An object is moving with speed ${v_0}$ towards a spherical mirror with radius of curvature $R$ , along the central axis of mirror. The speed of the image with respect to the mirror is ( $U$ is the distance of the object form mirror at any given time $t$ )
 A. $ + \left( {\dfrac{R}{{U - 2R}}} \right)v_0^2$
 B. $ - {\left( {\dfrac{R}{{2U - R}}} \right)^2}{v_0}$
 C. $ - {\left( {\dfrac{R}{{U - 2R}}} \right)^2}{v_0}$
 D. $ + \left( {\dfrac{R}{{2U - R}}} \right)v_0^2$

Hint: This problem can be solved by using mirror formula. The value of the focal length is always half of the radius of curvature. We need to differentiate the mirror formula w.r.t time. This can lead us to the solution to the above-mentioned problem.

Formula used:
1. Mirror Formula:
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ … (1)
Here, $f$ is the focal length of the mirror,
$u$ is the object distance and
$v$ is the image distance.
2. Focal length
$f = \dfrac{R}{2}$
where , R is Radius of curvature

Complete answer:
When we think about the case of mirror,
As we know the relation between focal length and radius of curvature, $f = \dfrac{R}{2}$ , putting the value in mirror formula,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{2}{R}$
(because u and v are time dependent and R is a constant )
By differentiating both side with respect to t in the above equation, we get,
$\dfrac{{ - 1dv}}{{{v^2}dt}} + \left( {\dfrac{{ - 1}}{{{u^2}}}} \right)\dfrac{{du}}{{dt}} = 0$
As given in the question that, $\dfrac{{du}}{{dt}} = + {v_0}$ (rate of change of distance is speed)by putting the value in above equation,
$\dfrac{{dv}}{{dt}} = \dfrac{{ - {v^2}}}{{{u^2}}} \times ( + {v_0}) = \dfrac{{ - {v^2}{v_0}}}{{{u^2}}}$
As in above we get that,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{2}{R}$
Since, taking the value of v from above equation,
$v = \dfrac{{Ru}}{{2u - R}}$
As putting v’s value in the differential equation,
$\dfrac{{dv}}{{dt}} = - {\left( {\dfrac{{Ru}}{{2u - R}}} \right)^2} \times \dfrac{{{v_0}}}{{{u^2}}}$
By doing further solution of the above equation, we get,
$\dfrac{{dv}}{{dt}} = - {\left( {\dfrac{R}{{2u - R}}} \right)^2}{v_0}$
Therefore, we get the correct answer as $ - {\left( {\dfrac{R}{{2U - R}}} \right)^2}{v_0}$ .
Hence, the correct option is (B).


Note: Before proceeding to solve this problem make sure to know the quotient rule of differentiation. Many make mistakes in this step. According to the Quotient Rule, the derivative of a quotient is the denominator times the numerator's derivative minus the numerator times the denominator's derivative, all divided by the denominator's square.