Science And Technology Solutions Solutions for Class 9 Science Chapter 11 - Reflection Of Light

Answer:

a.

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.

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:

Answer:

Answer:

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
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v}}=\frac{{\displaystyle 1}}{{\displaystyle -15}}-\frac{{\displaystyle 1}}{{\displaystyle -25}}=-\frac{2}{75} \\ \Rightarrow v=-37.5 \mathrm{cm} \end{gathered}$$
The screen should be placed at a distance of 37.5 cm in front of the concave mirror. 
Now, using magnification formula, we have
$$\begin{gathered} \frac{h_{\mathrm{i}}}{h_{\mathrm{o}}}=-\frac{v}{u} \\ \Rightarrow h_{\mathrm{i}}=-\frac{{\displaystyle -37.5}}{{\displaystyle -25}}\times 7=-10.5 \mathrm{cm} \end{gathered}$$
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
$$\begin{gathered} \frac{h_{\mathrm{i}}}{h_{\mathrm{o}}}=-\frac{v}{u} \\ \Rightarrow v=-u\times \frac{{\displaystyle h_{\mathrm{i}}}}{{\displaystyle h_{\mathrm{o}}}}=-\frac{u}{2} .....[\mathrm{using} (\mathrm{i}\left)\right] \end{gathered}$$
Now, using mirror formula, we have
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \Rightarrow \frac{1}{18}=-\frac{2}{u}+\frac{1}{u} \\ \Rightarrow u=-18 \mathrm{cm} \end{gathered}$$
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
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v}}=\frac{{\displaystyle 1}}{{\displaystyle -10}}-\frac{{\displaystyle 1}}{{\displaystyle -30}}=-\frac{2}{30} \\ \Rightarrow v=-15 \mathrm{cm} \end{gathered}$$
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
$$\begin{gathered} \frac{1}{f}=\frac{1}{v\text{'}}+\frac{1}{u} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v\text{'}}}=\frac{{\displaystyle 1}}{{\displaystyle -10}}-\frac{{\displaystyle 1}}{{\displaystyle -20}}=-\frac{1}{20} \\ \Rightarrow v\text{'}=-20 \mathrm{cm} \end{gathered}$$

Therefore, the length of the image = v'$-$ v = 20 $-$ 15 = 5 cm

Answer:

Let us name the mirrors as M₁, M₂ and M₃ 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.