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A mirror of radious of curvature 20cm and an object which is placed at a distance of 15cm are both moving with velocity 1ms^-1 and 10ms^-1. The velocity of image in this situation is
Option: 1
40 cm/s

Option: 2
44 cm/s

Option: 3
45 cm/s

Option: 4
30 cm/s

Answers (1)

best_answer

As we learn

Relation between object and image velocity if object is moving along principal axis -

$\underset{V_{im}}{\rightarrow}$ $=$ $\frac{-v^{2}}{u^{2}}$ $\underset{V_{om}}{\rightarrow}$

$\underset{V_{im}}{\rightarrow}$ = $\underset{V_{i}}{\rightarrow}$ - $\underset{V_{m}}{\rightarrow}$

$\underset{V_{om}}{\rightarrow}$ = $\underset{V_{o}}{\rightarrow}$ - $\underset{V_{m}}{\rightarrow}$

- wherein

$\underset{V_{o}}{\rightarrow}$ = velocity of light

$\underset{V_{i}}{\rightarrow}$ = velocity of image

$\underset{V_{m}}{\rightarrow}$ = velocity of mirror

$\underset{V_{om}}{\rightarrow}$ = velocity of object w.r.t. mirror

$\underset{V_{im}}{\rightarrow}$ = velocity of image w.r.t. mirror

$f =\frac{R}{2}=-10 cm$

u= -15 cm

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

or

$\frac{1}{v}=\frac{1}{-10cm}+\frac{1}{15}= \frac{-3+2}{-30}$

v = -30 cm

$\overrightarrow{v_{i}}- \overrightarrow{v_{m}}=\frac{v^{2}}{u^{2}}(\overrightarrow{v_{0}}- \overrightarrow{v_{m}})$

$\overrightarrow{v_{0}}= -10\hat{i}, \overrightarrow{v_{m}}=1\hat{i}$

$\overrightarrow{v_{i}}- 1\hat{i}=-\frac{(-30)^{2}}{(-15)^{2}}\left [ (-10\hat{i}-(+\hat{i}) \right ]$

$=-4 (-11\hat{i}) +44\hat{i}$

$\overrightarrow{v_{i}}=45\hat{i}$

$\therefore$ velocity of image is 45 cm/s

Posted by

Gaurav

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