Question

A concave mirror produces three times enlarged image of an object placed at $10\mathrm{cm}$ in front of it. Calculate the focal length of the mirror.

Answer 1

Step 1: Given that

Size of image(${\mathrm{h}}_{\mathrm{i}}$) = Three times the size of the object(${\mathrm{h}}_{\mathrm{o}}$)

${\mathrm{h}}_{\mathrm{i}}$ = $3{\mathrm{h}}_{\mathrm{o}}$

$\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}=3$

Object distance($\mathrm{u}$) = $-10\mathrm{cm}$

{Note: All the distance to the left of the pole of a spherical mirror is negative($-$) and the distances taken to the right of the pole are positive($+$).}

Step 2: Formula used:

1. Magnification of a mirror($\mathrm{m}$)= $=\frac{\mathrm{height}\mathrm{of}\mathrm{image}\left({\mathrm{h}}_{\mathrm{i}}\right)}{\mathrm{height}\mathrm{of}\mathrm{object}\left({\mathrm{h}}_{\mathrm{o}}\right)}=-\frac{\mathrm{v}}{\mathrm{u}}$

(Note: Negative magnification suggests a real and inverted image whereas a positive magnification suggests a virtual and erect image.)

2. Object distance($\mathrm{u}$), image distance($\mathrm{v}$) and focal length($\mathrm{f}$) of a mirror are related as:

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

Step 3: Calculation of focal length ($\mathrm{f}$) of the concave mirror:

$$\begin{gathered} \mathrm{m}=\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}=-\frac{\mathrm{v}}{\mathrm{u}} \\ 3=-\frac{\mathrm{v}}{\mathrm{u}} \\ \mathrm{v}=-3\mathrm{u} \\ \mathrm{v}=-3\times 10\mathrm{cm} \\ \mathrm{v}=-30\mathrm{cm} \end{gathered}$$

Now,

$$\begin{gathered} \frac{1}{-30\mathrm{cm}}+\frac{1}{-10\mathrm{cm}}=\frac{1}{\mathrm{f}} \\ -\frac{1}{30}-\frac{1}{10}=\frac{1}{\mathrm{f}} \\ \mathrm{Or}, \\ \frac{1}{\mathrm{f}}=\frac{-1-3}{30} \\ \frac{1}{\mathrm{f}}=\frac{-4}{30} \\ \mathrm{f}=\frac{-30}{4} \\ \mathrm{f}=-\frac{15}{2}\mathrm{cm} \end{gathered}$$

Thus, the focal length of the concave mirror is $\frac{15}{2}\mathrm{cm}$.