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#### Page No 173:

#### Question 1:

What happens when a ray of light falls normally (or perpendiculary) on the surface of a plane mirror?

#### Answer:

When a ray of light falls normally (or perpendicularly) on the surface of a plane mirror, it gets reflected along the same path because the angles of incidence and reflection are both equal to zero.

#### Page No 173:

#### Question 2:

A ray of light is incident on a plane mirror at an angle of 30°. What is the angle of reflection?

#### Answer:

The angle of reflection is 30⁰ as the angle of incidence is equal to the angle of reflection, in accordance with the first law of reflection.

#### Page No 173:

#### Question 3:

A ray of light strikes a plane mirror at an angle of 40° to the mirror surface. What will be the angle of reflection?

#### Answer:

The angle of reflection is 50^{0}, in accordance with the first law of reflection, which states that the angle of incidence is equal to the angle of reflection.

Here, the angle of incidence = 90^{o} - 40^{o} = 50^{o}( the angle of incidence is the angle between the incident ray and the normal).

#### Page No 173:

#### Question 4:

A ray of light is incident normally on a plane mirror. What will be the

(*a*) angle of incidence?

(*b*) angle of reflection?

#### Answer:

(a) The angle of incidence is 0^{o} because the incident ray is parallel to the normal.

(b) The angle of reflection is 0^{o} in accordance with the first law of reflection, which states that the angle of incidence is equal to the angle of reflection.

#### Page No 173:

#### Question 5:

What type of image is formed:

(*a*) in a plane mirror?

(*b*) on a cinema screen?

#### Answer:

(a) The image formed in a plane mirror is virtual because it is formed inspite of there not being an actual meeting of the light rays.

(b) The image formed on a cinema screen is real because it is formed due to an actual meeting of the light rays.

#### Page No 173:

#### Question 6:

What kind of mirror is required for obtaining a virtual image of the same size as the object?

#### Answer:

A plane mirror is required in order to obtain a virtual image of the same size as the object, as this is one of the properties of a plane mirror.

#### Page No 173:

#### Question 7:

What is the name of the phenomenon in which the right side of an object appears to be the left side of the image in a plane mirror?

#### Answer:

Lateral inversion is the phenomenon in which the right side of an object appears to be the left side in its image in a plane mirror.

#### Page No 173:

#### Question 8:

Name the phenomenon responsible for the following effect:

When we sit in front of a plane mirror and write with our right hand, if appears in the mirror that we are writing with the left hand.

#### Answer:

The phenomenon of lateral inversion is the reason for the given effect.

#### Page No 173:

#### Question 9:

If an object is placed at a distance of 10 cm in from of a plane mirror, how far would it be from its image?

#### Answer:

It will be 20 cm away from its image. This is because the distance of the plane mirror from the object is equal to its distance from the image.

#### Page No 173:

#### Question 10:

Which property of light makes a pencil cast a shadow when it is held in front of a light source?

#### Answer:

One property of light that makes a pencil cast a shadow when it is held in front of it is that light travels in a straight line.

#### Page No 173:

#### Question 11:

The image seen in a plane mirror cannot be formed on a screen. What name is given to this type of image?

#### Answer:

An image that cannot be formed on a screen is called a virtual image.

#### Page No 173:

#### Question 12:

Fill in the following blank with a suitable word:

When light is reflected, the angles of incidence and reflection are ............ .

#### Answer:

When light is reflected, the angles of incidence and reflection are __equal__.

#### Page No 173:

#### Question 13:

State whether the following statement is true of false:

A student says that we can see an object because light from our eyes is reflected back by the object.

#### Answer:

False.We can see an object because light, on being reflected from the object, is scattered and diffused.

#### Page No 173:

#### Question 14:

Where is the image when you look at something in a mirror?

#### Answer:

When we look at something in a mirror, the image formed is behind it.

#### Page No 173:

#### Question 15:

A ray of light strikes a plane mirror such that its angle of incidence is 30°. What angle does the reflected ray make with the mirror surface?

#### Answer:

The reflected ray makes an angle of 60^{o} with the mirror surface because:

angle of incidence = angle of reflection = 30^{o}, and

angle of the reflected ray with the mirror surface = 90^{o} - angle of reflection = 90^{o} - 30^{o} = 60^{o} .

#### Page No 173:

#### Question 16:

What is the difference between a real image and a virtual image? Give one example of each type of image.

#### Answer:

S. No. | Real Image | Virtual Image |

1 | An image that can be projected on a screen is called a real image. | An image that cannot be projected on a screen is called a virtual image. |

2 | The light rays actually meet at the point of image formation. | The light rays do not actually meet at the point of image formation, but appear to meet there. |

3 | Example: the image formed on the screen in a cinema hall. | Example: the image formed by a plane mirror. |

#### Page No 173:

#### Question 17:

The letter F is placed in front of a plane mirror:

(*a*) How would its image look like when seen in a plane mirror?

(*b*) What is the name of the phenomenon involved?

#### Answer:

(a) The image of the letter F will appear laterally inverse, like this:

(b) The phenomenon involved is lateral inversion.

#### Page No 173:

#### Question 18:

What is lateral inversion? Explain by giving a suitable example.

#### Answer:

When an object is placed in front of a plane mirror, the right side of the object appears as the left side in its image; and the left side of the object appears as the right side in the image. This change of sides of an object and its mirror image is called lateral inversion. For example, the right hand of a person appears as the left hand in the image formed by a plane mirror.

#### Page No 173:

#### Question 19:

Write the word AMBULANCE as it would appear when reflected in a plane mirror. Why is it sometimes written in this way (as its mirror image) on the front of an ambulance?

#### Answer:

The word ambulance would appear as when reflected in a plane mirror. It is written this way so as to help a person driving a vehicle ahead of the ambulance to read it as AMBULANCE when he/she sees the rear view mirror. This is because the rear view mirror forms a laterally inverted image.

#### Page No 173:

#### Question 20:

What are the important differences between looking at a photograph of your face and looking at yourself in a plane mirror?

#### Answer:

Photograph | Image in a Plane Mirror |

A photograph is made when a real image is projected on a photographic film. | The image formed by a plane mirror is virtual and cannot be taken on a photograph. |

The image is smaller in size compared to the actual object. | The image is the same size as the object. |

#### Page No 174:

#### Question 21:

(*a*) A wall reflects light and a mirror also reflects light. What difference is there in the way they reflect light?

(*b*) Which type of reflection of light leads to the formation of images?

#### Answer:

(a)

S.No. | Wall Reflected Light | Mirror Reflected Light |

1 | Light rays are reflected in all directions due to the rough surface. | Light rays are reflected in one direction only due to the smooth surface. |

(b) Regular reflection from smooth surfaces like mirrors leads to the formation of images.

#### Page No 174:

#### Question 22:

What is the difference between regular reflection of light and diffuse reflection of light? What type of reflection of light takes place from:

(*a*) a cinema screen

(*b*) a plane mirror

(*c*) a cardboard

(*d*) still water surface of a lake

#### Answer:

Regular Reflection | Diffused Reflection |

In regular reflection, light is reflected in one direction only because of the smooth surface of the plane. | In diffused reflection, light is reflected in all directions because of the rough surface of the plane. |

S.No. | Type of Surface | Type of Reflection |

a | Cinema screen | Diffused reflection |

b | Plane mirror | Regular reflection |

c | Cardboard | Diffused reflection |

d | Still water surface of a lake | Regular reflection |

#### Page No 174:

#### Question 23:

What can you see in a completely dark room? If you switch on an electric bulb in this dark room as a light source, explain how you could now see:

(*a*) the electric bulb

(*b*) a piece of white paper.

#### Answer:

We cannot see anything in a dark room as there is no light present.

(a) We see the electric bulb because it produces light that reaches our eyes directly.

(b) We see a piece of white paper due to the diffused reflection of light from the surface of the paper.

#### Page No 174:

#### Question 24:

(*a*) A boy with a mouth 5 cm wide stands 2 m away from a plane mirror. Where is his image and how wide is the image of his mouth?

(*b*) The boy walks towards the mirror at a speed of 1 m/s. At what speed does his image approach him?

#### Answer:

(a) The image formed is 2m behind the mirror and 5 cm wide. The reason being, a plane mirror forms an image of the same size as the object, and at the same distance behind the mirror as that of the object from the mirror.

(b) The image approaches him at a speed of 2 m/s . The reason being, the image moves at a speed of 1 m/s and

the relative speed of the image will be equal to the sum of the speeds of the image and the boy.

#### Page No 174:

#### Question 25:

(*a*) An extended object in the form of an arrow pointing upward has been placed in front of a plane mirror. Draw a labelled ray-diagram to show the formation of its image.

(*b*) State the uses of plane mirrors.

#### Answer:

(a)

(b) Uses of a plane mirror:

- Plane mirrors are used to see ourselves. Example, mirrors in a bathroom.
- Plane mirrors are fixed on the inner walls of certain shops ( like jewellery shops) to make them look bigger.
- Plane mirrors are fitted at blind turns of some busy roads so that drivers can see the vehicles coming from the other side and prevent accidents.
- Plane mirrors are used in periscopes.

#### Page No 174:

#### Question 26:

What is meant by 'reflection of light'? Define the following terms used in the study of reflection of light by drawing a labelled ray-diagram:

(*a*) Incident ray

(*b*) Point of incidence

(*c*) Normal

(*d*) Reflected ray

(*e*) Angle of incidence

(*f*) Angle of reflection

#### Answer:

The process of sending back light rays that fall on the surface of an object is called reflection of light. (Figure)

(a) Incident ray: The ray of light that falls on a mirror is called the incident ray. In the figure, AO is the incident ray of light. The incident ray gives the direction of the light falling on the mirror.

(b) Point of incidence: The point at which the incident ray falls on a mirror is called the point of incidence. In the figure, the point O on the surface of the mirror is the point of incidence.

(c) Normal: The normal is a line at right angles to the mirror surface at the point of incidence. In the figure, the line ON is the normal to the mirror surface at point O.

(d) Reflected ray: The ray of light that is sent back by the mirror is called the reflected ray. In the figure, OB is the reflected ray of light. The reflected ray of light shows the direction in which the light travels after being reflected from the mirror.

(e) Angle of incidence: The angle of incidence is the angle made by the incident ray with the normal at the point of incidence. In the figure, the angle AON is the angle of incidence. The angle of incidence is denoted by the letter i.

(f) Angle of reflection: The angle of reflection is the angle made by the reflected ray with the normal at the point of incidence. In the figure, the angle NOB is the angle of reflection. The angle of reflection is denoted by the letter r.

#### Page No 174:

#### Question 27:

State and explain the laws of reflection of light at a plane surface (like a plane mirror), With the help of a labelled ray-diagram. Mark the angles of 'incidence' and 'reflection' clearly on the diagram. If the angle of reflection is 47.5°, what will be the angle of incidence?

#### Answer:

a) The two laws of reflection are:

- The angle of incidence is always equal to the angle of reflection . If the angle of incidence is i and the angle of reflection is r, then, according to the first law of reflection, $\angle \mathrm{i}=\angle \mathrm{r}$ .

- The second law of reflection states that the incident ray, the reflected ray and the normal (at the point of incidence), all lie in the same plane. For example, in the figure, the incident ray AO, the reflected ray OB and the normal ON, all lie in the same plane.

#### Page No 174:

#### Question 28:

With the help of a labelled ray-diagram, describe how a plane mirror forms an image of a point source of light placed in front of it. State the characteristics of the image formed in a plane mirror.

#### Answer:

Consider a small object (a point source of light), placed in front of a plane mirror MM' (figure 1). The mirror will form an image I of the object O. This process of image formation is explained as follows:

The object O gives out light rays in all directions. Now, a ray of light OA, coming from the object O, is incident on the plane mirror at point A. OA gets reflected in the direction AX in accordance with the law of reflection, which states that the angle of reflection r_{1}_{ }equals the angle of incidence i_{1}. Another ray of light OB, coming from the object O, strikes the mirror at point B. OB gets reflected in the direction BY, thus, making the angle of reflection r_{2} equal to the angle of incidence i_{2}.

The two reflected rays AX and BY are divergent and cannot meet on the left side. Let's produce the reflected rays AX and BY backwards. They meet at a point I behind the mirror. When the reflected rays AX and BY enter the eye of a person at position E, the eye sees the rays in the direction in which they enter. So, the person looking into the mirror from position E sees the reflected rays as if they are coming from the point I behind the mirror. Thus, the point I is the image of the object O formed by the plane mirror.

The image produced by the plane mirror is virtual, laterally inverse and of the same size as the object.

#### Page No 174:

#### Question 29:

(*a*) Explain why, though both a plane mirror and a sheet of paper reflect light but we can see the image of our face in a plane mirror but not in a sheet of paper.

(*b*) The image in a plane mirror is virtual and laterally inverted. What does this statement mean?

(*c*) Write all the capital letters of the alphabet which look the same in a plane mirror.

#### Answer:

(a) This is because there occurs regular reflection from a plane mirror, which has a smooth surface. Since the particles of the smooth surface are facing one direction, a beam of parallel light rays falling on it is reflected as a beam of parallel light rays in one direction only. These rays meet when produced backwards to form a virtual image of the light source.

However, in the case of a rough paper, a parallel beam of incident light is reflected in different directions (diffused reflection). So, the light rays don't meet to form an image of the object.

(b) This statement implies that the image formed by a plane mirror cannot be produced on a screen. Thus, it is a virtual image. Further, in the image formed, the right side of an object appears as the left side and vice-versa. This is called lateral inversion.

(c) All the capital letters of the alphabet that look the same in a plane mirror are W, X, V, A, H, M, O, I and T.

#### Page No 174:

#### Question 30:

The angle of reflection is equal to the angle of incidence:

(*a*) always

(*b*) sometimes

(*c*) under special conditions

(*d*) never

#### Answer:

(a) Always,

In accordance with the first law of reflection, which states that the angle of reflection is equal to the angle of incidence.

#### Page No 174:

#### Question 31:

The angle between an incident ray and the plane mirror is 30°. The total angle between the incident ray and reflected ray will be:

(*a*) 30°

(*b*) 60°

(*c*) 90°

(*d*) 120°

#### Answer:

(b) 120^{o}

Since, angle of incidence = 90^{o} - angle between plane mirror and incident ray = 90^{o} - 30^{o} = 60^{o}

and according to first law of reflection, angle of incidence = angle of reflection = 60^{o}

Total angle between incident ray and reflected ray = 60^{o}^{ }+ 60^{o}^{ }= 120^{o}^{ .}

#### Page No 174:

#### Question 32:

A ray of light is incident on a plane mirror making an angle of 90° with the mirror surface. The angle of reflection for this ray of light will be:

(*a*) 45°

(*b*) 90°

(*c*) 0°

(*d*) 60°

#### Answer:

(c) 0^{o}

since angle of incidence = 0^{o} .

According to the first law of reflection, the angle of incidence is equal to the angle of reflection.

#### Page No 174:

#### Question 33:

The image of an object formed by a plane mirror is:

(*a*) virtual

(*b*) real

(*c*) diminished

(*d*) upside-down

#### Answer:

(a) Virtual

The reason being, the image cannot be projected on a screen.

#### Page No 174:

#### Question 34:

The image formed by a plane mirror is :

(*a*) virtual, behind the mirror and enlarged.

(*b*) virtual, behind the mirror and of the same size as the object.

(*c*) real, at the surface of the mirror and enlarged.

(*d*) real, behind the mirror and of the same size as the object.

#### Answer:

(b) Virtual, behind the mirror and of the same size as the object

The image formed by a plane mirror is virtual, behind the mirror and of the same size because it cannot be projected on a screen.

#### Page No 174:

#### Question 35:

The figure given alongside shows the image of a clock as seen a plane mirror. The correct time is:

Figure

(*a*) 2.25

(*b*) 2.35

(*c*) 6.45

(*d*) 9.25

#### Answer:

(d) 9.25,

since the image formed by a plane mirror is laterally inverted.

#### Page No 175:

#### Question 36:

A man stands 10 m in front of a large plane mirror. How far must he walk before he is 5 m away from his image?

#### Answer:

Man should walk 7.5 m towards the mirror.

The reason being, the image formed by a plane mirror is the same distance behind the mirror as it is between the object and the mirror. A distance of 5m between man and his image means that the distance between him and the mirror = $\frac{5}{2}$= 2.5 m.

Thus, the distance he should walk = 10 - 2.5 = 7.5 m.

#### Page No 175:

#### Question 37:

An object is placed 20 cm in front of a plane mirror. The mirror is moved 2 cm towards the object. The distance between the positions of the original and final images seen in the mirror is:

(*a*) 2 cm

(*b*) 4 cm

(*c*) 10 cm

(*d*) 22 cm

#### Answer:

(b) 4 cm

Distance between original image and final image = distance the mirror moved + same distance the image move= 2 + 2 = 4cm.

#### Page No 175:

#### Question 38:

A man sits in an optician's chair looking into plane mirror which is 2 m away from him and views the image of a chart which faces the mirror and is 50 cm behind his head. How far away from his eyes does the chart appear to be?

#### Answer:

The image of the chart will appear 4.5 m away from the eye.

The image of the chart will form at a distance of 2 m + 0.5 m = 2.5 m behind the mirror. The reason being, the image formed by a plane mirror is at the same distance behind it as it is between the object and the mirror.

Thus,

distance of the chart's image from the eye = distance of man from the mirror + distance of image formed behind the mirror = 2 + 2.5 = 4 m.

#### Page No 175:

#### Question 39:

A ray of light strikes a plane mirror* PQ *at an angle of incidence of 30°, is reflected from the plane mirror and then strikes a second plane mirror *QR* placed at right angles to the first mirror. The angle of reflection at the second mirror is:

(*a*) 30°

(*b*) 45°

(*c*) 60°

(*d*) 90°

Draw a ray-diagram to illustrate your answer.

#### Answer:

(c) 60^{o}

Here,

$\angle \mathrm{ABN}=\angle \mathrm{NBC}(\mathrm{angle}\mathrm{of}\mathrm{incidence}=\mathrm{angle}\mathrm{of}\mathrm{reflection})\phantom{\rule{0ex}{0ex}}\angle \mathrm{BCO}=\angle \mathrm{NBC}(\mathrm{alternate}\mathrm{angles})\phantom{\rule{0ex}{0ex}}\angle \mathrm{MCB}={90}^{\mathrm{O}}-\angle \mathrm{BCO}(\angle \mathrm{MCB}=\mathrm{angle}\mathrm{of}\mathrm{incidence}\mathrm{for}\mathrm{mirror}\mathrm{QR})\phantom{\rule{0ex}{0ex}}={60}^{\mathrm{o}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\angle \mathrm{MCB}=\angle \mathrm{MCD}={60}^{\mathrm{o}}(\angle \mathrm{MCD}=\mathrm{angle}\mathrm{of}\mathrm{reflection}\mathrm{from}\mathrm{mirror}\mathrm{QR}\mathrm{and}\mathrm{angle}\mathrm{of}\mathrm{incidence}=\mathrm{angle}\mathrm{of}\mathrm{reflection})\phantom{\rule{0ex}{0ex}}$

#### Page No 175:

#### Question 40:

Explain how to read the following message which was found on some blotting paper:

Figure

#### Answer:

Place a plane mirror in front of the message and see the image of the message. The mirror now reads as 'meet me at midnight'. The phenomenon of lateral inversion is used here.

#### Page No 178:

#### Question 1:

Name the spherical mirror which has:

(*a*) virtual principal focus.

(*b*) real principal focus.

#### Answer:

The spherical mirror which has

(*a*) a virtual principal focus is a convex mirror

(*b*) a real principal focus is a concave mirror

#### Page No 178:

#### Question 2:

Out of convex mirror and concave mirror, whose focus is situated behind the mirror?

#### Answer:

The focus of the convex mirror is situated behind the mirror, as the latter has a virtual focus.

#### Page No 178:

#### Question 3:

Find the focal length of a concave mirror whose radius of curvature is 32 cm.

#### Answer:

Concave mirror has a radius of curvature of 32 cm.

Therefore, R = -32 cm

Hence, the focal length of a concave mirror 'f' = $\frac{-32}{2}$ = -16 cm

#### Page No 178:

#### Question 4:

If the focal length of a convex mirror is 25 cm, what is its radius of curvature?

#### Answer:

The focal length of a convex mirror is 25 cm, i.e. 'f' is 25 cm.

So, the radius of curvature of the convex mirror, 'R' = 2f = $2\times 25$ = 50 cm

#### Page No 179:

#### Question 5:

Fill in the following blanks with suitable words:

(*a*) Parallel rays of light are reflected by a concave mirror to a point called the ..........

(*b*) The focal length of a concave mirror is the distance from the ......... to the mirror.

(*c*) A concave mirror .......... rays of light whereas convex mirror ............ rays of light

(*d*) For a convex mirror, parallel rays of light appear to diverge from a point called the ......... .

#### Answer:

(a) Parallel rays of light are reflected by a concave mirror to a point called the **focus****.**

(b) The focal length of a concave mirror is the distance from the __ focus__ to the mirror.

(c) A concave mirror

**converges****rays of light whereas convex mirror**

**rays of light.**

__diverges__(d) For a convex mirror, parallel rays of light appear to diverge from a point called the

**.**

__focus__#### Page No 179:

#### Question 6:

What is a spherical mirror? Distinguish between a concave mirror and a convex mirror.

#### Answer:

Difference between convex and concave mirror are as follows:

Convex Mirror | Concave Mirror |

1.In a convex mirror, reflection of light takes place on the bulged-out surface. 2. Convex mirror has virtual focus. 3. Convex mirror diverges the rays of light incident on it. |
1.In a concave mirror, reflection of light takes place on the bent surface . 2. Concave mirror has real focus. 3. Concave mirror converges the rays of light incident on it. |

#### Page No 179:

#### Question 7:

Name the two types of spherical mirrors. What type of mirror is represented by the:

(*a*) back side of a shining steel spoon?

(*b*) front side of a shining steel spoon?

#### Answer:

The types of spherical mirrors represented by the

(*a*) back side of a shining steel spoon is convex mirror, and

(*b*) front side of a shining steel spoon is concave mirror

#### Page No 179:

#### Question 8:

What is the relation between the focal length and radius of curvature of a spherical mirror (concave mirror or convex mirror)? Calculate the focal length of a spherical mirror whose radius of curvature is 25 cm.

#### Answer:

The relation between the focal length and radius of curvature of a spherical mirror is as follows:

Radius of curvature = 2 x focal length

R = 2f

Given, the radius of curvature 'R' = 25 cm

Then, the focal length 'f' =$\frac{25}{2}$ = 12.5 cm

#### Page No 179:

#### Question 9:

Explain with a suitable diagram, how a concave mirror converges a parallel beam of light rays. Mark clearly the pole, focus and centre of curvature of concave mirror in this diagram.

#### Answer:

When a parallel beam of light, also parallel to the principle axis, is incident on a concave mirror, it reflects from the mirror and meets at a point called the focus (F) of the concave mirror. So, a concave mirror has a real focus.

__ Ray diagram-__ A concave mirror converges a parallel beam of light rays.

#### Page No 179:

#### Question 10:

Describe with a suitable diagram, how a convex mirror diverges a parallel beam of light rays. Mark clearly the pole, focus and centre of curvature of convex mirror in this diagram.

#### Answer:

When a parallel beam of light, also parallel to the principle axis, is incident on a convex mirror, it will reflect from the mirror, and the reflected rays will appear to come from a point which is called the focus (F) of the convex mirror. So, a convex mirror has a virtual focus.

__ Ray diagram-__ A convex mirror diverges a parallel beam of light rays.

#### Page No 179:

#### Question 11:

Define (*a*) centre of curvature (*b*) radius of curvature (*c*) pole (*d*) principal axis, and (*e*) aperture, of a spherical mirror with the help of a labelled diagram.

#### Answer:

**(a) Centre of curvature (C)** – It is the centre of the sphere from which the mirror is formed.

**(b)**– It is the radius of the sphere from which the mirror is formed.

__Radius of curvature (R)__R = PC

**(c) Pole (P)** – Pole is the centre of the curved mirror surface, MM’.

**(d)**– It is the line joining P and C.

__Principal axis__**(e)**– It is the effective length of the mirror, i.e. MM’.

__Aperture__#### Page No 179:

#### Question 12:

(*a*) Define (*i*) principal focus of a concave mirror, and (*ii*) focal length of a concave mirror.

(*b*) Draw diagram to represent the action of a concave mirror on a beam of parallel light rays. Mark on this diagram principal axis, focus F, centre of curvature C, pole P and focal length *f*, of the concave mirror.

#### Answer:

(a) **Definitions-**

(i) __ Principal focus (F) of a concave mirror __– Focus of a concave mirror is the point at which the incident rays, parallel to the principal axis, actually meet after reflecting from the concave mirror.

(i) __ Focal length (f) __– Focal length of a concave mirror may be defined as the distance between its focus and pole.

(b) **Ray Diagram-**

** **

#### Page No 179:

#### Question 13:

(*a*) What is meant by (*i*) principal focus of a convex mirror, and (*ii*) focal length of a convex mirror?

(*b*) Draw diagram to show the action of convex mirror on a beam of parallel light rays. Mark on this diagram principal axis, focus F, centre of curvature C, pole P and focal length *f*, of the convex mirror.

#### Answer:

(a) **Definitions-**

(i) ** Principal focus (F) of a convex mirror** - Focus is the point at which the incident rays, parallel to the principal axis, appear to have come from, after reflection.

(i)

**Focal length of a convex mirror may be defined as the distance between its pole and focus.**

__Focal length (f) -__(b)

**Ray Diagram-**

#### Page No 179:

#### Question 14:

In a convex spherical mirror, reflection of light takes place at:

(*a*) a flat surface

(*b*) a bent-in surface

(*c*) a bulging-out surface

(*d*) an uneven surface

#### Answer:

(*c*) a bulging-out surface

In a convex spherical mirror, reflection of light takes place at the bulged-out surface.

#### Page No 179:

#### Question 15:

A diverging mirror is

(*a*) a plane mirror

(*b*) a convex mirror

(*c*) a concave mirror

(*d*) a shaving mirror

#### Answer:

(b) convex mirror

A convex mirror diverges the rays of light incident on it; hence, it acts as a diverging mirror.

#### Page No 179:

#### Question 16:

If *R* is the radius of curvature of a spherical mirror and *f* is its focal length, then:

(*a*) *R* = *f*

(*b*) *R* = 2*f*

(*c*) $R=\frac{f}{2}$

(*d*) *R* = 3*f*

#### Answer:

(*b*) *R* = 2*f*

If R is the radius of curvature of a spherical mirror and f is its focal length, then R = 2f. |

#### Page No 179:

#### Question 17:

The focal length of a spherical mirror of radius of curvature 30 cm is:

(*a*) 10 cm

(*b*) 15 cm

(*c*) 20 cm

(*d*) 30 cm

#### Answer:

(*b*) 15 cm

The focal length of a spherical mirror is half of its radius of curvature.

#### Page No 179:

#### Question 18:

If the focal length of a spherical mirror is 12.5 less cm, its radius of curvature will be:

(*a*) 25 cm

(*b*) 15 cm

(*c*) 20 cm

(*d*) 35 cm

#### Answer:

(*a*) 25 cm

$\mathrm{R}=2\times 12.5\phantom{\rule{0ex}{0ex}}=25\mathrm{cm}$

#### Page No 179:

#### Question 19:

A communications satellite in orbit sends a parallel beam of signals down to earth. If these signals obey the same laws of reflection as light and are to be focussed onto a small receiving aerial, what should be the best shape of the metal 'dish' used to collect them?

#### Answer:

#### Page No 179:

#### Question 20:

When a spherical mirror is held towards the sun and its sharp image is formed on a piece of carbon paper for some time, a hole is burnt in the carbon paper.

(*a*) What is the nature of spherical mirror?

(*b*) Why is a hole burnt in the carbon paper?

(*c*) At which point of the spherical mirror the carbon paper is placed?

(*d*) What name is given to the distance between spherical mirror and carbon paper?

(*e*) What is the advantage of using a carbon paper rather than a white paper?

#### Answer:

(a) The spherical mirror is concave.

(b) A concave mirror converges light rays, and in this case it will converge the incoming parallel rays of the sun at its focus. Since the carbon paper was kept at the focus of the concave mirror, the hole was burnt into it.

(c) The carbon paper was kept at the focus of the spherical mirror.

(d) The distance between the mirror and the carbon paper is the focal length.

(e) A carbon paper is a good absorber of sunlight; hence, it burnt quickly.

#### Page No 189:

#### Question 1:

For what position of an object, a concave mirror forms a real image equal in size to the object?

#### Answer:

When an object is placed at the centre of curvature of a concave mirror, it forms a real image of a size equal to that of the object.

#### Page No 189:

#### Question 2:

Where should an object be placed in front of the concave mirror so as to obtain its virtual, erect and magnified image?

#### Answer:

The object should be placed within the focus (between the pole and the focus) and in front of the concave mirror in order to obtain a virtual, erect and magnified image.

#### Page No 189:

#### Question 3:

For which positions of the object does a concave mirror produce an inverted, magnified and real image?

#### Answer:

When an object is placed at the focus or between the focus and centre of curvature of a concave mirror, the image produced is inverted, magnified and real.

#### Page No 189:

#### Question 4:

If an object is placed at the focus of a concave mirror, where is the image formed?

#### Answer:

If an object is placed at the focus of a concave mirror, the image is formed at infinity.

#### Page No 189:

#### Question 5:

If an object is at infinity (very large distance) in front of a concave mirror, where is the image formed?

#### Answer:

If an object is at infinity and in front of a concave mirror, the image is formed at the focus.

#### Page No 189:

#### Question 6:

For what position of an object, a real and diminished image is formed by a concave mirror?

#### Answer:

When an object is kept beyond C, a real and diminished image is formed by a concave mirror.

#### Page No 189:

#### Question 7:

Copy this figure in your answer book and show the direction of the light ray after reflection:

Figure

#### Answer:

Because a light ray passing through the focus of a concave mirror becomes parallel to the principal axis after reflection, the resultant diagram is as shown:

#### Page No 189:

#### Question 8:

Draw the following diagram in your answer book and show the formation of image of the object AB with the help of suitable rays:

Figure

#### Answer:

Using the following rules for image formation:

- A ray of light passing through the centre of curvature of a concave mirror is reflected along the same path, and
- A ray of light passing through the focus of a concave mirror becomes parallel to the principal axis after reflection,

#### Page No 189:

#### Question 9:

Draw the following diagram in your answer book and show the formation of image with the help of suitable rays:

Figure

#### Answer:

Using the following rules for image formation:

- A ray of light that is parallel to the principal axis of a concave mirror passes through its focus after reflection, and
- A ray of light passing through the focus of a concave mirror becomes parallel to the principal axis after reflection,

#### Page No 189:

#### Question 10:

Which type of mirror could be used as a dentist's mirror?

#### Answer:

A concave mirror could be used as a dentist's mirror as it produces a magnified image of a tooth.

#### Page No 189:

#### Question 11:

Which kind of mirror is used in the headlights of a car? Why is it used for this purpose?

#### Answer:

A concave mirror is used in the headlights of a car. A light bulb placed at the focus of a concave mirror reflector produces a strong, parallel-sided beam of light. This is because the diverging light rays of the bulb are collected by the concave reflector and then reflected.

#### Page No 189:

#### Question 12:

Explain why, a ray of light passing through the centre of curvature of a concave mirror gets reflected back along the same path.

#### Answer:

A ray of light passing through the centre of curvature of a concave mirror gets reflected along the same path. This is because, it strikes the mirror normally or perpendicularly.

#### Page No 189:

#### Question 13:

What is the minimum number of rays required for locating the image formed by a concave mirror for an object? Draw a ray diagram to show the formation of a virtual image by a concave mirror.

#### Answer:

A minimum of two rays are required for locating the image of an object formed by a concave mirror.

When an object is placed between the focal point and the pole of a concave mirror, a virtual image is formed(as shown in the figure).

#### Page No 189:

#### Question 14:

With the help of a ray diagram, determine the position, nature and size of the image formed of an object placed at the centre of curvature of a concave mirror.

#### Answer:

When an object is placed at the centre of curvature of a concave mirror, the image formed will be as shown:

The image of an object placed at the centre of curvature of a concave mirror is real, inverted and of the same size as the object.

#### Page No 189:

#### Question 15:

Describe with the help of a diagram, the nature, size and position of the image formed when an object is placed beyond the centre of curvature of a concave mirror.

#### Answer:

The image formed by an object placed beyond the centre of curvature of a concave mirror is as follows:

The above shows that the image formed by the concave mirror is

- between the focus and centre of curvature
- real and inverted
- smaller than the object (or diminished)

#### Page No 189:

#### Question 16:

If an object is placed at a distance of 8 cm from a concave mirror of focal length 10 cm, discuss the nature of the image formed by drawing the ray diagram.

#### Answer:

As shown in the diagram, the image formed is

- behind the mirror
- virtual and erect
- larger than the object

#### Page No 189:

#### Question 17:

Draw a ray diagram showing how a concave mirror can be used to produce a real, inverted and diminished image of an object.

#### Answer:

When an object is placed beyond the centre of curvature of a concave mirror, a real, inverted and diminished image of the object is formed.

#### Page No 190:

#### Question 18:

Which mirror is used as a torch reflector? Draw a labelled diagram to show how a torch reflector can be used to produce a parallel beam of light. Where is the bulb placed in relation to the torch reflector?

#### Answer:

A concave mirror is used as a torch reflector.

Light bulb placed at the focus of a torch reflector:

When a light bulb is placed at the focus of a concave mirror reflector, the diverging light rays of the bulb are collected by the reflector. These rays are then reflected to produce a strong, parallel-sided beam of light.

#### Page No 190:

#### Question 19:

State where an object must be placed so that the image formed by a concave mirror is:

(*a*) erect and virtual.

(*b*) at infinity.

(*c*) the same size as the object.

#### Answer:

(a) For the image formed by a concave mirror to be erect and virtual, the object must be placed between the pole and the focus.

(b) For the image formed by a concave mirror to be at infinity, the object must be placed at the focus.

(c) For the image formed by a concave mirror to be of the same size as the object, the object must be placed at the centre of curvature.

#### Page No 190:

#### Question 20:

With the help of a labelled ray diagram, describe how a converging mirror can be used to give an enlarged upright image of an object.

#### Answer:

A concave mirror gives an enlarged and upright image when an object is placed between the pole and the focus (object between P and F):

In the diagram, AB is the object and A'B' is the virtual and magnified image of AB.

#### Page No 190:

#### Question 21:

Make labelled ray diagrams to illustrate the formation of:

(*a*) a real image by a converging mirror.

(*b*) a virtual image by a converging mirror.

Mark clearly the pole, focus, centre of curvature and position of object in each case.

#### Answer:

(a) A real image is formed by a converging mirror when an object is placed anywhere between the focus and infinity, including the focus and infinity:

Here, P: pole, F: focus, C: centre of curvature, AB: object and A'B': image.

(b) A virtual image is formed by a converging lens when an object is placed between the pole and the focus:

Here, P: pole, F: focus, C: centre of curvature, AB: object and A'B': image

#### Page No 190:

#### Question 22:

Briefly describe how you would find the focal length of a concave mirror quickly but approximately.

#### Answer:

For an object placed at the focus, the image formed by a concave mirror is at infinity. The image is highly magnified and real. So, we will move the object at a different point and check the image size and the point of formation. The point of focus of the concave mirror will be the point where the image formed is highly magnified and at infinity.

#### Page No 190:

#### Question 23:

Which type of mirror is used in a solar furnace? Support your answer with reason.

#### Answer:

A concave mirror is used in a solar furnace. A furnace placed at the focus of a concave reflector gets heated up because the reflector focuses the sun's heat rays on it.

#### Page No 190:

#### Question 24:

Name the type of mirror used by dentists. How does it help?

#### Answer:

Concave mirrors are used by dentists to see a magnified image of a tooth. The reason being, a concave mirror produces an enlarged image of an object(here, teeth) that is placed within its focus.

#### Page No 190:

#### Question 25:

Explain why, concave mirrors are used as shaving mirrors.

#### Answer:

Concave mirrors are used as shaving mirrors to view a magnified image of the face. The reason being, an enlarged image of the face is seen when it is held within the focus of a concave mirror.

#### Page No 190:

#### Question 26:

Give two uses of concave mirrors. Explain why you would choose concave mirrors for these uses.

#### Answer:

Two uses of concave mirrors are:

- Shaving mirrors
- By dentists, to see the teeth of patients

#### Page No 190:

#### Question 27:

(*a*) Draw ray-diagrams to show the formation of images when the object is placed in front of a concave mirror (converging mirror):

(i) between its pole and focus

(ii) between its centre of curvature and focus

Describe the nature, size and position of the image formed in each case.

(*b*) State one use of concave mirror based on the formation of image as in case (*i*) above.

#### Answer:

(i) When an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, erect, magnified and behind the mirror. This is illustrated as follows:

Here, the image A'B' of the object AB, placed between the pole and the focus of a concave mirror, is virtual and erect.

(ii) When an object is placed between the centre of curvature and the focus of a concave mirror, the image formed is real, inverted, and enlarged. This is illustrated as follows:

Here, the image A'B' of the object AB, placed between the centre of curvature and the focus of a concave mirror, is real and inverted.

(iii) The concave mirror can be used in the following, on the bases of image formation as demonstrated in case (i):

- As a shaving mirror
- As a make-up mirror
- As a dentist's mirror

#### Page No 190:

#### Question 28:

(*a*) Give two circumstances in which a concave mirror can form a magnified image of an object placed in front of it. Illustrate your answer by drawing labelled ray diagrams for both.

(*b*) Which one of these circumstances enables a concave mirror to be used as a shaving mirror?

#### Answer:

(a) The two circumstances in which a concave mirror can form a magnified image of an object that is placed in front of it are:

- When an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, erect and magnified, as shown in the figure. Here, the object AB is placed between the pole and the focus of a concave mirror. Its image A'B' is formed behind the mirror and is virtual, erect and magnified.
- When an object is placed between the centre of curvature and the focus of a concave mirror, the image formed is real, inverted and enlarged, as shown in figure 2. Here, the object AB is placed between the centre of curvature and the focus of a concave mirror. Its image A'B' is formed beyond the centre of curvature and is real, inverted and enlarged.

(b) From the above mentioned cases, only case 1, where the object is placed between the pole and the focus of a concave mirror, can be used as a shaving mirror. This is because, the image formed is virtual, enlarged and erect.

#### Page No 190:

#### Question 29:

The real image formed by a concave mirror is larger than the object when the object is:

(*a*) at a distance equal to radius of curvature

(*b*) at a distance less than the focal length

(*c*) between focus and centre of curvature

(*d*) at a distance greater than radius of curvature

#### Answer:

(*c*) Between the focus and the centre of curvature

For an object placed between the focus and the centre of curvature, the real image formed by a concave mirror is larger than the object.

#### Page No 190:

#### Question 30:

The real image formed by a concave mirror is smaller than the object if the object is:

(*a*) between centre of curvature and focus

(*b*) at a distance greater than radius of curvature

(*c*) at a distance equal to radius of curvature

(*d*) at a distance equal to focal length

#### Answer:

(b) at a distance greater than the radius of curvature

The real image formed by a concave mirror is smaller than the object if the object is placed at a distance greater than the radius of curvature.

#### Page No 190:

#### Question 31:

The image formed by a concave mirror is virtual, erect and magnified. The position of object is:

(*a*) at focus

(*b*) between focus and centre of curvature

(*c*) at pole

(*d*) between pole and focus

#### Answer:

(*d*) Between the pole and the focus

The reason being, the image formed by a concave mirror is virtual, erect and magnified. The position of the object is between the pole and the focus.

#### Page No 190:

#### Question 32:

The image formed by a concave mirror is real, inverted and of the same size as the object. The position of the object must then be:

(*a*) at the focus

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

(*c*) at the centre of curvature

(*d*) beyond the centre of curvature

#### Answer:

(*c*) At the centre of curvature

The reason being, the image formed by a concave mirror is real, inverted and of the same size as the object. The position of the object must then be at the centre of curvature.

#### Page No 190:

#### Question 33:

The image formed by a concave mirror is real, inverted and highly diminished (much smaller than the object). The object must be:

(*a*) between pole and focus

(*b*) at focus

(*c*) at the centre of curvature

(*d*) at infinity

#### Answer:

(*d*) At infinity

The reason being, the image formed by a concave mirror is real, inverted and highly diminished (much smaller than the object). Therefore, the object must be at infinity.

#### Page No 190:

#### Question 34:

The angle of incidence for a ray of light passing through the centre of curvature of a concave mirror is:

(*a*) 45°

(*b*) 90°

(*c*) 0°

(*d*) 180°

#### Answer:

(*c*) 0°

The reason being, a ray of light passing through the centre of curvature of a concave mirror strikes the mirror normally or perpendicularly.

#### Page No 191:

#### Question 35:

In the concave reflector of a torch, the bulb is placed:

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

(*b*) at the focus of reflector

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

(*d*) at the centre of curvature of reflector

#### Answer:

(*b*) At the focus of the reflector

When a light bulb is placed at the focus of a concave mirror reflector, the diverging light rays of the bulb are collected by the reflector and then reflected, to produce a strong, parallel-sided beam of light.

#### Page No 191:

#### Question 36:

The focal length of a small concave mirror is 2.5 cm. In order to use this concave mirror as a dentist's mirror, the distance of tooth from the mirror should be:

(*a*) 2.5 cm

(*b*) 1.5 cm

(*c*) 4.5 cm

(*d*) 3.5 cm

#### Answer:

(b) 1.5 cm

When an object is placed between the pole and the focus of a concave mirror, an enlarged image is formed.

#### Page No 191:

#### Question 37:

An object is 100 mm in front of a concave mirror which produces an upright (erect image). The radius of curvature of the mirror is:

(*a*) less than 100 mm

(*b*) between 100 mm and 200 mm

(*c*) exactly 200 mm

(*d*) more than 200 mm

#### Answer:

(*d*) More than 200 mm

The reason being, a concave mirror forms an erect image when an object is placed between the focus and the pole. Further, the radius of curvature is twice the distance between the focus and the pole. This gives a radius of curvature greater than 200mm.

#### Page No 191:

#### Question 38:

A virtual, erect and magnified image of an object is to be produced with a concave mirror of focal length 12 cm. Which of the following object distance should be chosen for this purpose?

(i) 10 cm

(ii) 15 cm

(iii) 20 cm

Give reason for your choice

#### Answer:

The object must be kept at a distance of __10 cm__ from the concave mirror in order to produce a virtual, erect and magnified image. The reason being, a concave mirror forms a virtual, erect and magnified image when an object is placed between the focus and the pole.

#### Page No 191:

#### Question 39:

A concave mirror has a focal length of 25 cm. At which of the following distance should a person hold his face from this concave mirror so that it may act as a shaving mirror?

(*a*) 45 cm

(*b*) 20 cm

(*c*) 25 cm

(*d*) 30 cm

Give reason for your choice.

#### Answer:

A person must hold his face at a distance of __20 cm__ , away from the concave mirror. A concave mirror forms an erect, virtual and magnified image when an object is placed between the focus and the pole. Therefore, it acts as a shaving mirror.

#### Page No 191:

#### Question 40:

An object is placed at the following distances from a concave mirror of focal length 15 cm, turn by turn:

(*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:

Here, radius of curvature = 2 x focal length = 2 x 15 cm = 30 cm

(i) At position (c), i.e., 20 cm from the concave mirror, the image formed will be magnified and real. The reason being, a real and magnified image is formed by the concave mirror when an object is placed between the focus and the centre of curvature.

(ii) At position (d), i.e., 10 cm from the concave mirror, the image formed will be magnified and virtual. The reason being, a virtual and magnified image is formed by the concave mirror when an object is placed between the focus and the pole.

(iii) At position (a), i.e., 35 cm from the concave mirror, the image formed will be diminished and real. The reason being, a diminished and real image is formed by the concave mirror when an object is placed beyond the centre of curvature.

(iv) At position (b), i.e., 30 cm from the concave mirror, the image formed will be of the same size as the object. The reason being, an image of the same size as the object is formed by the concave mirror when the object is placed at the centre of curvature.

#### Page No 192:

#### Question 1:

According to the "New Cartesian Singh Convention" for mirrors, what sign has been given to the focal length of:

(i) a concave mirror?

(ii) a convex mirror?

#### Answer:

(i) The sign given to the focal length of a concave mirror (according to the New Cartesian Sign Convention) is negative. This is because the focus of a concave mirror is in front of the mirror, on the left side.

(ii) The sign given to the focal length of a convex mirror (according to the New Cartesian Sign Convention) is positive. The is because the focus of a convex mirror is behind the mirror, on the right side.

#### Page No 192:

#### Question 2:

Which type of mirror has:

(*a*) positive focal length?

(*b*) negative focal length?

#### Answer:

(a) A convex mirror has a positive focal length because the focus of a convex mirror is behind the mirror and on the right side.

(b) A concave mirror has a negative focal length because the focus of a concave mirror is in front of the mirror and on the left side.

#### Page No 192:

#### Question 3:

What is the nature of a mirror having a focal length of, +10 cm?

#### Answer:

A mirror having a focal length of +10 cm is a convex mirror(a convex mirror has a positive focal length, according to the 'new cartesian sign convention').

#### Page No 192:

#### Question 4:

What kind of mirror can have a focal length of, −20 cm?

#### Answer:

A concave mirror can have a focal length of -20 cm. The reason being, a concave mirror has a negative focal length(according to the 'new cartesian sign convention').

#### Page No 192:

#### Question 5:

Complete the following sentence:

All the distances are measured from the .......... of a spherical mirror.

#### Answer:

All distances are measured from the __pole__ of a spherical mirror.

#### Page No 192:

#### Question 6:

What sign (+ve or −ve) has been given to the following on the basis of Cartesian Sign Convention?

(*a*) Height of a real image.

(*b*) Height of a virtual image.

#### Answer:

(a) A negative sign has been given to the height of a real image.

(Because a real image is formed below the principal axis, it's given a negative sign).

(b) A positive sign is given to the height of a virtual image.

(Because a virtual image is formed above the principal axis, it's given a positive sign).

#### Page No 192:

#### Question 7:

Describe the New Cartesian Sign Convention used in optics. Draw a labelled diagram to illustrate this sign convention.

#### Answer:

According to the new cartesian sign convention :

- All distances are measured from the pole of the mirror as origin.
- Distances measured in the same direction as that of incident light are taken as positive.
- Distances measured against the direction of incident light are taken as negative.
- Distances measured upward and perpendicular to the principal axis are taken as positive.
- Distances measured downward and perpendicular to the principal axis are taken as negative.

#### Page No 193:

#### Question 8:

Giving reasons, state the 'signs' (positive or negative) which can be given to the following:

(*a*) object distance (*u*) for a concave mirror or convex mirror

(*b*) image distance (*v*) for a concave mirror

(*c*) image distance (*v*) for a convex mirror

#### Answer:

(a) Object distance (*u*) for a concave or convex mirror is given a negative sign. This is because, the object is always placed to the left of the mirror.

(*b*)

- The image distance (
*v*) for a concave mirror is given a positive sign if the image formed is behind the mirror and to the right of it. - The image distance (
*v*) for a concave mirror is given a negative sign if the image formed is in front of the mirror and to the left of it.

*c*) The image distance (

*v*) for a convex mirror is always given a positive sign because the image is always formed behind the mirror and to the right of it.

#### Page No 193:

#### Question 9:

According to New Cartesian Sign Convention:

(*a*) focal length of concave mirror is positive and that of convex mirror is negative

(*b*) focal length of both concave and convex mirrors is positive

(*c*) focal length of both concave and convex mirrors is negative

(*d*) focal length of concave mirror is negative and that of convex mirror is positive

#### Answer:

(d) The focal length of a concave mirror is negative and that of a convex mirror is positive.

This is because the focus of a concave mirror is in front of the mirror, on the left side, and the focus of a convex mirror is behind the mirror, on the right side.

#### Page No 193:

#### Question 10:

One of the following does not apply to a concave mirror. This is:

(*a*) focal length is negative

(*b*) image distance can be positive or negative

(*c*) image distance is always positive

(*d*) height of image can be positive or negative

#### Answer:

(*c*) The image distance is always positive with the exception of one case, when the object is placed between the pole and the focus. In all other cases, the image is formed in front of the mirror and on the left side.

#### Page No 198:

#### Question 1:

If the magnification of a body of size 1 m is 2, what is the size of the image?

#### Answer:

Size of body (h) = 1 m

Size of image (h') = ?

Magnification (m) = 2

We know that,

m = $\frac{\mathrm{h}\text{'}}{\mathrm{h}}$

$2=\frac{\mathrm{h}\text{'}}{1}$

h' = 2 m

Size of the image is 2 m.

#### Page No 198:

#### Question 2:

What is the position of the image when an object is placed at a distance of 20 cm from a concave mirror of focal length 20 cm?

#### Answer:

When an object is placed at a distance of 20 cm from a concave mirror of focal length 20 cm, the position of the image is at infinity. The reason being, the image formed by a concave mirror is at infinity when an object is placed at its focus.

#### Page No 198:

#### Question 3:

What is the nature of image formed by a concave mirror if the magnification produced by the mirror (*a*) +4, and (*b*) −2?

#### Answer:

(a) The image formed by a concave mirror, with a magnification of +4, is __virtual and erect. __This is because, the height of both the object and the image is above the principal axis.

(b)The image formed by a concave mirror, with a magnification of -2, is __real and inverted.__This is because, the height of the object is above the principal axis and that of the image, below it.

#### Page No 198:

#### Question 4:

State the relation between object distance, image distance and focal length of a spherical mirror (concave mirror or convex mirror).

#### Answer:

The relation between the object distance (u), the image distance (v) and the focal length (f) of a spherical mirror (a concave or convex mirror) is:

$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$.

#### Page No 198:

#### Question 5:

Write the mirror formula. Give the meaning of each symbol which occurs in it.

#### Answer:

The formula for a mirror is:

$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$

where, v = distance of image from mirror

u = distance of object from mirror

f = focal length of the mirror

#### Page No 198:

#### Question 6:

What is the ratio of the height of an image to the height of an object known as?

#### Answer:

The ratio of the height of an image to the height of an object is known as linear magnification.

#### Page No 198:

#### Question 7:

Define linear magnification produced by a mirror.

#### Answer:

The ratio of the height of an image (h') to the height of an object (h) is known as linear magnification (m).

That is,

m = $\frac{\mathrm{h}\text{'}}{\mathrm{h}}$

where, h^{'} = height of image

h = height of object

#### Page No 198:

#### Question 8:

Write down a formula for the magnification produced by a concave mirror.

(*a*) in terms of height of object and height of image

(*b*) in terms of object distance and image distance

#### Answer:

The formula for the magnification (m) produced by a concave mirror,

(a)In terms of the height of an object (h) and the height of an image (h') is as follows:

Magnification (m) = $\frac{\mathrm{height}\mathrm{of}\mathrm{image}}{\mathrm{height}\mathrm{of}\mathrm{object}}$ = $\frac{\mathrm{h}\text{'}}{\mathrm{h}}$

(b) In terms of the object distance (u) and the image distance (v) is as follows:

Magnification (m) = $-\frac{\mathrm{image}\mathrm{distance}}{\mathrm{object}\mathrm{distance}}=-\frac{\mathrm{v}}{\mathrm{u}}$

#### Page No 198:

#### Question 9:

Describe the nature of image formed when the object is placed at a distance of 20 cm from a concave mirror of focal length 10 cm.

#### Answer:

Object distance (u) = - 20 m ( negative, due to sign convention)

Focal length (f) = -10 m ( negative, due to sign convention)

Image distance = ?

From the mirror formula, we know that:

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

or $\frac{1}{v}+\frac{1}{-20}=\frac{1}{-10}$

$or\frac{1}{v}-\frac{1}{20}=\frac{-1}{10}\phantom{\rule{0ex}{0ex}}or\frac{1}{v}=\frac{-1}{10}+\frac{1}{20}\phantom{\rule{0ex}{0ex}}or\frac{1}{v}=-\frac{1}{20}orv=-20m$

Here, the negative sign means that the image is formed on the left side of the mirror. Thus, it is a real and inverted image, formed at a distance of 20 m from the mirror.

#### Page No 198:

#### Question 10:

Fill in the following blanks with suitable words:

(*a*) If the magnification has a plus sign, than image is ......... and.........

(*b*) If the magnification has a minus sign, than the image is ......... and .......

#### Answer:

(a) If the magnification has a plus sign, then the image is __virtual__ and __erect__.

(b) If the magnification has a minus sign, then the image is __real__ and __inverted__.

#### Page No 198:

#### Question 11:

An object is placed at a distance of 10 cm from a concave mirror of focal length 20 cm.

(*a*) Draw a ray diagram for the formation of image.

(*b*) Calculate the image distance.

(*c*) State two characteristics of the image formed.

#### Answer:

(a) Ray Diagram-

(b) Image Distance-

*u*' = −10 cm

Focal length of the concave mirror '

*f*' = −20 cm

*v*'.

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Therefore

Thus, the distance of the image '*v*' is 20 cm.

(c) Two characteristics of the image formed is as follows:

(1) It is virtual and erect.

(2) It is magnified.

#### Page No 198:

#### Question 12:

If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm, find the position, nature and height of the image.

#### Answer:

Distance of the object from the mirror '*u*' = −36 cm

Height of the object '*h _{o}*' = 10 cm

Focal length of the mirror '

*f*' = −12 cm

We have to find the distance of the image '

*v*'.

Height of the image

*h*= ?

_{i}Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{-12}=\frac{1}{v}+\frac{1}{-36}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{36}-\frac{1}{12}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1-3}{36}=\frac{-2}{36}\phantom{\rule{0ex}{0ex}}v=-18\mathrm{cm}$

Distance of the image '

*v*' = −18 cm

Using the magnification formula, we get

$m=\frac{{h}_{i}}{{h}_{o}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}\frac{{h}_{i}}{10}=\frac{-(-18)}{-36}=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{{h}_{i}}{10}=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}{h}_{i}=-5\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Height of the image '

*h*' = −5 cm

_{i}Again using the magnification formula, we get

$m=\frac{-v}{u}=\frac{-(-18)}{36}\phantom{\rule{0ex}{0ex}}m=-\frac{1}{2}$

Thus, the image is real, inverted and small in size.

#### Page No 198:

#### Question 13:

At what distance from a concave mirror focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall?

#### Answer:

Height of the object '*h** _{o}*' = 2 cm

Focal length of the mirror '*f*' = −10 cm

Height of the image '*h _{i}*' = 6 cm

We have to find the distance of the object from the mirror '

*u*'.

Using the magnification formula, we get

$m=\frac{{h}_{i}}{{h}_{o}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{6}{2}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}m=3=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}\mathrm{thus}v=-3u$

Now, using the mirror formula, we get

Thus, the distance of the object from the mirror '*u*' is −6.67 cm.

#### Page No 198:

#### Question 14:

When an object is placed at a distance of 15 cm from a concave mirror, its image is formed at 10 cm in front of the mirror. Calculate the focal length of the mirror.

#### Answer:

Distance of the object from the mirror '*u*' = −15 cm

Distance of the image '*v*' = −10 cm

We have to find the focal length of the mirror '*f*'.

Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{-10}+\frac{1}{-15}=-\frac{1}{10}-\frac{1}{15}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=-\frac{3}{30}-\frac{2}{30}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=-\frac{5}{30}\phantom{\rule{0ex}{0ex}}f=-6\mathrm{cm}$

Thus, the focal length of the concave mirror '*f*' is 6 cm.

#### Page No 198:

#### Question 15:

An object 3 cm high is placed at a distance of 8 cm from a concave mirror which produces a virtual image 4.5 cm high:

(i) What is the focal length of the mirror?

(ii) What is the position of image?

(iii) Draw a ray-diagram to show the formation of image.

#### Answer:

Distance of the object from the mirror '*u*' = -8 cm

Height of the object '*h*_{o}' = 3 cm

Height of the image '*h*_{i}' = 4.5 cm

We have to find the focal length of the mirror '*f*' and distance of the image '*v*'.

Using the magnification formula, we get

$m=\frac{{h}_{i}}{{h}_{0}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{4.5}{3}=\frac{-v}{-8}\phantom{\rule{0ex}{0ex}}v=\frac{4.5\times 8}{3}=12\mathrm{cm}$

Therefore, the distance of the image '*v*' is 12 cm behind the mirror.

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{12}+\frac{1}{-8}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{2}{24}-\frac{3}{24}=\frac{-1}{24}\phantom{\rule{0ex}{0ex}}f=-24\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

*f*' is 24 cm.

#### Page No 198:

#### Question 16:

A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror:

(i) Calculate the image distance.

(ii) What is the focal length of the mirror?

#### Answer:

A concave mirror is a converging mirror.

*u*' = -20 cm

*h*' = 1 cm

_{o}*h*' = -4 cm

_{i}We have to find the distance of the image '

*v*' and focal length of the mirror '

*f*'.

$m=\frac{{h}_{i}}{{h}_{o}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}\frac{-4}{1}=\frac{-v}{-20}\phantom{\rule{0ex}{0ex}}v=\frac{20\times -4}{1}=-80\mathrm{cm}$

Thus, the distance of the image '*v*' is 80 cm.

Now, using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{-80}+\frac{1}{-20}=-\frac{1}{80}-\frac{4}{80}=-\frac{5}{80}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=-\frac{1}{16}\phantom{\rule{0ex}{0ex}}f=-16\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Thus, the focal length of the concave mirror is 16 cm.

#### Page No 198:

#### Question 17:

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of image.

#### Answer:

Distance of the object from the mirror '*u*' = −27 cm

Height of the object '*h _{o}*' = 7 cm

Focal length of the mirror '

*f*' = −18 cm

We have to find the distance of the image '

*v*' and height of the image '

*h*'.

_{i}Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\Rightarrow \frac{1}{-18}=\frac{1}{v}+\frac{1}{-27}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{-18}=\frac{1}{v}-\frac{1}{27}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{27}-\frac{1}{18}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-1}{54}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow v=-54\mathrm{cm}$

Thus, the distance of the image '

*v*' is 54 cm.

Now, using the magnification formula, we get

$m=\frac{{h}_{i}}{{h}_{o}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}\frac{{h}_{i}}{7}=-\frac{(-54)}{(-27)}=-\frac{2}{1}\phantom{\rule{0ex}{0ex}}{h}_{i}=-14\mathrm{cm}$

Therefore, the height of the image '

*h*' is 14 cm.

_{i}Again, using the magnification formula, we get

$m=-\frac{v}{u}=-\frac{(-54)}{(-27)}\phantom{\rule{0ex}{0ex}}m=-2\phantom{\rule{0ex}{0ex}}$

Thus, the image is real, inverted and large in size.

#### Page No 199:

#### Question 18:

An object 3 cm high is placed at a distance of 10 cm in front of a converging mirror of focal length 20 cm. Find the position, nature and size of the image formed.

#### Answer:

Distance of the object from the mirror '*u*' = −10 cm

Height of the object '*h _{o}*' = 3 cm

Focal length of the mirror '

*f*' = −20 cm

We have to find the distance of the image '

*v*' and the height of the image '

*h*'.

_{i}Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{-20}=\frac{1}{v}+\frac{1}{-10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{-20}+\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow v=20\mathrm{cm}$

Thus, the distance of the image 'v' is 20 cm other side of cancave mirror.

Now, using the magnification formula, we get

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

$\frac{{h}_{i}}{{h}_{o}}=\frac{-20}{10}$

$\frac{{h}_{i}}{3}=-2$

*h*

_{i }= −6 cm

Therefore, the height of the image '

*h*' is 6 cm.

_{i}Again, using the magnification formula, we get

$m=\frac{-v}{u}$

$m=\frac{-20}{10}\phantom{\rule{0ex}{0ex}}m=-2\mathrm{cm}$

Thus, the image is virtual, erect.

#### Page No 199:

#### Question 19:

A concave mirror has a focal length of 4 cm and an object 2 cm tall is placed 9 cm away from it. Find the nature, position and size of the image formed.

#### Answer:

Distance of the object from the mirror '*u*' = −9cm

Height of the object '*h _{o}*' = 2 cm

Focal length of the mirror '

*f*' = −4 cm

We have to find the distance of the image '

*v*' and height of the image '

*h*'.

_{i}Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{-4}=\frac{1}{v}+\frac{1}{-9}\phantom{\rule{0ex}{0ex}}\frac{1}{-4}=\frac{1}{v}-\frac{1}{9}\phantom{\rule{0ex}{0ex}}\frac{1}{9}-\frac{1}{4}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\frac{4-9}{36}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\frac{-5}{36}=\frac{1}{v}=\frac{-1}{7.2}\phantom{\rule{0ex}{0ex}}v=7.2\mathrm{cm}$

Therefore, distance of the image '

*v*' is 7.2 cm.

Now, using the magnification formula, we get

$m=\frac{{h}_{i}}{{h}_{o}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}\frac{{h}_{i}}{{h}_{o}}=-\frac{(-7.2)}{-9}\phantom{\rule{0ex}{0ex}}\frac{{h}_{i}}{2}=-\frac{(-7.2)}{-9}\phantom{\rule{0ex}{0ex}}{h}_{i}=\frac{2\times (-7.2)}{9}\phantom{\rule{0ex}{0ex}}{h}_{i}=-1.6\mathrm{cm}$

Thus, the height of the image '

*h*' is 1.6 cm.

_{i}Again, using the magnification formula, we get

$m=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{-(-7.2)}{-9}\phantom{\rule{0ex}{0ex}}m=-0.8$

Thus, the image is real, inverted and small in size.

#### Page No 199:

#### Question 20:

When an object is placed 20 cm from a concave mirror, a real image magnified three times is formed. Find:

(a) the focal length of the mirror.

(*b*) Where must the object be placed to give a virtual images three times the height of the object?

#### Answer:

Given,

*u*' = -20 cm

Magnification '

*m*' = −3

(a)We have to find the focal length of the mirror.

$-3=\frac{-v}{-20}$

*v *= −60 cm

*v*' is

Using the mi

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{f}=-\frac{1}{60}-\frac{1}{20}$

$\frac{1}{f}=-\frac{1}{60}-\frac{3}{60}=-\frac{4}{60}$

$\frac{1}{f}=\frac{-1}{15}$

*f *= −15 cm

Thus, the focal length of the concave mirror is 15 cm.

*m*' = 3

$3=-\frac{v}{u}$

*v*' is *u *

#### Page No 199:

#### Question 21:

A dentist's mirror has a radius of curvature of 3 cm. How far must it be placed from a small dental cavity to give a virtual image of the cavity that is magnified five times?

#### Answer:

*u*

Now, using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{-1.5}=\frac{1}{-5u}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{-1.5}=\frac{1}{-5u}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-1}{1.5}=\frac{1}{-5u}+\frac{5}{5u}=\frac{4}{5u}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{\left(-1.5\right)\times 4}{5}=\frac{-6}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow u=-1.2\mathrm{cm}$

The dentist should place the object at a distanc of 1.2 cm from the mirror.

#### Page No 199:

#### Question 22:

A large concave mirror has a radius of curvature of 1.5 m. A person stands 10 m in front of the mirror. Where is the person's image?

#### Answer:

Given,

The radius of curvature of the mirror 'R' = -1.5m

Focal length of the mirror $\text{'}\mathrm{f}\text{'}=\frac{1}{\mathrm{R}}=\frac{1}{-1.5}=-0.75\mathrm{m}$

Distance of the object 'u' =-10 m

We have to find the distance of the image 'v'.

Using the mirror formula, we get

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

$\frac{1}{0.75}=\frac{1}{v}+\frac{1}{-10}\phantom{\rule{0ex}{0ex}}-\frac{100}{75}=\frac{1}{v}-\frac{1}{10}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{v}=\frac{1}{10}-\frac{100}{75}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{v}=\frac{1}{10}-\frac{4}{3}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{v}=\frac{3}{30}-\frac{40}{30}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{v}=\frac{-37}{30}\phantom{\rule{0ex}{0ex}}\mathrm{or}v=\frac{-30}{37}\phantom{\rule{0ex}{0ex}}\mathrm{or}v=0.81\mathrm{m}$

Thus, the person's image will be formed at a distance of 0.81 m from the mirror.

#### Page No 199:

#### Question 23:

An object of 5.0 cm size is placed at a distance of 20.0 cm from a converging mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed to get the sharp image? Also calculate the size of the image.

#### Answer:

Distance of the object from the mirror 'u' = -20 cm

_{o}' = 5 cm

_{i}'.

$\mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{magnification}\mathrm{formula},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}m=\frac{-v}{u}=\frac{{h}_{i}}{{h}_{o}}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{{h}_{i}}{5}=\frac{-(-60)}{-20}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{{\mathrm{h}}_{\mathrm{i}}}{5}=-3\phantom{\rule{0ex}{0ex}}\mathrm{or}{\mathrm{h}}_{\mathrm{i}}=-3\times 5=-15\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}15\mathrm{cm},\mathrm{and}\mathrm{the}\mathrm{negative}\mathrm{sign}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}\mathrm{below}\mathrm{the}\mathrm{principal}\mathrm{axis}.$

#### Page No 199:

#### Question 24:

A concave mirror produces three times enlarged virtual image of an object placed at 10 cm in front of it. Calculate the radius of curvature of the mirror.

#### Answer:

Given,

Distance of the object 'u' = -10 cm

Magnification 'm' = 3

We have to find the focal length 'f' and the radius of curvature 'R'.

Using the magnification formula, we get

$m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}3=\frac{-\mathrm{v}}{-10}\phantom{\rule{0ex}{0ex}}v=30\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{image}\text{'}\mathrm{v}\text{'}\mathrm{is}30\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}Now,u\mathrm{sin}gthe\mathrm{mirror}\mathrm{formula},weget\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{30}+\frac{1}{-10}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{f}=\frac{1}{30}-\frac{3}{30}=\frac{-2}{30}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{\mathrm{f}}=-\frac{1}{15}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{f}=-15\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{mirror}\text{'}\mathrm{f}\text{'}\mathrm{is}-15\mathrm{cm}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{mirror}\text{'}\mathrm{R}\text{'}\mathrm{will}\mathrm{be}2f=2\times -15\phantom{\rule{0ex}{0ex}}=-30\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{mirror}\mathrm{is}30\mathrm{cm}.$

#### Page No 199:

#### Question 25:

A bright object 50 mm high stands on the axis of a concave mirror of focal length 100 mm and at a distance of 300 mm from the concave mirror. How big will the image be?

#### Answer:

Distance of the object from the mirror 'u' = -300 mm

_{o}' = 50 mm

_{i}'.

#### Page No 199:

#### Question 26:

How far should an object be placed from the pole of a converging mirror of focal length 20cm to form a real image of the size exactly $\frac{1}{4}$th the size of the object?

#### Answer:

Given,

Focal length of the concave mirror (f) = 20 cm

Magnification (m) = -1/4

Using the magnification formula, we get

$m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}\frac{-1}{4}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}v=\frac{u}{4}\phantom{\rule{0ex}{0ex}}U\mathrm{sin}gthe\mathrm{mirror}\mathrm{formula},weget\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{-20}=\frac{4}{u}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{-1}{20}=\frac{5}{u}\phantom{\rule{0ex}{0ex}}u=-100\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{object}\mathrm{should}\mathrm{be}\mathrm{placed}\mathrm{in}\mathrm{front}\mathrm{of}theco\mathrm{ncave}\mathrm{mirror}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}100\mathrm{cm}.$

#### Page No 199:

#### Question 27:

When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is, $-\frac{1}{2}$. Where should the object be placed to get a magnification of, $-\frac{1}{5}$?

#### Answer:

Given,

Magnification $\left(m\right)=\frac{-1}{2}$

Therefore, using the magnification formula, we get

$m=\frac{-v}{u}$

*v = *−*mu*

$v=-(-\frac{1}{2}\times -50)$

$v=-25\mathrm{cm}$

Now, using the mirror formula, we get

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{-mu}+\frac{1}{u}$

$\frac{1}{f}=\frac{1}{-25}+\frac{1}{-50}$

$\mathrm{o}\mathrm{r}\frac{1}{f}=-\frac{2}{50}-\frac{1}{50}=-\frac{3}{50}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{or}f=-\frac{50}{3}\mathrm{cm}$

Now,

Magnification = $m=-\frac{1}{5}$

Therefore, using the mirror formula, we get

$\frac{1}{f}=\frac{1}{-mu}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{-3}{50}=\frac{-5}{-5}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{-3}{50}=\frac{5}{u}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{-3}{50}=\frac{6}{u}\phantom{\rule{0ex}{0ex}}u=-\frac{50\times 6}{3}=-100\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{object}\mathrm{should}\mathrm{be}\mathrm{placed}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}100\mathrm{cm}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{magnefiction}\mathrm{of}-\frac{1}{5}.$

#### Page No 199:

#### Question 28:

An object is placed (*a*) 20 cm, (*b*) 4 cm, in front of a concave mirror of focal length 12 cm. Find the nature and position of the image formed in each case.

#### Answer:

**Case (a)**

Distance of the object from the mirror (u) = -20 cm

The focal length of the concave mirror (f) = -12 cm

We have to find the distance of the image (v).

Using the magnification formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\mathrm{So}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{-12}=\frac{1}{v}+\frac{1}{-20}\phantom{\rule{0ex}{0ex}}\frac{1}{-12}=\frac{1}{v}-\frac{1}{20}\phantom{\rule{0ex}{0ex}}or\frac{1}{20}-\frac{1}{12}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}or\frac{3-5}{60}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}or\frac{-2}{60}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}orv=\frac{60}{-2}\phantom{\rule{0ex}{0ex}}v=-30\mathrm{cm}\phantom{\rule{0ex}{0ex}}T\mathrm{he}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distace}\mathrm{of}30\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}Again,\phantom{\rule{0ex}{0ex}}M\mathrm{agn}i\mathrm{fication}\left(m\right)=\frac{-\mathrm{v}}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}m=\frac{-(-30)}{-20}=-1.5\phantom{\rule{0ex}{0ex}}Therefore,t\mathrm{he}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{real},\mathrm{inverted}\mathrm{and}\mathrm{enlarged}.\phantom{\rule{0ex}{0ex}}$

Thus the image is real, inverted and enlarged.

**Case B**

Distance of the object from the mirror (u) = -4 cm

The focal length of the concave mirror (f) = -12 cm

We have to find the distance of the image (v).

Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{-12}=\frac{1}{v}+\frac{1}{-4}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{4}-\frac{1}{12}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{3-1}{12}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{2}{12}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\mathrm{v}=6\mathrm{cm}\phantom{\rule{0ex}{0ex}}Therefore,d\mathrm{istance}\mathrm{of}\mathrm{image}\left(\mathrm{v}\right)\mathrm{is}6\mathrm{cm}.\phantom{\rule{0ex}{0ex}}Now,u\mathrm{sin}gthe\mathrm{magnif}ication\mathrm{formula},weget\phantom{\rule{0ex}{0ex}}m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{-6}{-4}=1.5\phantom{\rule{0ex}{0ex}}Thus,\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{virtual}\mathrm{and}\mathrm{er}e\mathrm{ct}.$

#### Page No 199:

#### Question 29:

A concave mirror produces a real image 1 cm tall of an object 2.5 mm tall placed 5 cm from the mirror. Find the position of the image and the focal length of the mirror.

#### Answer:

Distance of the object from the mirror (u) = -5 cm

Height of the image (h_{i}) = -1 cm

Height of the object (h_{o}) = 2.5 mm = 0.25 cm

We have to find the image distance (v) and the focal length of the mirror (f).

Using the magnification formula, we get

$m=\frac{{h}_{i}}{{h}_{o}}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{-1}{0.25}=\frac{-\mathrm{v}}{-5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{v}=\frac{-5}{0.25}=-20\mathrm{cm}$

Thus, the distance of the image is -20 cm.

It means that the image will form 20 cm in front of the mirror.

Now, using the mirror formula, we get

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

$\frac{1}{f}=\frac{1}{-20}+\frac{1}{-5}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{f}=-\frac{1}{20}-\frac{1}{5}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{f}=-\frac{1}{20}-\frac{4}{20}=\frac{-5}{20}=\frac{-1}{4}\phantom{\rule{0ex}{0ex}}f=-4\mathrm{cm}$

Thus, the focal length of the mirror is 4 cm.

#### Page No 199:

#### Question 30:

A man holds a spherical shaving mirror of radius of curvature 60 cm, and focal length 30 cm, at a distance of 15 cm, from his nose. Find the position of image, and calculate the magnification.

#### Answer:

The focal length of the concave mirror (f) = -30 cm

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

#### Page No 199:

#### Question 31:

(*a*) An object is placed just outside the principal focus of concave mirror. Draw a ray diagram to show how the image is formed, and describe its size, position and nature.

(*b*) If the object is moved further away from the mirror, what changes are there in the position and size of the image?

(*c*) An object is 24 cm away from a concave mirror and its image is 16 cm from the mirror. Find the focal length and radius of curvature of the mirror, and the magnification of the image.

#### Answer:

(a) Ray Diagram-

(b) The image will move towards the mirror, and its size will gradually decrease.

(c) Distance of object (u) = -24 cm

Distance of image (v) = -16 cm

We have to find the focal length of the mirror (f) and the magnification of the image (m).

Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

#### Page No 199:

#### Question 32:

Linear magnification produced by a concave mirror may be:

(*a*) less than 1 or equal to 1

(*b*) more than 1 or equal to 1

(*c*) less than 1, more than 1 or equal to 1

(*d*) less than 1 or more than 1

#### Answer:

(c) less than 1, more than 1 or equal to 1

Linear magnification produced by a concave mirror may be less than 1, more than 1 or equal to 1.

#### Page No 199:

#### Question 33:

Magnification produced by a convex mirror is always:

(*a*) more than 1

(*b*) less than 1

(*c*) equal to 1

(*d*) more or less than 1

#### Answer:

(b) less than 1

Magnification produced by a convex mirror is always less than 1. This is because the size of the image formed by a convex mirror is smaller than the object.

#### Page No 199:

#### Question 34:

Magnification produced by a plane mirror is:

(*a*) less than one

(*b*) greater than one

(*c*) zero

(*d*) equal to one

#### Answer:

(*d*) equal to one

Magnification produced by a plane mirror is equal to one. This is because the size of the image formed by a plane mirror is same as the size of the object.

#### Page No 199:

#### Question 35:

In order to obtain a magnification of −2 (minus 2) with a concave mirror, the object should be placed:

(*a*) between pole and focus

(*b*) between focus and centre of curvature

(*c*) at the centre of curvature

(*d*) beyond the centre of curvature

#### Answer:

(b) between focus and centre of curvature

In order to obtain a magnification of −2 (minus 2) with a concave mirror, the object should be placed between the focus and the centre of curvature.

#### Page No 199:

#### Question 36:

A concave mirror produces magnification of +4. The object is placed:

(*a*) at the focus

(*b*) between focus and centre of curvature

(*c*) between focus and pole

(*d*) beyond the centre of curvature

#### Answer:

(*c*) between focus and pole

A concave mirror produces a magnification of +4 when the object is placed between the focus and the pole.

#### Page No 200:

#### Question 37:

If a magnification of, −1 (minus one) is to be obtained by using a converging mirror, then the object has to be placed:

(*a*) between pole and focus

(*b*) at the centre of curvature

(*c*) beyond the centre of curvature

(*d*) at infinity

#### Answer:

(*b*) at the centre of curvature

If a magnification of −1 (minus one) is to be obtained by using a converging mirror, the object needs to be placed at the centre of curvature so that an image of same size as the object can be formed.

#### Page No 200:

#### Question 38:

In order to obtain a magnification of, −0.6 (minus 0.6) with a concave mirror, the object must be placed:

(*a*) at the focus

(*b*) between pole and focus

(*c*) between focus and centre of curvature

(*d*) beyond the centre of curvature

#### Answer:

(d) beyond the centre of curvature

In order to obtain a magnification of −0.6 (minus 0.6) with a concave mirror, the object needs to be placed beyond the centre of curvature, as at this point a diminished image will be formed.

#### Page No 200:

#### Question 39:

An object is placed at a large distance in front of a concave mirror of radius of curvature 40 cm. The image will be formed in front of the mirror at a distance:

(*a*) 20 cm

(*b*) 30 cm

(*c*) 40 cm

(*d*) 50 cm

#### Answer:

(*c*) 20 cm

#### Page No 200:

#### Question 40:

In order to obtain a magnification of, −1.5 with a concave mirror of focal length 16 cm, the object will have to be placed at a distance

(*a*) between 6 cm and 16 cm

(*b*) between 32 cm and 16 cm

(*c*) between 48 cm and 32 cm

(*d*) beyond 64 cm

#### Answer:

(*b*) between 32 cm and 16 cm

To obtain a magnification of -1.5, the object needs to be placed between the focus and the centre of curvature.

#### Page No 200:

#### Question 41:

Linear magnification (*m*) produced by a rear view mirror fitted in vehicles:

(*a*) is equal to one

(*b*) is less than one

(*c*) is more than one

(*d*) can be more less than one depending on the position of object

#### Answer:

(b) is less than one

Linear magnification (*m*) produced by a rear view mirror, installed in vehicles, is less than one.

#### Page No 200:

#### Question 42:

Between which two points of concave mirror should an object be placed to obtain a magnification of:

(*a*) −3

(*b*) +2.5

(*c*) −0.4

#### Answer:

(*a*) An object, in front of a concave mirror, should be placed between the latter's centre of curvature (C) and focus (F) to obtain a magnification of −3.

(*b*) An object, in front of a concave mirror, should be placed between the latter's focus (F) and the pole (P) to obtain a magnification of +25.

(*c*) An object, in front of a concave mirror, should be placed beyond the latter's centre of curvature (C) to obtain a magnification of −0.4.

#### Page No 200:

#### Question 43:

At what distance from a concave mirror of focal length 10 cm should an object be placed so that:

(*a*) its real image is formed 20 cm from the mirror?

(*b*) its virtual image is formed 20 cm from the mirror?

#### Answer:

$Given,\phantom{\rule{0ex}{0ex}}Itisaconcavemirror.\phantom{\rule{0ex}{0ex}}f=-10\mathrm{cm}\phantom{\rule{0ex}{0ex}}v=-20\mathrm{cm}\phantom{\rule{0ex}{0ex}}S\mathrm{ince}the\mathrm{image}\mathrm{is}\mathrm{real}and\mathrm{form}s\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror},theequationwillbe\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{-10}=\frac{1}{-20}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{1}{-10}+\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{-20+10}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{-10}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=-\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow u=-20\mathrm{cm}\phantom{\rule{0ex}{0ex}}Therefore,t\mathrm{he}\mathrm{object}\mathrm{should}\mathrm{be}\mathrm{placed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}20\mathrm{cm}fromthemirror\mathrm{to}\mathrm{form}a\mathrm{real}\mathrm{image}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

(b)

$Given,\phantom{\rule{0ex}{0ex}}f=-10\mathrm{cm}\phantom{\rule{0ex}{0ex}}v=+20\mathrm{cm}\phantom{\rule{0ex}{0ex}}S\mathrm{ince}the\mathrm{image}\mathrm{is}\mathrm{virtual}and\mathrm{form}s\mathrm{behind}\mathrm{the}\mathrm{mirror},theequationwillbe\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{-10}=\frac{1}{20}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{1}{10}-\frac{1}{20}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{-20-10}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{-30}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{-3}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow u=-\frac{20}{3}$

Therefore, the object must be placed at a distance of $\frac{20}{3}$cm from the mirror to form the virtual image.

#### Page No 200:

#### Question 44:

If a concave mirror has a focal length of 10 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.

#### Answer:

Given,

Focal length (f) of the concave mirror = -10 cm

__Case-1 __

The image is real, and its magnification (m) is -2.

Using the magnification formula, we get

$m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow -2=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow v=2u$

Now, using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{-10}=\frac{1}{2u}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{-10}=\frac{1}{2u}+\frac{2}{2u}=\frac{3}{2u}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{3\times (-10)}{2}=-15\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{object}\mathrm{should}\mathrm{be}\mathrm{placed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}15\mathrm{cm}\mathrm{from}\mathrm{the}\mathrm{concave}\mathrm{mirror}.$

__Case-2__

The image is virtual and has a magnification 'm' of 2.

Using the magnification formula, we get

$\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$

$\Rightarrow 2=\frac{-\mathrm{v}}{\mathrm{u}}$

*v=*-2u

Now, using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\Rightarrow \frac{1}{-10}=\frac{1}{-2\mathrm{u}}+\frac{1}{\mathrm{u}}$

$\Rightarrow \frac{1}{-10}=\frac{-1}{2u}+\frac{2}{2u}=\frac{1}{2u}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{-10}{2}=-5\mathrm{cm}$

Thus, the object should be placed at a distance of 5 cm in front of the concave mirror.

#### Page No 200:

#### Question 45:

A mirror forms an image which is 30 cm from an object and twice its height.

(*a*) Where must the mirror be situated?

(*b*) What is the radius of curvature?

(*c*) Is the mirror convex or concave?

#### Answer:

(a) Given, v - u = 30 cm

Magnification 'm' = 2

$m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow 2=\frac{-\mathrm{v}}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\Rightarrow v=-2u\phantom{\rule{0ex}{0ex}}\mathrm{By}\mathrm{putting}\mathrm{the}\mathrm{va}\mathrm{lue}\mathrm{of}\text{'}\mathrm{v}\text{'}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},weget\phantom{\rule{0ex}{0ex}}-2u-u=30\phantom{\rule{0ex}{0ex}}\Rightarrow -3u=30\phantom{\rule{0ex}{0ex}}u=\frac{-30}{3}=-10\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{object}\mathrm{is}\mathrm{situated}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}10\mathrm{cm}\mathrm{from}\mathrm{the}\mathrm{lens}.\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{mirror}\mathrm{is}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}10\mathrm{cm}\mathrm{from}\mathrm{the}\mathrm{object}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{image}\text{'}\mathrm{v}\text{'}=-2\times -10=20\mathrm{cm}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)U\mathrm{sin}gthe\mathrm{mirror}\mathrm{formula},weget\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{20}+\frac{1}{-10}\Rightarrow \frac{1}{f}=\frac{-10}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{10-20}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{-1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow f=-20\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{curvature}R=2f\phantom{\rule{0ex}{0ex}}R=2\times -20\phantom{\rule{0ex}{0ex}}R=-40\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{lens}\mathrm{is}40\mathrm{cm}.\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\mathrm{Since}\mathrm{the}\mathrm{focal}\mathrm{length}\mathrm{shows}\mathrm{a}\mathrm{negative}\mathrm{value},\mathrm{the}\mathrm{given}\mathrm{mirror}\mathrm{is}\mathrm{concave}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 205:

#### Question 1:

What type of image/images are formed by:

(*a*) a convex mirror?

(*b*) a concave mirror?

#### Answer:

(a) A convex mirror always forms virtual, erect and small-sized images.

(b) A concave mirror can form

1) virtual, erect and large-sized images

2) real and inverted images of any size (small, same size or large)

#### Page No 205:

#### Question 2:

Which mirror has a wider field of view?

#### Answer:

A convex mirror has a wider field of view because it is a diverging mirror and forms small and erect images of a large number of objects at the same time.

#### Page No 205:

#### Question 3:

If you want to see an enlarged image of your face, state whether you will use a concave mirror or a convex mirror?

#### Answer:

#### Page No 205:

#### Question 4:

Which mirror always produces a virtual, erect and diminished image of an object?

#### Answer:

A convex mirror always produces a virtual, erect and diminished image of any object.

#### Page No 205:

#### Question 5:

An object is placed at a long distance in front of a convex mirror of radius of curvature 30 cm. State the position of its image.

#### Answer:

When an object is placed in front of a convex mirror, the image is always formed behind the latter, and is virtual, erect and smaller than the object.

#### Page No 205:

#### Question 6:

Name the spherical mirror which can produce a real and diminished image of an object.

#### Answer:

Only a concave mirror can produce a real and diminished image of an object.

#### Page No 205:

#### Question 7:

Name the spherical mirror which can produce a virtual and diminished image of an object.

#### Answer:

Only a convex mirror can produce a virtual and diminished image of an object.

#### Page No 205:

#### Question 8:

One wants to see a magnified image of an object in a mirror. What type of mirror should one use?

#### Answer:

A person should use a concave mirror to see a magnified image of any object.

#### Page No 205:

#### Question 9:

Name the mirror which can give:

(*a*) an erect and enlarged image of an object.

(*b*) an erect and diminished image of an object.

#### Answer:

The mirror which can give

(*a*) an erect and enlarged image of an object is concave mirror

(*b*) an erect and diminished image of an object is convex mirror

#### Page No 205:

#### Question 10:

State whether the following statement is true or false:

A diverging mirror is used as a rear-view mirror.

#### Answer:

A diverging mirror, i.e. convex mirror, is used as a rear-view mirror, as it has a wider field of view and always forms virtual, erect and small-sized images of innumerable objects at the same time.

#### Page No 205:

#### Question 11:

What type of mirror should be used:

(*a*) as a shaving mirror?

(*b*) as a shop security mirror?

#### Answer:

#### Page No 205:

#### Question 12:

Which type or mirror is usually used as a rear-view mirror in motor cars?

#### Answer:

#### Page No 205:

#### Question 13:

What kind of mirrors are used in big shopping centres to watch the activities of the customers?

#### Answer:

#### Page No 205:

#### Question 14:

A ray of light of going towards the focus of a convex mirror becomes parallel to the principal axis after reflection from the mirror. Draw a labelled diagram to represent this situation.

#### Answer:

**Ray Diagram-**

A ray of light travelling towards the focus of a convex mirror becomes parallel to the principal axis after reflection on the mirror.

#### Page No 205:

#### Question 15:

Fill in the following blank with a suitable word:

A ray of light which is parallel to the principal axis of a convex mirror, appears to be coming from ............ after reflection from the mirror.

#### Answer:

A ray of light which is parallel to the principal axis of a convex mirror, appears to be coming from ** focus** after reflection from the mirror.

#### Page No 205:

#### Question 16:

Why does a driver prefer to use a convex mirror as a rear-view mirror in a vehicle?

#### Answer:

A driver prefers to use a convex mirror as a rear-view mirror in his vehicle because this type of mirror has a wider field of view. Besides this, the convex mirror always forms virtual, erect and small sized images of the large number of objects at the same time.

#### Page No 205:

#### Question 17:

Why can you not use a concave mirror as a rear-view mirror in vehicles?

#### Answer:

#### Page No 205:

#### Question 18:

Where would the image be formed by a convex mirror if the object is placed:

(*a*) between infinity and pole of the mirror?

(*b*) at infinity?

Draw labelled ray-diagrams to show the formation of image in both the case.

#### Answer:

Following are the ray-diagrams to show the formation of image, by a convex mirror, of the same object but in different locations:

(*a*) between infinity and pole of the mirror

The image will be virtual, magnified and behind the mirror when the object is between the infinity and focus of the mirror.

(*b*) at infinity

The image will be virtual, diminished and behind the mirror when the object is at infinity.

#### Page No 206:

#### Question 19:

The shiny outer surface of a hollow sphere of aluminium of radius 50 cm is to be used as a mirror:

(*a*) What will be the focal length of this mirror?

(*b*) Which type of spherical mirror will it provide?

(*c*) State whether this spherical mirror will diverge or converge light rays.

#### Answer:

#### Page No 206:

#### Question 20:

What is the advantage of using a convex mirror as rear-view mirror in vehicles as compared to a plane mirror? Illustrate your answer with the help of labelled diagrams.

#### Answer:

#### Page No 206:

#### Question 21:

Give two uses of a convex mirror. Explain why you would choose convex mirror for these uses.

#### Answer:

(a) Convex mirror is used for vigilance purpose in big shopping centres to keep a watch on the activities of the customers, as this mirror has a wider field of view and always forms virtual, erect and small-sized images of the large number of objects at the same time.

(b) Convex mirror is also used as a rear-view mirror in vehicles, as it has a wider field of view and always forms virtual, erect and small-sized images of the large number of objects at the same time.

#### Page No 206:

#### Question 22:

What would your image look like if you stood close to a large:

(*a*) convex mirror?

(*b*) concave mirror?

#### Answer:

(*a*) In case we stand close to a large convex mirror, our image will be smaller than us.

(*b*) In case we stand close to a large concave mirror, our image will be larger than us.

#### Page No 206:

#### Question 23:

Which of the following are concave mirrors and which convex mirrors?

Shaving mirrors, Car headlight mirror, Searchlight mirror, Driving mirror, Dentist's inspection mirror, Touch mirror, Staircase mirror in a double-decker bus, Make-up mirror, Solar furnace mirror, Satellite TV dish, Shop security mirror.

#### Answer:

Concave mirror is used as a shaving mirror, car headlight mirror, searchlight mirror, dentist's inspection mirror, touch mirror, make-up mirror, solar furnace mirror and satellite TV dish.

#### Page No 206:

#### Question 24:

How will you distinguish between a plane mirror, a concave mirror and a convex mirror without touching them?

#### Answer:

#### Page No 206:

#### Question 25:

If a driver has one convex and one plane rear-view mirror, how would the images in each mirror appear different?

#### Answer:

Image formed in a plane mirror | Image formed in a convex mirror |

1. Image is of the same size as the object. 2. The distance of the image from the mirror is equal to the distance of the object from the mirror. 3. It has a small field of view. |
1. Image is smaller than the object. 2. Distance of the image from the mirror is not equal to the distance of the object from the mirror. 3.Iit has a large field of view. |

#### Page No 206:

#### Question 26:

(*a*) Draw a labelled ray diagram to show the formation of image of an object by a convex mirror. Mark clearly the pole, focus and centre of curvature on the diagram.

(*b*) What happens to the image when the object is moved away from the mirror gradually?

(*c*) State three characteristics of the image formed by a convex mirror.

#### Answer:

(a)

(b) When the object is moved away from the mirror, the size of the image decreases.

(c) Image formed by a convex mirror is always

2) erect, and

3) diminished, i.e. smaller than the object

#### Page No 206:

#### Question 27:

(*a*) Draw a labelled ray diagram to show the formation of image in a convex-mirror when the object is at infinity. Mark clearly the pole and focus of the mirror in the diagram.

(*b*) State three characteristics of the image formed in this case.

(*c*) Draw diagram to show how a convex mirror can be used to give a large field of view.

#### Answer:

(a)

2) erect, and

(c) The following ray diagram shows that a convex mirror can be used to produce a large field of view. This is because when an object is in front of the convex mirror, irrespective of its distance (other than infinity), a virtual, erect and diminished image of the former is obtained. So, a convex mirror can produce image of a large number of objects at the same time.

#### Page No 206:

#### Question 28:

The image formed by a spherical mirror is virtual. The mirror will be:

(*a*) concave

(*b*) convex

(*c*) either concave or convex

(*d*) metallic

#### Answer:

(*b*) convex

A convex mirror always forms a virtual image.

#### Page No 206:

#### Question 29:

Whatever be the position of the object, the image formed by a mirror is virtual, erect and smaller than the object. The mirror then must be:

(*a*) plane

(*b*) concave

(*c*) convex

(*d*) either concave or convex

#### Answer:

(*c*) convex

It should be a convex mirror. This is because when an object is in front of a convex mirror, irrespective of its distance, a virtual, erect and diminished image of the object is obtained.

#### Page No 206:

#### Question 30:

The mirror used by a dentist to examine the teeth of a person is:

(*a*) convex

(*b*) concave

(*c*) plane

(*d*) any one of the above

#### Answer:

(*b*) concave

A concave mirror is used by a dentist to examine the teeth of a person.

#### Page No 206:

#### Question 31:

If the image formed is always virtual, the mirror can be:

(*a*) concave or convex

(*b*) concave or plane

(*c*) convex or plane

(*d*) only convex

#### Answer:

(*c*) convex or plane

Both convex and plane mirrors always form virtual images.

#### Page No 206:

#### Question 32:

A concave mirror cannot be used as:

(*a*) a magnifying mirror

(*b*) a torch reflector

(*c*) a dentist's mirror

(*d*) a real view mirror

#### Answer:

(d) a rear view mirror

A concave mirror cannot be used as a rear view mirror because it forms inverted images of distant objects.

#### Page No 206:

#### Question 33:

A boy is standing in front of and close to a special mirror. He finds the image of his head bigger than normal, the middle part of his body of the same size, and his legs smaller than normal. The special mirror is made up of three types of mirrors in the following order from top downwards:

(*a*) Convex, Plane, Concave

(*b*) Plane, Convex, Concave

(*c*) Concave, Plane, Convex

(*d*) Convex, Concave, Plane

#### Answer:

*c*) concave, plane, convex

A concave mirror forms an image larger than the object.

Aconvex mirror forms an image smaller than the object.

A plane mirror forms an image similar to the size of the object.

#### Page No 207:

#### Question 34:

The mirror which can form a magnified image of an object is:

(*a*) convex mirror

(*b*) plane mirror

(*c*) concave mirror

(*d*) both convex and concave mirror

#### Answer:

(c) concave mirror

A concave mirror forms a magnified image of an object.

#### Page No 207:

#### Question 35:

A real image of an object is to be obtained. The mirror required for this purpose is:

(*a*) convex

(*b*) concave

(*c*) plane

(*d*) either convex or concave

#### Answer:

(b) concave

A concave mirror forms a real image of an object.#### Page No 207:

#### Question 36:

Consider two statements A and B given below:

A : real image is always inverted

B : virtual image is always erect

Out of these two statements:

(*a*) only A is true

(*b*) only B is true

(*c*) both A and B are true

(*d*) none is true

#### Answer:

(*c*) both A and B are true

A real image is always inverted and a virtual image is always erect.

#### Page No 207:

#### Question 37:

The diagrams show the appearance of a fork when placed in front of and close to two mirrors A and B, turn by turn.

Figure

(*a*) Which mirror is convex?

(*b*) Which mirror is concave?

Give reasons for your choice.

#### Answer:

(*a*) Mirror B is convex, as it forms a small image.

(*b*) Mirror A is concave, as it forms a large image.

#### Page No 207:

#### Question 38:

The diagram shows a dish antenna which is used to receive television signals from a satellite. The antenna (signal detector) is fixed in front of the curved dish.

Figure

(*a*) What is the purpose of the dish?

(*b*) Should it be concave or convex?

(*c*) Where should the antenna be positioned to receive the strongest possible signals?

(*d*) Explain what change you would expect in the signals if a larger dish was used.

#### Answer:

(b) The dish should be concave.

(c) The antenna should be positioned at the focus of the concave dish to receive the strongest possible signals.

(d) If a larger dish was used, the aperture of the concave mirror would have been bigger; therefore, the signals received by the antenna would have been stronger.

#### Page No 207:

#### Question 39:

A man standing in front of a special mirror finds his image having a very small head, a fat body and legs of normal size. What is the shape of:

(*a*) top part of the mirror?

(*b*) middle part of the mirror?

(*c*) bottom part of the mirror?

Give reasons for your choice.

#### Answer:

*a*) the top part of the mirror is convex, as it forms virtual, erect and small-sized image.

(

*b*) the middle of the mirror is concave, as it forms virtual, erect and large-sized image.

(

*c*) the bottom of the mirror is a plane mirror, as it forms virtual, erect and image of the same size as the object.

#### Page No 207:

#### Question 40:

Two big mirrors A and B are fitted side by side on a wall. A man is standing at such a distance from the wall that he can see the erect image of his face in both the mirrors. When the man starts walking towards the mirrors, he finds that the size of his face in mirror A goes on increasing but that in mirror B remains the same.

(*a*) mirror A is concave and mirror B is convex

(*b*) mirror A is plane and mirror B is concave

(*c*) mirror A is concave and mirror B is plane

(*d*) mirror A is convex and mirror B is concave

#### Answer:

*c*) mirror A is concave and mirror B is plane

Image formed by a plane mirror is virtual, erect and of the same size as the object.

Image formed by a concave mirror is virtual,erect and larger than the object.

#### Page No 209:

#### Question 9:

(*a*) Draw a diagram to represent a convex mirror. On this diagram mark principal axis, principal focus F and the centre of curvature C if the focal length of convex mirror is 3 cm.

(*b*) An object 1 cm tall is placed 30 cm in front of a convex mirror of focal length 20 cm. Find the size and position of the image formed by the convex mirror.

#### Answer:

(a) Ray Diagram-

The mirror is a diverging mirror, i.e. convex mirror.

_{o}'

_{ }= 1 cm

_{i}' and its magnification 'm'.

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{-30}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{30}+\frac{1}{20}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{60}+\frac{3}{60}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{60}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{60}{5}=12\mathrm{cm}$

#### Page No 209:

#### Question 10:

A shop security mirror 5.0 m from certain items displayed in the shop produces on-tenth magnification.

(*a*) What is the type of mirror?

(*b*) What is the radius of curvature of the mirror?

#### Answer:

Distance of the object (u) = -5 m

Using the magnification formula, we get

*R*= 1.1

#### Page No 209:

#### Question 1:

An object is kept at a distance of 5 cm in front of a convex mirror of focal length 10 cm. Calculate the position and magnification of the image and state its nature.

#### Answer:

Distance of the object from the mirror 'u' = -5 cm

Focal length of the mirror 'f '= 10 cm

We have to find the distance of the image 'v'.

Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{10}=\frac{1}{v}+\frac{1}{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{10}=\frac{1}{v}-\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{10}+\frac{1}{5}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{10}+\frac{2}{10}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{v}=\frac{10}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{v}=3.3\mathrm{cm}\phantom{\rule{0ex}{0ex}}T\mathrm{hus},\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{image}\text{'}\mathrm{v}\text{'}\mathrm{is}3.3\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{magnification}\mathrm{formula},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{-3.3}{-5}\phantom{\rule{0ex}{0ex}}\mathrm{m}=0.66$

Thus, the image is virtual, erect and small in size.

#### Page No 209:

#### Question 2:

An object is placed at a distance of 10 cm from a convex mirror of focal length 5 cm.

(i) Draw a ray-diagram showing the formation image

(ii) State two characteristics of the image formed

(iii) Calculate the distance of the image from mirror.

#### Answer:

(i) Ray Diagram-

(ii) Following are the characteristics of the image formed:

(2) It is smaller than the object.

Focal length of the convex mirror is 5 cm.

We have to find the distance of the image 'v'.

Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{5}=\frac{1}{v}+\frac{1}{-10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{5}=\frac{1}{v}-\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{5}+\frac{1}{10}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{10}+\frac{1}{10}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{v}=\frac{10}{3}\Rightarrow \mathrm{v}=3.3\mathrm{cm}$

Thus, the distance of the image is 3.3 cm behind the mirror.

#### Page No 209:

#### Question 3:

An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.

#### Answer:

Distance of the object (u) = $-$6 cm

Focal length of the convex mirror (f) = 12 cm

We have to find the position of the image (v) and the nature of the image.

Using the mirror formula, we get:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{12}=\frac{1}{v}+\frac{1}{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{12}=\frac{1}{v}-\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{12}+\frac{1}{6}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{12}+\frac{2}{12}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{3}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{v}=4\mathrm{cm}$

Thus, the distance of the image (v) is 4 cm behind the mirror.

Now, using the magnification formula, we get:

$\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}\phantom{\rule{0ex}{0ex}}\mathrm{m}=\frac{-4}{-6}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\mathrm{m}=0.6$

Thus, the image is virtual, erect and small.

#### Page No 209:

#### Question 4:

An object placed 20 cm in front of a mirror is found to have an image 15 cm (*a*) in front of it, (*b*) behind the mirror. Find the focal length of the mirror and the kind of mirror in each case.

#### Answer:

$\frac{\mathit{1}}{f}\mathit{=}\frac{\mathit{1}}{v}\mathit{+}\frac{\mathit{1}}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{15}+\frac{1}{-20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{4}{60}-\frac{3}{60}=\frac{1}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow f=60\mathrm{cm}$

Thus, the mirror is convex and has a focal length of 60 cm.

#### Page No 209:

#### Question 5:

An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.

#### Answer:

The mirror is a diverging mirror, i.e. convex mirror.

_{o}'

_{ }= 2.5 cm

_{i}' and its magnification 'm'.

$\frac{\mathit{1}}{\mathit{f}}\mathit{=}\frac{\mathit{1}}{\mathit{v}}\mathit{+}\frac{\mathit{1}}{\mathit{u}}\phantom{\rule{0ex}{0ex}}\frac{\mathit{1}}{\mathit{20}}\mathit{=}\frac{\mathit{1}}{\mathit{v}}\mathit{+}\frac{\mathit{1}}{\mathit{-}\mathit{25}}\phantom{\rule{0ex}{0ex}}\frac{1}{20}+\frac{1}{25}=\frac{1}{\mathrm{v}}\phantom{\rule{0ex}{0ex}}\frac{5}{100}+\frac{4}{100}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{9}{100}\phantom{\rule{0ex}{0ex}}v=\frac{100}{9}=11.1\mathrm{cm}$

Now, using the magnification formula, we get

$m=\frac{-\mathrm{v}}{\mathrm{u}}=\frac{-11.1}{-25}\phantom{\rule{0ex}{0ex}}\mathrm{m}=0.44$

Therefore, the image is virtual, erect and smaller in size.

Again,

Therefore, the height of the image is 1.1cm.

#### Page No 209:

#### Question 6:

A convex mirror used as a rear-view mirror in a car has a radius of curvature of 3 m. If a bus is located at a distance of 5 m from this mirror, find the position of image. What is the nature of the image?

#### Answer:

The mirror is convex.

Radius of curvature of the mirror 'R' = 3 m

Using the mirror formula, we get

$\Rightarrow \frac{1}{1.5}=\frac{1}{v}+\frac{1}{-5}$

$\Rightarrow \frac{10}{15}+\frac{1}{5}=\frac{1}{v}$

$\Rightarrow \frac{10}{15}+\frac{3}{15}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{13}{15}$

$\Rightarrow \mathrm{v}=\frac{15}{13}=1.15\mathrm{m}$

Thus, the distance of the image 'v' is 1.15 m behind the mirror.

Now, using the magnification formula, we get

$m=\frac{-v}{u}$

$m=\frac{-\left(1.15\right)}{-5}=\frac{115}{500}=0.223$

Thus, the image is virtual, erect and smaller in size.

#### Page No 209:

#### Question 7:

A diverging mirror of radius of curvature 40 cm forms an image which is half the height of the object. Find the object and image positions.

#### Answer:

The mirror is convex.

#### Page No 209:

#### Question 8:

The radius of curvature of a convex mirror used as a rear view mirror in a moving car is 12.0 m. A truck is coming from behind it at a distance of 3.54 m. Calculate (*a*) position, and (*b*) size of the image relative to the size of the truck. What will be the nature of the image?

#### Answer:

The mirror is convex.

Radius of curvature of the mirror 'R' = 12 cm

Thus, the distance of the image 'v' is 0.77 m behind the mirror.

Thus, the image is virtual, erect and smaller in size.

#### Page No 210:

#### Question 11:

An object is placed 15 cm from (*a*) a converging mirror, and (*b*) a diverging mirror, of radius of curvature 20 cm. Calculate the image position and magnification in each case.

#### Answer:

__Case -1__

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{-10}=\frac{1}{-15}+\frac{1}{\mathrm{v}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{-10}+\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\mathrm{v}}=-\frac{1}{10}+\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\mathrm{v}}=\frac{-15}{150}+\frac{10}{150}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\mathrm{v}}=\frac{-5}{150}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{v}=-30\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{form}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}30\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{converging}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}\mathrm{magnification}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{-(-30)}{-15}\phantom{\rule{0ex}{0ex}}m=-2\phantom{\rule{0ex}{0ex}}\mathrm{magnification}=-2$

Thus the image is real, inverted and large in size.

__Case -2__

$U\mathrm{sing}the\mathrm{mirror}\mathrm{formula},weget\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{-15}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{10}=\frac{1}{-15}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{10}+\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{15}{150}+\frac{10}{150}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{25}{150}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow v=6\mathrm{cm}\phantom{\rule{0ex}{0ex}}There\mathrm{fore},\mathrm{the}\mathrm{image}\mathrm{willl}\mathrm{form}6\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{the}\mathrm{magnification}\mathrm{formula},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\mathrm{m}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{-6}{-15}\phantom{\rule{0ex}{0ex}}m=0.4$

Thus, the image is virtual, erect and small in size.

#### Page No 210:

#### Question 12:

An object 20 cm from a spherical mirror gives rise to a virtual image 15 cm behind the mirror. Determine the magnification of the image and the type of mirror used.

#### Answer:

The distance of the object 'u' = -20 cm

Using the magnification formula, we get

$m=\frac{-15}{-20}$

$m=0.75$

Thus, the image is virtual, erect and smaller in size.

The mirror used here is convex, as it forms a smaller virtual image

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