Consider a concave mirror and a convex lens (refractive index= 1.5) of focal length 10cm each, separated by a distance of 50cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification $M_{1}$ . When the set-up is kept in a medium of refractive index 7/6, the magnification becomes $M_{2}$ . The magnitude $|\frac{M_{2}}{M_{1}}|$ is

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answer is 7.

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Detailed Solution

1st case:
Reflection from mirror
$\begin{array}{l} \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \Rightarrow \frac{1}{- 10} = \frac{1}{v} + \frac{1}{(- 15)} \\ \Rightarrow v = - 30 \end{array}$

$\begin{array}{l} F o r l e n s \\ \frac{1}{10} = \frac{1}{v} - \frac{1}{- 20} \\ \Rightarrow v = 20 M_{1} = m_{1} \times m_{2} = \frac{v_{1}}{u_{1}} \times \frac{v_{2}}{u_{2}} = \frac{- 30}{- 15} \times \frac{20}{- 20} = 2 \times 1 = 2 (i n a i r) \end{array}$

2nd case :

For mirror

$\begin{array}{l} v = - 30 \\ F o r l e n s \\ \frac{1}{v} - \frac{1}{- 20} = (\frac{3 / 2}{7 / 6} - 1) (\frac{2}{R}) \\ \Rightarrow \frac{1}{v} + \frac{1}{20} = (\frac{2}{7}) (\frac{2}{10}) = \frac{4}{70} \\ (∵ \frac{1}{f_{a i r}} = (μ - 1) (\frac{2}{R}) \Rightarrow \frac{1}{10} = (1.5 - 1) (\frac{2}{R}) \Rightarrow R = 10 c m) \\ \frac{1}{v} = \frac{4}{70} - \frac{1}{20} \\ \Rightarrow v = 140 M_{2} = m_{1} m_{2} = (2) \times (\frac{140}{- 20}) = - 14 \\ H e n c e, |\frac{M_{2}}{M_{1}}| = \frac{14}{2} = 7 \end{array}$

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