Question

A point object at $15\text{cm}$ from a concave mirror of radius of curvature $20\text{cm}$ is made to oscillate along the principal axis with amplitude $2\text{mm}.$ The amplitude of its image will be

• A. $2\text{mm}$
• B. $4\text{mm}$
• C. $8\text{mm}$
• D. $16\text{mm}$

Answer 1

The correct option is C $8\text{mm}$

Given,

$u=15\text{cm};R=20\text{cm};f=10\text{cm}$


Using mirror formula,

$\frac{1}{v}+\frac{1}{(-15)}=\frac{1}{(-10)}$

$\frac{1}{v}=\frac{1}{15}-\frac{1}{10}$

$\Rightarrow v=-30\text{cm}$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}......\left(1\right)$

Differentiating (1) w.r.t time we get,

$-\frac{1}{v^2}dv-\frac{1}{u^2}du=0$

$dv=-\frac{v^2}{u^2}du$

$dv=-{\left(\frac{-30}{-15}\right)}^2\times du$

Here, $du=2\text{mm}$

$\Rightarrow \left|\begin{array}{c}dv\end{array}\right|=4\times 2=8\text{mm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.