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#### Page No 127:

#### Question 1:

a. Explain the difference between a plane mirror, a concave mirror and a convex mirror with respect to the type and size of the images produced.

1.Torch light

3. Flood light

#### Answer:

a.

Mirror |
Type of image |
Size of image |

Plane mirror | Virtual and erect | Same size as that of object |

Concave mirror | Real and inverted as well as virtual and erect depending on the position of object w.r.t. the mirror | Magnified, diminished as well as same size as that of object depending on the position of object w.r.t. the mirror |

Convex mirror | Always virtual and erect | Always diminished |

b.

1.Torch light: The source of light is placed at the focus of the mirror to obtain a parallel beam of light.

2. Projector lamp: â€‹The source of light is placed at the centre of curvature of the mirror.

3. Flood light: The source of light is placed just beyond the centre of curvature of the mirror to obtain bright beam of light.

c. Concave mirror is also known as converging mirror because of its ability of focus the parallel rays at a single point anywhere on its focal plane. This property of concave mirror is used in solar devices. Parallel rays from the Sun falling on the concave mirror of a solar device are focussed at the particular point to generate a lot of heat required for the proper functioning of the device.

d. Convex mirrors have the ability to cover wider field of view by forming diminished and virtual image of the objects. Thus, the mirrors fitted on the outside of cars are convex so that larger view of the road can be observed by the driver sitting in the car.

e. The rays from the Sun are parallel. When these rays fall on the concave mirror, they are converged to a single point on the focal plane of the mirror. Thus a lot of heat is produced at that point. Now, if a paper is placed at that point to get the image of the Sun, the paper gets burnt because of huge amount of heat concentrated at that point.

f. When spherical mirror breaks, their nature does not change. This means if a concave mirror breaks into pieces, the pieces will act as concave mirros only. Same happens in case of convex mirrors.

#### Page No 127:

#### Question 2:

What sign conventions are used for reflection from a spherical mirror?

#### Answer:

**Sign Conventions for spherical Mirrors:**

**I.** Objects are always placed to the left of the mirror i.e. light must fall on the mirror from left to right.

**II**. All distances are measured from the pole of the mirror.

**III.** Distances along the direction of the incident ray (along positive *x*- axis) are taken as positive, while distances along the direction of the reflected ray (along negative *x*-axis) are taken as negative.

**IV.** Heights measured perpendicular to and above the principal axis (along positive *y*-axis) are taken as positive.

**V**. Heights measured perpendicular to and below the principal axis (along negative *y*-axis) are taken as negative.

These sign conventions can also represented in the following diagram:

#### Page No 127:

#### Question 3:

Draw ray diagrams for the cases of images obtained in concave mirrors as described in the table on page 122.

#### Answer:

The light rays coming from infinity are parallel. When parallel light rays are incident on the reflecting surface of a concave mirror, they tend to meet at its focus after reflection. In this case, the image is formed at the focus, and is point-sized. |
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II. When the object is behind the centre of curvature |
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In this case, the image is formed between the focus (F) and the centre of curvature (C). This image is real, inverted and diminished. |
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III. When the object is at the centre of curvature |
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In this case, the image is formed at the centre of curvature. This image is real, inverted and of the same size as the object. | ||||

IV. When the object is between the centre of curvature (C) and the focus (F) |
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In this case, the image is formed behind the centre of curvature. This image is real, inverted and magnified. | ||||

V. When the object is at the focus (F) |
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In this case, the image is formed at infinity. This image is real, inverted and highly enlarged. | ||||

VI. When the object is placed between the focus (F) and the pole (P) |
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In this case, the image is formed behind the mirror. This image is virtual, erect and magnified. |

#### Page No 127:

#### Question 4:

#### Answer:

Device |
Type of mirror |

Periscope | Plane mirror |

Floodlights | Concave mirror |

Shaving mirror | Concave mirror |

Kaleidoscope | Plane mirror |

Street lights | Convex mirror |

Head lamps of a car | Concave mirror |

#### Page No 127:

#### Question 5:

a. An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal length of the mirror is 15 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image?

#### Answer:

a. Given,

*h*_{o }= 7 cm

*u* = $-$25 cm

*f* = $-$15 cm

Let* v* be the distance of the screen from the mirror so as to get a clear image. Now, using mirror formula, we have

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v}}=\frac{{\displaystyle 1}}{{\displaystyle -15}}-\frac{{\displaystyle 1}}{{\displaystyle -25}}=-\frac{2}{75}\phantom{\rule{0ex}{0ex}}\Rightarrow v=-37.5\mathrm{cm}$

The screen should be placed at a distance of 37.5 cm in front of the concave mirror.

Now, using magnification formula, we have

$\frac{{h}_{\mathrm{i}}}{{h}_{\mathrm{o}}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{\mathrm{i}}=-\frac{{\displaystyle -37.5}}{{\displaystyle -25}}\times 7=-10.5\mathrm{cm}$

The size of the image is 10.5 cm and the negative sign shows that the image is real and inverted.

b. *f* = 18 cm

*h _{i}*

_{ }= $\frac{{h}_{\mathrm{o}}}{2}$

$\Rightarrow \frac{{h}_{\mathrm{i}}}{{h}_{\mathrm{o}}}=\frac{1}{2}$ .....(i)

*u*= ?

Using magnification formula, we have

$\frac{{h}_{\mathrm{i}}}{{h}_{\mathrm{o}}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow v=-u\times \frac{{\displaystyle {h}_{\mathrm{i}}}}{{\displaystyle {h}_{\mathrm{o}}}}=-\frac{u}{2}.....[\mathrm{using}(\mathrm{i}\left)\right]$

Now, using mirror formula, we have

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

Hence, the distance of the object from the mirror is 18 cm.

c.

__For end A,__

*u*= $-$30 cm

*f*= $-$10 cm

Let the image of the end of stick at point A is formed at a distance

*v*from the the mirror. Therefore, using mirror formula, we have

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

__For end B,__

*u*= $-$20 cm

*f*= $-$10 cm

Let the image of the end of stick at point B is formed at a distance

*v'*from the the mirror. Therefore, using mirror formula, we have

$\frac{1}{f}=\frac{1}{v\text{'}}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v\text{'}}}=\frac{{\displaystyle 1}}{{\displaystyle -10}}-\frac{{\displaystyle 1}}{{\displaystyle -20}}=-\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow v\text{'}=-20\mathrm{cm}$

Therefore, the length of the image =

*v*'$-$

*v*= 20 $-$ 15 = 5 cm

#### Page No 127:

#### Question 6:

Three mirrors are created from a single sphere. Which of the following - pole, centre of curvature, radius of curvature, principal axis - will be common to them and which will not be common?

#### Answer:

Let us name the mirrors as M_{1}, M_{2} and M_{3 }as shown below.

From the figure, we see that

- pole (P) is not common
- centre of curvature (C) is common
- principal axis (PA) is not common
- radius of curvature (
*R*) is common. Radius of curvature is the radius of the sphere. In the given figure, length CP is same for all mirrors.

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