Question

A particle is moving at a constant speed V from a large distance towards a concave mirror of radius R along its principal axis. Find the speed of the image formed by the mirror as a function of the distance x of the particle from the mirror.

Answer 1

Given,
Radius of the concave mirror is R
Therefore focal length of the mirror, $f=\frac{R}{2}$
Velocity of the particle, $V=\frac{\mathrm{d}x}{\mathrm{d}t}$
Object distance, u = −x

Using mirror equation,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
On putting the respective values we get,
$$\begin{gathered} \frac{1}{v}+\frac{1}{-x}=-\frac{2}{R} \\ \Rightarrow \frac{1}{v}=-\frac{2}{R}+\frac{1}{x}=\frac{R-2x}{Rx} \\ \therefore v=\frac{Rx}{R-2x} \end{gathered}$$

Velocity of the image is given by V₁
$$\begin{gathered} V_1=\frac{\mathrm{dv}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{Rx}{R-2x}\right] \\ =\frac{\left[\frac{\mathrm{d}}{\mathrm{d}t}\left(Rx\right)\left(R-2x\right)\right]-\left[\frac{\mathrm{d}}{\mathrm{d}t}\left(R-2x\right)\left(Rx\right)\right]}{{\left(R-2x\right)}^2} \\ =\frac{R\left[\left(\frac{\mathrm{dx}}{\mathrm{d}t}\right)\left(R-2x\right)\right]-\left[2\frac{\mathrm{dx}}{\mathrm{d}t}x\right]}{{\left(R-2x\right)}^2} \\ =\frac{R\left[\left(V\right)\left(R-2x\right)\right]-\left[2V\times 0\right]}{{\left(R-2x\right)}^2} \\ =\frac{V{R}^2}{(2x-R{)}^2} \end{gathered}$$