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 $$t$$)
A.
$$ + \left( {\frac{R}{{U - 2R}}} \right)v_0^2$$
B.
$$ - {\left( {\frac{R}{{R - 2U}}} \right)^2}{v_0}$$
C.
$$ - {\left( {\frac{R}{{2U - 2R}}} \right)^2}{v_0}$$
D.
$$ + \left( {\frac{R}{{2U - 2}}} \right)v_0^2$$
Answer :
$$ - {\left( {\frac{R}{{R - 2U}}} \right)^2}{v_0}$$
Solution :
For concave mirror
$$\eqalign{
& \frac{2}{R} = \frac{1}{v} + \frac{1}{u}\,\,{\text{or}}\,\,\frac{2}{{ - R}} = \frac{1}{v} + \frac{1}{{ - u}} \cr
& \therefore \frac{1}{v} = \frac{1}{U} - \frac{2}{R} = \frac{{R - 2U}}{{UR}} \cr
& {\text{or}}\,\,v = \left[ {\frac{{RU}}{{R - 2U}}} \right] \cr} $$
In spherical mirror, image velocity
$${v_i} = - \left[ {\frac{{{v^2}}}{{{u^2}}}} \right]{v_0} = - {\left[ {\frac{{RU}}{{R - 2U}}} \right]^2}\frac{{{v_0}}}{{{U^2}}} = - {\left[ {\frac{R}{{R - 2U}}} \right]^2}{v_0}$$