Question

An object moving at a speed of $5m/s$ towards a concave mirror of focal length $f=1m$ is at a distance of $9m$. The average speed of the image is:

• A. $\frac{1}{5}m/s$
• B. $\frac{1}{10}m/s$
• C. $\frac{5}{9}m/s$
• D. $\frac{2}{5}m/s$

Answer 1

The correct option is B $\frac{1}{10}m/s$

Given,

Focal length, $f=-1m$

Object distance, $u=-9m$

Image distance, $v=?$

We know that

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

$\frac{1}{-1}=\frac{1}{-9}-\frac{1}{v}$

$\frac{1}{v}=\frac{-10}{9}$

$v=\frac{-9}{10}$

Negative sign indicates that it is on the left side of the mirror.

Consider a time of $1$ second.

Object distance $=5m/s$

Hence, $Distance=Speed\times Time=5\times 1=5m$ is travelled by object in $1$ second

So, $u=9-5=4m$

Now, we need to find the new image distance using the formula

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

$\frac{1}{-1}=\frac{1}{-4}+\frac{1}{-v}$

$v=\frac{-4}{5}$

Negative sign indicates that it is on the left side of the mirror.

Change in image distance $=\frac{-9}{10}-\frac{-4}{5}=\frac{1}{10}m$

Average speed of image $=\frac{Distancemovedbyimage}{time}=\frac{1}{10}m/s$