#### Page No 173:

#### Answer:1

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:

#### Answer:2

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:

#### Answer:3

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:

#### Answer:4

(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:

#### Answer:5

(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:

#### Answer:6

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:

#### Answer:7

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:

#### Answer:8

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

#### Page No 173:

#### Answer:9

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:

#### Answer:10

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:

#### Answer:11

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

#### Page No 173:

#### Answer:12

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

#### Page No 173:

#### Answer:13

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

#### Page No 173:

#### Answer:14

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

#### Page No 173:

#### Answer:15

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:

#### Answer:16

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:

#### Answer:17

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

(b) The phenomenon involved is lateral inversion.

#### Page No 173:

#### Answer:18

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:

#### Answer:19

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:

#### Answer:20

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:

#### Answer:26

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:

#### Answer:27

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:

#### Answer:28

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:

#### Answer:29

(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:

#### Answer:30

(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:

#### Answer:31

(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:

#### Answer:32

(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:

#### Answer:33

(a) Virtual

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

#### Page No 174:

#### Answer:34

(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:

#### Answer:35

(d) 9.25, since the image formed by a plane mirror is laterally inverted.

#### Page No 174:

#### Answer:21

(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:

#### Answer:22

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:

#### Answer:23

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:

#### Answer:24

(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:

#### Answer:25

(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 175:

#### Answer:36

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:

#### Answer:37

(b) 4 cm

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

#### Page No 175:

#### Answer:38

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 + 0.5m = 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 = 4m.

#### Page No 175:

#### Answer:39

(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:

#### Answer:40

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:

#### Answer:1

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:

#### Answer:2

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

#### Page No 178:

#### Answer:3

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:

#### Answer:4

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:

#### Answer:11

**(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:

#### Answer:12

(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:

#### Answer:13

(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:

#### Answer:14

(*c*) a bulging-out surface

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

#### Page No 179:

#### Answer:15

(b) convex mirror

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

#### Page No 179:

#### Answer:16

(*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:

#### Answer:17

(*b*) 15 cm

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

#### Page No 179:

#### Answer:18

(*a*) 25 cm

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

#### Page No 179:

#### Answer:5

(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:

#### Answer:19

#### Page No 179:

#### Answer:20

(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 179:

#### Answer:6

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:

#### Answer:7

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:

#### Answer:8

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:

#### Answer:9

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:

#### Answer:10

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 189:

#### Answer:1

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:

#### Answer:2

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:

#### Answer:3

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:

#### Answer:4

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

#### Page No 189:

#### Answer:5

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

#### Page No 189:

#### Answer:6

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

#### Page No 189:

#### Answer:7

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:

#### Answer:8

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:

#### Answer:9

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:

#### Answer:10

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

#### Page No 189:

#### Answer:11

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:

#### Answer:12

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:

#### Answer:13

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:

#### Answer:14

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:

#### Answer:15

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:

#### Answer:16

As shown in the diagram, the image formed is

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

#### Page No 189:

#### Answer:17

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:

#### Answer:27

(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:

#### Answer:28

(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:

#### Answer:29

(*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:

#### Answer:30

(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:

#### Answer:31

(*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:

#### Answer:32

(*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:

#### Answer:33

(*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:

#### Answer:34

(*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 190:

#### Answer:18

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:

#### Answer:19

(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:

#### Answer:20

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:

#### Answer:21

(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:

#### Answer:22

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:

#### Answer:23

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:

#### Answer:24

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:

#### Answer:25

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:

#### Answer:26

Two uses of concave mirrors are:

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

#### Page No 191:

#### Answer:35

(*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:

#### Answer:36

(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:

#### Answer:37

(*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:

#### Answer:38

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:

#### Answer:39

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:

#### Answer:40

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:

#### Answer:1

(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:

#### Answer:2

(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:

#### Answer:3

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:

#### Answer:4

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:

#### Answer:5

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

#### Page No 192:

#### Answer:6

(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:

#### Answer:7

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:

#### Answer:9

(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:

#### Answer:10

(*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 193:

#### Answer:8

(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 198:

#### Answer:1

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:

#### Answer:2

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:

#### Answer:3

(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:

#### Answer:4

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:

#### Answer:5

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:

#### Answer:6

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

#### Page No 198:

#### Answer:7

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:

#### Answer:8

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:

#### Answer:9

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:

#### Answer:10

(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:

#### Answer:11

(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:

#### Answer:12

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

Height of the object '*h _{o}*'â€‹ = 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*'â€‹ = −5 cm

_{i}Again using the magnification formula, we getâ€‹

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

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

#### Page No 198:

#### Answer:13

â€‹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:

#### Answer:14

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:

#### Answer:15

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:

#### Answer:16

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:

#### Answer:17

â€‹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:

#### Answer:31

(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:

#### Answer:33

(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:

#### Answer:34

(*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:

#### Answer:35

(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:

#### Answer:36

(*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 199:

#### Answer:18

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:

#### Answer:19

â€‹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:

#### Answer:20

â€‹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:

#### Answer:21

*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:

#### Answer:22

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:

#### Answer:23

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:

#### Answer:24

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:

#### Answer:25

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

_{o}' = 50 mm

_{i}'.

#### Page No 199:

#### Answer:26

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:

#### Answer:27

â€‹

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:

#### Answer:28

**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:

#### Answer:29

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:

#### Answer:30

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

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

#### Page No 200:

#### Answer:37

(*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:

#### Answer:38

(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:

#### Answer:39

(*c*) 20 cm

#### Page No 200:

#### Answer:40

(*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:

#### Answer:41

(b) is less than one

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

#### Page No 200:

#### Answer:42

(*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:

#### Answer:43

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

(b)

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

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:

#### Answer:44

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

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

Now, using the mirror formula, we get

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

__Case-2__

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

Using the magnification formula, we get

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

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

v=-2u

Now, using the mirror formula, we get

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

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

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

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

#### Page No 200:

#### Answer:45

(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:

#### Answer:1

(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:

#### Answer:2

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:

#### Answer:3

#### Page No 205:

#### Answer:4

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

#### Page No 205:

#### Answer:5

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:

#### Answer:6

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

#### Page No 205:

#### Answer:7

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

#### Page No 205:

#### Answer:8

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

#### Page No 205:

#### Answer:9

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:

#### Answer:10

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:

#### Answer:11

#### Page No 205:

#### Answer:12

#### Page No 205:

#### Answer:13

#### Page No 205:

#### Answer:14

**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:

#### Answer:15

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:

#### Answer:16

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:

#### Answer:17

#### Page No 205:

#### Answer:18

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:

#### Answer:26

(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:

#### Answer:27

(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:

#### Answer:28

(*b*) convex

A convex mirror always forms a virtual image.

#### Page No 206:

#### Answer:29

(*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:

#### Answer:30

(*b*) concave

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

#### Page No 206:

#### Answer:31

(*c*) convex or plane

Both convex and plane mirrors always form virtual images.

#### Page No 206:

#### Answer:32

(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:

#### Answer:33

*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 206:

#### Answer:19

#### Page No 206:

#### Answer:20

#### Page No 206:

#### Answer:21

(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:

#### Answer:22

(*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:

#### Answer:23

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:

#### Answer:24

#### Page No 206:

#### Answer:25

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 207:

#### Answer:34

(c) concave mirror

A concave mirror forms a magnified image of an object.

#### Page No 207:

#### Answer:35

(b) concave

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

#### Answer:36

(*c*) both A and B are true

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

#### Page No 207:

#### Answer:37

(*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:

#### Answer:38

(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:

#### Answer:39

*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:

#### Answer:40

*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:

#### Answer:9

(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:

#### Answer:10

Distance of the object (u) = -5 m

Using the magnification formula, we get

*R*= 1.1

#### Page No 209:

#### Answer:1

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:

#### Answer:2

(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:

#### Answer:3

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:

#### Answer:4

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

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

#### Page No 209:

#### Answer:5

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:

#### Answer:6

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:

#### Answer:7

The mirror is convex.

#### Page No 209:

#### Answer:8

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:

#### Answer:11

__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:

#### Answer:12

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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