Question

A point object is placed at a distance $\frac{5R}{3}$ from the pole of a concave mirror. The radius of curvature of the mirror is $R$. The point object oscillates with an amplitude of $1\text{mm}$ perpendicular to the principal axis. Then, the amplitude of oscillation of the image is

• A. $\frac{3}{7}\text{mm}$
• B. $\frac{2}{7}\text{mm}$
• C. $\frac{4}{3}\text{mm}$
• D. $\frac{11}{7}\text{mm}$

Answer 1

The correct option is A $\frac{3}{7}\text{mm}$

Given that
$u=-\frac{5R}{3};f=-\frac{R}{2}$
Amplitude of oscillation, $A_O=1\text{mm}$

By using mirror formula,
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\Rightarrow \frac{1}{v}=-\frac{2}{R}-\frac{1}{\left(\frac{-5R}{3}\right)}$

$\Rightarrow \frac{1}{v}=-\frac{2}{R}+\frac{3}{5R}$

$\Rightarrow \frac{1}{v}=\frac{-7}{5R}\Rightarrow v=\frac{-5R}{7}$

Now using formula of magnification, $m=\frac{-v}{u}$
$\Rightarrow m=-\frac{(-\frac{5R}{7})}{-\frac{5R}{3}}$
$\Rightarrow m=-\frac{3}{7}$

So, Amplitude of image$\left(A_I\right)$ $=|m|\times$ amplitude of object$\left(A_O\right)$
$\Rightarrow A_I=\frac{3}{7}\times 1$

$\Rightarrow A_I=\frac{3}{7}\text{mm}$
Hence, option $\left(a\right)$ is correct.