Question

A particle moves towards a concave mirror of focal length $30cm$ along its axis and with a constant speed of $4cm/sec$. What is the speed of its image when the partical is at $90cm$ from the mirror?

• A. $2cm/s$
• B. $8cm/s$
• C. $1cm/s$
• D. $4cm/s$

Answer 1

The correct option is C $1cm/s$

Given,

$f=30cm$

$u=90cm$

From mirror formula,

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

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\Rightarrow \frac{1}{v}=-\frac{1}{30}+\frac{1}{90}$

$\Rightarrow v=-45cm$

Velocity of the image,

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

$=-{\left(\frac{-45}{-90}\right)}^2\times 4$

$\Rightarrow \frac{dv}{dt}=-1cm/sec$

$=1cm/sec$