A $U$ -shaped wire is placed before a concave mirror of curvature $R$ . One bend $^{'} B^{'}$ is at the centre of curvature as shown in figure. Separation between vertical limbs of the structure is $R$ , where $R$ is the radius of curvature of the mirror. Find the ratio of the lengths of the images of the nearer and the farther limbs.

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Solution

$\frac{1}{v} + \frac{1}{u} = \frac{2}{R}$

$\Rightarrow \frac{1}{v} - \frac{1}{u} = \frac{2}{R}$

$v = - R$

Magnification $m_{1} = \frac{v}{u} = \frac{- R}{- R} = 1$

$L_{1} = m_{1} L_{0}$

$L_{1} = L_{0}$

For $(A)$

$\frac{1}{v} + \frac{1}{- 2 R} = \frac{2}{- R}$

$\frac{1}{v} = \frac{1}{2 R} = \frac{2}{R} = \frac{1 - 4}{2 R} = \frac{- 3}{2 R}$

$v = \frac{- 2 R}{3}$

$m_{2} = \frac{v}{u} = \frac{- 2 R}{3 (- 2 R)} = \frac{1}{3}$

$L_{2} = m_{2} L_{0}$

$= \frac{1}{3} L_{0}$

$∴ \frac{L_{1}}{L_{2}} = \frac{3}{1}$

$= 3 : 1$

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