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?

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2. 5

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Solution

The correct option is C
2. 5

The distance of object from the pole is 60cm and focal length is $\frac{1}{2} \times c e n t r e o f c u r v a t u r e$ =
$\frac{1}{2} \times 40 = 20 c m$

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 $= 20 \times 2 = 40 c m$

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

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

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

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

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

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

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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 20ms. Find the average speed of the image after a time interval of 2 sec?

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?