First slide

The lens

Question

In the situation as shown in the figure, the focal length of the thin plane concave lens is 20 cm. A point object P is at distance of 20 cm from the lens. Find the velocity of image formed by plane mirror at the instant as shown in the figure.

Difficult

A

$(- 2 \hat{i} - 3 \hat{j}) c m s^{- 1}$
B

$(- 3 \hat{j}) c m s^{- 1}$
C

$(- 2 \hat{i} + 3 \hat{j}) c m s^{- 1}$
D

$(+ 3 \hat{j}) c m s^{- 1}$

Solution

Image formation due to concave lens, $u = - 20 c m, f = - 20 c m$
$∵ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Or $\frac{1}{v} + \frac{1}{20} = - \frac{1}{20}$
$∴ v = - 10 c m$
$∵ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Or $- \frac{1}{v^{2}} \frac{d v}{d t} + \frac{1}{u^{2}} \frac{d u}{d t} = 0$
Or $\frac{d v}{d t} = \frac{v^{2}}{u^{2}} \frac{d u}{d t}$
Or $v_{x} =$ component of velocity along optical axis = $\frac{100}{400} \times 10 \cos 37^{0} = \frac{1}{4} \times 8 = 2 c m s^{- 1}$
$∵ \frac{y'}{y} = m$
Or $y' = m y$
$∴ \frac{d y'}{d t} = m \frac{d y}{d t} + y \frac{d m}{d t}$
Here, y = 0 ( $∵$ object is on optical axis)
$∴ \frac{d y'}{d t} = v_{y} = \frac{v}{u} \times 10 \sin 37^{o} = \frac{- 10}{- 20} \times 6 = 3 c m s^{- 1}$
For Plane Mirror,
Horizontal component :
$v_{I} - v_{M} = - (v_{o} - v_{M}) \Rightarrow v_{I} - 1 = - (2 - 1) \Rightarrow v_{I} = 0$
Vertical Component :
$v_{I} - v_{M} = + (v_{o} - v_{M}) \Rightarrow v_{I} - 0 = + (3 - 0) \Rightarrow v_{I} = + 3$
Hence,
$v_{i m a g e} = (+ 3 \hat{j}) c m s^{- 1}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App