Question

A particle is moving at a constant speed $v$ from a large distance towards concave mirror of radius $R$ along the principal axis. If the speed of the images as a function of the distance $x$ of the particle from the mirror is given as $\frac{R^{2}v}{{(R-bx)}^2}$. Find $b$

Answer 1

Let y represent the image distance and x the object distance from the mirror.

$\Rightarrow \frac{1}{y}+\frac{1}{-x}=-\frac{2}{R}$

$\Rightarrow \frac{1}{y}=\frac{-2}{R}+\frac{1}{x}=-\frac{-2x+R}{Rx}$

$y=\frac{Rx}{R-2x}................\left(1\right)$

Differentiating equation (1)

$\frac{dy}{dt}=\frac{R\frac{dx}{dt}}{(R-2x)}+\frac{2Rx\frac{dx}{dt}}{{(R-2x)}^2}$

$\frac{dy}{dt}=\frac{[R(R-2x)+2Rx]\frac{dx}{dt}}{{(R-2x)}^2}=\frac{R^{2}v}{{(R-2x)}^2}$