Question

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

• A. $2mm$
• B. $4mm$
• C. $8mm$
• D. $16mm$

Answer 1

The correct option is D $16mm$

$u=-(15-0.2)cm=-14.8cm$

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

$\Rightarrow \frac{1}{v}=\frac{1}{-10}-\frac{1}{14.8}=\frac{1}{14.8}=\frac{1}{10}$

$\Rightarrow v=\frac{108}{4.8}cm=-30.83cm$

When it is at $B"$

$u=-(15+0.2)cm=-15.2cm$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}+\frac{1}{+15.2}$

$=\frac{1}{15.2}=\frac{1}{10}=-\frac{5.2}{152}$

$v^{\prime }=-\frac{152}{5.2},△v=|v^{\prime }-v|$

$=-22.23=1.6cm$

$=16mm$

Amplitude of $V=\frac{16}{2}=8cm$.