Question

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 from mirror at any given time $t$)

• A. $+\left(\frac{R}{U-2R}\right)v_0^2$
• B. $-\left(\frac{R}{2U-R}\right)v_0$
• C. $-{\left(\frac{R}{U-2R}\right)}^2{v}_0$
• D. $+\left(\frac{R}{2U-R}\right)v_0^2$

Answer 1

The correct option is C $-{\left(\frac{R}{U-2R}\right)}^2{v}_0$

In case of mirror $\frac{1}{v}+\frac{1}{u}=\frac{2}{R}asf=\frac{R}{2}$

differentiate both sides $w.r.t"t"$

$\frac{-1}{v^2}\frac{dv}{dt}+\left(\frac{-1}{u^2}\right)\frac{du}{dt}=0$

as given is question $\frac{du}{dt}=+v_0$

$\therefore \frac{dv}{dt}=\frac{{-v}^2}{u^2}\times (+v_0)=\frac{{-v}^2{v}_0}{u^2}as\frac{1}{v}+\frac{1}{u}=\frac{2}{R}\therefore v=\frac{Ru}{2u-R}\therefore \frac{dv}{dt}=-{\left(\frac{Ru}{2u-R}\right)}^2\times \frac{v_0}{u^2}=-{\left(\frac{R}{2u-R}\right)}^2{v}_0\therefore \frac{dv}{dt}=-{\left(\frac{R}{2u-R}\right)}^2{v}_0$