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 $\frac{R ^{2} v}{( R - b x ) ^{2}}$ . Find $b$

Question

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Answer 1

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

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

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

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

Differentiating equation (1)

$\frac{d y}{d t} = \frac{R \frac{d x}{d t}}{( R - 2 x )} + \frac{2 R x \frac{d x}{d t}}{( R - 2 x ) ^{2}}$

$\frac{d y}{d t} = \frac{[ R ( R - 2 x ) + 2 R x ] \frac{d x}{d t}}{( R - 2 x ) ^{2}} = \frac{R ^{2} v}{( R - 2 x ) ^{2}}$

Answer 2

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{- 2 x + R}{R x}

y = \frac{R x}{R - 2 x} . . . . . . . . . . . . . . . . (1)

Differentiating equation (1)

\frac{d y}{d t} = \frac{R \frac{d x}{d t}}{( R - 2 x )} + \frac{2 R x \frac{d x}{d t}}{( R - 2 x ) ^{2}}

\frac{d y}{d t} = \frac{[ R ( R - 2 x ) + 2 R x ] \frac{d x}{d t}}{( R - 2 x ) ^{2}} = \frac{R ^{2} v}{( R - 2 x ) ^{2}}