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Q.

A luminous point is moving at speed $v_{0}$ towards a spherical mirror, along its axis. Then the speed at which the image of this point object is moving is given by (with R = radius of curvature and u = object distance)

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a

$- v_{0}$

b

$- v_{0} \frac{R}{(2 u - R)}$

c

$- v_{0} (\frac{2 u - R}{R})$

d

$- v_{0} {(\frac{R}{2 u - R})}^{2}$

answer is D.

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Detailed Solution

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f} d i f f e r e n t i a t i i n g - \frac{1}{u^{2}} \frac{d u}{d t} - \frac{1}{v^{2}} \frac{d v}{d t} = 0 \frac{d v}{d t} = - v_{0} \frac{v^{2}}{u^{2}} (\frac{d u}{d t} = v_{0}) \frac{v}{u} = \frac{f}{u - f} f = R / 2$

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A luminous point is moving at speed v0towards a spherical mirror, along its axis. Then the speed at which the image of this point object is moving is given by (with R = radius of curvature and u = object distance)