JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Geometrical Optics

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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

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}}

Answer: 7

Solution

For the mirror (focal length unchanged): using

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

with

f = - 10

cm,

u = - 15

cm gives

v = - 30

cm. Magnification

m_{1} = - \frac{v}{u} = - 2

. This image acts as object for the lens at distance

50 - 30 = 20

cm. For the lens in air with

f = 10

cm: magnification

m_{2} = \frac{f}{f - u} = \frac{10}{10 - 20} = - 1

. Total magnification

M_{1} = m_{1} \times m_{2} = 2

.

In medium with

μ_{0} = 7/6

: lens focal length changes to

f^{'} = f \frac{μ - 1}{μ / μ _{0} - 1} = 10 \times \frac{0.5}{1.5 \times 6/7 - 1} = 10 \times \frac{0.5}{2/7} = \frac{70}{4}

cm. Mirror magnification unchanged:

m_{1} = - 2

. Lens magnification:

m_{2}^{'} = \frac{70/4}{70/4 - 20} = \frac{70/4}{- 10/4} = - 7

. Total

M_{2} = 14

. Therefore

\frac{M _{2}}{M _{1}} = 7

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer

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JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Geometrical Optics

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