Question

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

• A. 10
• B. 20
• C. 2. 5
• D. 5

Answer 1

The correct option is C

2. 5

The distance of object from the pole is 60cm and focal length is $\frac{1}{2}\times centreofcurvature$ =
$\frac{1}{2}\times 40=20cm$

Applying sign convention, u = -60, f = +20, v = ?

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

$\frac{1}{v}-\frac{1}{60}=\frac{1}{20}$

v = 15 cm. Positive sign shows that the image is virtual.

Now after 2 sec, the distance covered by object is speed $\times$ time $=20\times 2=40cm$

The distance of the object from the pole is 20cm.

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

$\frac{1}{v}-\frac{1}{20}=\frac{1}{20}$

$\frac{1}{v}=\frac{40}{20\times 20}$
$v=10cm$

So, image moves from 15cm to 10cm in 2 sec.

Hence,
average speed = $\frac{5}{2}=2.5m/s$