Solve the following examples

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?

b. A convex mirror has a focal length of 18 cm. The image of an object kept in front of the mirror is half the height of the object. What is the distance of the object from the mirror?

c. A 10 cm long stick is kept in front of a concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?

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