Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm 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 M1. When the set-up is kept in a medium of refractive index 7/6, the magnification becomes M2. The magnitude is

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Physics

Curved Surface Refraction

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Question

Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm 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 M₁. When the set-up is kept in a medium of refractive index 7/6, the magnification becomes M₂. The magnitude $| \frac{M_{2}}{M_{1}} |$ is

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For Mirror

u = –15; f = –10

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} + \frac{1}{15} = \frac{- 3 + 2}{5} = \frac{1}{- 30}$

v₁ = 20 cm

For Lens

u₂ = –20 cm; f = +10

$\frac{1}{v_{2}} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}$

v₂ = 20 cm

$m_{1} = - \frac{v_{1}}{u_{1}} = - \frac{30}{15} = - 2$

$m_{2} = \frac{v_{2}}{u_{2}} = \frac{20}{- 20} = - 1$

So, M₁ = m₁m₂ = +2

In medium focal length of lens will get changed

$\frac{1}{10} = (\frac{3}{2} - 1) {\frac{1}{R_{1}} - \frac{1}{R_{2}}} = \frac{1}{2} {\frac{1}{R_{1}} - \frac{1}{R_{2}}} \Rightarrow$ in air

$\frac{1}{f'} = (\frac{3 \times 6}{2 \times 7} - 1) {\frac{1}{R_{1}} - \frac{1}{R_{2}}} = \frac{4}{14} {\frac{1}{R_{1}} - \frac{1}{R_{2}}} \Rightarrow$ in medium

$\frac{f_{1}}{10} = \frac{1}{2} \times \frac{14}{4} = \frac{7}{4}$ so, $f_{1} = \frac{7}{4} \times 10 = \frac{35}{2}$ cm

So for lens u₂ = –20; $f_{2} = \frac{35}{2}$

$\frac{1}{v'_{2}} = \frac{2}{35} - \frac{1}{20} = \frac{8 - 7}{140} = \frac{1}{140}$

m₂' = $\frac{v'_{2}}{u_{2}} = \frac{140}{- 20}$ = –7

So, M₂ = –2 × –7 = 14

$So, | \frac{M_{2}}{M_{1}} | = \frac{14}{2} = 7$