Optics Question 791

Question: An object is moving with speed $ v _{0} $ towards a spherical mirror with radius of curvature R, along the central axis of the mirror. The speed of the image with respect to the mirror is (U is the distance of the object from mirror at any given time

Options:

A) $ +( \frac{R}{U-2R} )v _{0}^{2} $

B) $ -{{( \frac{R}{R-2U} )}^{2}}v _{0}^{{}} $

C) $ -{{( \frac{R}{2U-2R} )}^{2}}v _{0}^{{}} $

D) $ +( \frac{R}{2U-2} )v _{0}^{2} $

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Answer:

Correct Answer: B

Solution:

[b] For concave mirror $ \frac{2}{R}=\frac{1}{v}+\frac{1}{u}\text{ or }\frac{2}{-R}=\frac{1}{v}+\frac{1}{-u} $

$ \therefore \frac{1}{v}=\frac{1}{U}-\frac{2}{R}=\frac{R-2U}{UR}\text{ or }v=[ \frac{RU}{R-2U} ] $

In spherical mirror, image velocity $ v _{t}=-[ \frac{v^{2}}{u^{2}} ]v _{0}=-{{[ \frac{RU}{R-2U} ]}^{2}}\frac{v _{0}}{U^{2}}=-{{[ \frac{R}{R-2U} ]}^{2}}v _{0} $



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