm{d}}=1.5\times 1.6\phantom{\rule{0ex}{0ex}}=2.4$

Hence, the the absolute refractive index of diamond is 2.4.

#### Page No 84:

#### Question 25:

A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?

#### Answer:

In convex lens, when the object is kept between the optical centre and the focus, the virtual, erect and magnified image is formed. The same object, when kept between the focus and the centre of curvature will produce the real, inverted and magnified image. Therefore, the statement is correct. To obtain the magnified virtual as well as real image, the object must be kept between 0 to 20 cm and 20 cm to 40 cm respectively.

#### Page No 84:

#### Question 26:

Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?

#### Answer:

The image of the window pane is formed at a distance of 15 cm from the lens. The building which acts as an object is outside the window so it must be away from the widow pane, it means the distance of the object is increased from the lens. When the object distance is increased, the image distance is decreased and the image gets closer to focus. Therefore, the screen must be moved towards the lens to obtain the sharp image of the building.

The approximate value of the focal length is 15 cm, but to get the exact value, the given data is insufficient.

#### Page No 84:

#### Question 27:

How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?

#### Answer:

The power of the lens is the reciprocal of the focal length.

Mathematically, $\mathrm{Power}\left(P\right)=\frac{1}{\mathrm{focal}\mathrm{length}\left(f\right)}$

The high power of the lens indicates less focal length and more converging power. Therefore, the lens of focal length 20 cm is more converging.

#### Page No 84:

#### Question 28:

Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.

#### Answer:

When the two plane mirrors are perpendicular to each other, the incident ray and the reflected ray will always be parallel to each other, whatever may be angle of incidence. The same can be seen in the below diagram:

#### Page No 84:

#### Question 29:

Draw a ray diagram showing the path of rays of light when it enters with oblique incidence (i) from air into water; (ii) from water into air.

#### Answer:

(i) The ray diagram showing the path of rays of light when it enters with oblique incidence from air into water;

(ii) The ray diagram showing the path of rays of light when it enters with oblique incidence from water into air;

#### Page No 84:

#### Question 30:

Draw ray diagrams showing the image formation by a concave mirror when an object is placed

(a) between pole and focus of the mirror

(b) between focus and centre of curvature of the mirror

(c) at centre of curvature of the mirror

(d) a little beyond centre of curvature of the mirror

(e) at infinity

#### Answer:

The ray diagrams showing the image formation by a concave mirror when an object is placed,

(a) between pole and focus of the mirror,

(b) between focus and centre of curvature of the mirror,

(c) at centre of curvature of the mirror,

(d) a little beyond centre of curvature of the mirror,

(e) at infinity,

#### Page No 84:

#### Question 31:

Draw ray diagrams showing the image formation by a convex lens when an object is placed

(a) between optical centre and focus of the lens

(b) between focus and twice the focal length of the lens

(c) at twice the focal length of the lens

(d) at infinity

(e) at the focus of the lens

#### Answer:

The ray diagrams showing the image formation by a convex lens when an object is placed,

(a) between optical centre and focus of the lens,

(b) between focus and twice the focal length of the lens,

(c) at twice the focal length of the lens,

(d) at infinity,

(e) at the focus of the lens,

#### Page No 84:

#### Question 32:

Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.

#### Answer:

The two laws of refraction are:

__First Law__: The incident ray, the refracted ray and the normal all lie on the same plane.

__Second Law__: The ratio of the sine angle of incident ray and the refracted ray are constant. This law is also known as "Snell's Law".

In the above diagram, we can see that the incident ray, the refracted ray and the normal all lie on the same plane. It verifies the first law. Similarly, if we take the ratio of the sine angle of incident ray and the refracted ray, we will obtain a constant value. It verifies the second law.

#### Page No 85:

#### Question 33:

Draw ray diagrams showing the image formation by a concave lens when an object is placed

(a) at the focus of the lens

(b) between focus and twice the focal length of the lens

(c) beyond twice the focal length of the lens

#### Answer:

The ray diagram showing the image formation by concave lens when an object is placed,

(a) at the focus of the lens,

(b) between focus and twice the focal length of the lens,

(c) beyond twice the focal length of the lens,

#### Page No 85:

#### Question 34:

Draw ray diagrams showing the image formation by a convex mirror when an object is placed

(a) at infinity

(b) at finite distance from the mirror

#### Answer:

The ray diagrams showing the image formation by a convex mirror when an object is placed,

(a) at infinity,

(b) at finite distance from the mirror,

#### Page No 85:

#### Question 35:

The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?

#### Answer:

Given:

Height of image$\left({h}_{\mathrm{i}}\right)=3\times \mathrm{Height}\mathrm{of}\mathrm{object}\left({h}_{\mathrm{O}}\right)$

Image distance$\left(v\right)=80\mathrm{cm}$

Using the magnification formula,

$\frac{{h}_{\mathrm{i}}}{{h}_{\mathrm{o}}}=\frac{v}{u}$

The image is formed on the other side of the lens, so it must be real and inverted, substituting the values in the above equation,

$\frac{-3\times {h}_{\mathrm{o}}}{{h}_{\mathrm{o}}}=\frac{80}{u}\phantom{\rule{0ex}{0ex}}u=\frac{-80}{3}$

The negative sign shows that the object is in front of the lens and the unit is in cm.

The image is formed on the other side of the lens, so the image must be real and inverted. The image is formed on the other side of the lens, it is possible only in the case of convex lens.

#### Page No 85:

#### Question 36:

Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

#### Answer:

Given:

Focal length$\left(f\right)=20\mathrm{cm}$, Size of image$\left({h}_{\mathrm{i}}\right)=\frac{1}{3}\times {h}_{\mathrm{o}}(\mathrm{Size}\mathrm{of}\mathrm{object})$

Using the magnification formula,

$m=\frac{{h}_{\mathrm{i}}}{{h}_{\mathrm{o}}}=\frac{-v}{u}$

Substituting the values in the above equation,

$\frac{{\displaystyle \frac{{h}_{\mathrm{o}}}{3}}}{{h}_{\mathrm{o}}}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{3}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}v=\frac{-u}{3}$

Using the mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\frac{-3}{u}+\frac{1}{u}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}$

$\frac{-2}{u}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}u=-40\mathrm{cm}$

Negative sign shows that the object is infront of the mirror.

Substituting the above value in obtained image distance and object distance relation,

$v=\frac{-(-40)}{3}\phantom{\rule{0ex}{0ex}}=\frac{40}{3}\mathrm{cm}$

It shows that the image formed is virtual, erect and diminished. It is possible in case of convex mirror only. Therefore, it is the case of convex mirror.

#### Page No 85:

#### Question 37:

Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of –50 cm. What is the nature of the lens and its power used by each of them?

#### Answer:

Power of a lens indicates the converging effect of the lens. It is the reciprocal of the focal length of the lens.

Mathematically, $\mathrm{Power}\left(P\right)=\frac{1}{f}$

The unit of power is dioptre (*D*).

The nature and power of the lens in each case are:

When the focal length of the lens is 50 cm,

$\left(P\right)=\frac{1}{f}\phantom{\rule{0ex}{0ex}}=\frac{1}{0.5}\phantom{\rule{0ex}{0ex}}=2\mathrm{D}$

This lens has positive value of focal length. So, it is convex lens.

When the focal length of the lens is $-$50 cm,

$\left(P\right)=\frac{1}{f}\phantom{\rule{0ex}{0ex}}=\frac{1}{-0.5}\phantom{\rule{0ex}{0ex}}=-2\mathrm{D}$

This lens has negative value of focal length. So, it is concave lens.

#### Page No 85:

#### Question 38:

A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under

Position of candle = 12.0 cm

Position of convex lens = 50.0 cm

Position of the screen = 88.0 cm

(i) What is the focal length of the convex lens?

(ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?

(iii) What will be the nature of the image formed if he further shifts the candle towards the lens?

(iv) Draw a ray diagram to show the formation of the image in case

(iii) as said above.

#### Answer:

Given:

Position of candle = 12.0 cm, Position of the screen = 88.0 cm, Position of convex lens = 50.0 cm

Position of object$\left(u\right)=-(50-12)=-38\mathrm{cm}$, Position of image$\left(v\right)=88-50=38\mathrm{cm}$

(i)

Using the lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\frac{1}{38}-\frac{1}{-38}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\frac{2}{38}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}f=19\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

The focal length of the convex lens is 19 cm.

(ii)

Given: Position of object$\left(u\right)=-(50-31)=-19\mathrm{cm}$

Using the lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\frac{1}{v}-\frac{1}{-19}=\frac{1}{19}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=0\phantom{\rule{0ex}{0ex}}v=\mathrm{infinity}\phantom{\rule{0ex}{0ex}}$

(iii)

In the previous case, the candle was kept at the focus, if he further shifts the candle towards the lens. The virtual, erect and magnified image will be formed.

(iv)

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