Question

An object is placed in front of a convex mirror at $60m$ from the mirror. Its radius of curvature is $40m$. The object starts moving towards the mirror at $20m{s}^{-1}$. Find the average speed of the image after a time interval of 2 seconds?

• A. $10m{s}^{-1}$
• B. $20m{s}^{-1}$
• C. $2.5m{s}^{-1}$
• D. $5m{s}^{-1}$

Answer 1

The correct option is C $2.5m{s}^{-1}$

Given:
The distance of object $u$ from the pole is 60 m,
Radius of curvature $\left(r\right)=40m$
Let $f$ be the focal length and $v$ be the image distance
We know that,
$f=\frac{r}{2}=\frac{1}{2}\times 40=20m$.
Applying the sign conventions, $u=-60m$ and $f=+20m$
We know, $\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$
$⟹\frac{1}{v}$ - $\frac{1}{60}$ = $\frac{1}{20}$
$⟹v=15m$
Positive sign shows that the image is virtual. Speed of the object $=20m{s}^{-1}$ and time taken $=2s$
$\therefore$ Distance $=$ Speed $\times$ time
$⟹20\times 2=40m$
Therefore, now the position of object will be $u^{\prime }=60-40=20m$
Let $v^{\prime }$ be the new position of the image
Applying it in the mirror formula we get,
$\frac{1}{v^{\prime }}$ + $\frac{1}{u^{\prime }}$ = $\frac{1}{f}$
$⟹\frac{1}{v^{\prime }}$ - $\frac{1}{20}$ = $\frac{1}{20}$
$⟹\frac{1}{v^{\prime }}$ = $\frac{40}{20\times 20}$
$⟹v^{\prime }=10m$
So, the image moves from $15m$ to $10m$ in $2s$. Hence, its average speed = $\frac{15-10}{2}=2.5m{s}^{-1}$