First slide

The lens

Question

For the situation as shown in the figure, find the velocity of image formed by the mirror

Moderate

A

$10 c m s^{- 1}$ in the direction of velocity of lens
B

$10 c m s^{- 1}$ in the opposite direction of velocity of lens
C

Zero
D

$20 c m s^{- 1}$ in the direction of velocity of object

Solution

For first refraction through lens, $u = - 30 c m, f = + 20 c m$
$∵ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$∴ v = + 60 c m$
$v_{i m a g e l e n s} = \frac{v^{2}}{u^{2}} \frac{d u}{d t} = \frac{60^{2}}{{(- 30)}^{2}} (6 - 2)$
Or $v_{i m a g e} - v_{l e n s} = 16 c m s^{- 1}$
$∴ v_{i m a g e} = 16 + 2 = 18 c m s^{- 1}$ Image formed by lens will act as object for the mirror hence the object distance for the mirror is equal to $u^{}'$
For reflection through mirror, $u' = - (80 - 60) = - 20 c m, f' = - 10 c m$
$∴ v' = - 20 c m$
$∴ v_{i m a g e m i r r o r} = - \frac{{(- 20)}^{2}}{{(- 20)}^{2}} (18 - 4) = - 14 c m s^{- 1}$
$∴ v_{i m a g e} = - 14 + 4 = - 10 c m s^{- 1}$

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