Question 42: Ray Optics

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 $${7 \over 6}$$, the magnification becomes M2. The magnitude $$\left| {{{{M_2}} \over {{M_1}}}} \right|$$ is

Accepted Answer

Case I
Reflection from mirror
$${1 \over f} = {1 \over v} + {1 \over u} \Rightarrow {1 \over { - 10}} = {1 \over v} + {1 \over { - 15}}$$
$$ \Rightarrow v =  - 30$$
For lens $${1 \over f} = {1 \over v} - {1 \over u}$$
$${1 \over {10}} = {1 \over v} - {1 \over { - 20}}$$
$$v = 20$$
$$\left| {{M_1}} ight| = \left| {{{{v_1}} \over {{u_1}}}} ight|\left| {{{{v_2}} \over {{u_2}}}} ight|$$
$$ = \left( {{{30} \over {15}}} ight)\left( {{{20} \over {20}}} ight)$$
$$ = 2 	imes 1 = 2$$ (in air)
Case  II :
Now, consider the setup placed in a medium of refractive index $${\mu {'_1}}$$ = 7/6. The focal length of the mirror does not change. Thus, the distance of the image formed by the mirror and its magnification does not change. The focal length of the lens changes. The refractive index of the lens material is $$\mu$$
2
= 1.5. Apply lens maker's formula to get the new focal length of the lens
$${1 \over {f'}} = \left( {{{{\mu _2} - \mu {'_1}} \over {\mu {'_1}}}} ight)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} ight]$$
$$ = {{{\mu _2} - \mu {'_1}} \over {\mu {'_1}}}{{{\mu _1}} \over {{\mu _2} - {\mu _1}}}{{{\mu _2} - {\mu _1}} \over {{\mu _1}}}\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} ight]$$
$$ = \left( {{{{\mu _2} - \mu {'_1}} \over {{\mu _2} - {\mu _1}}}} ight)\left( {{{{\mu _1}} \over {\mu {'_1}}}} ight){1 \over f}$$
$$ = \left( {{{1.5 - 7/6} \over {1.5 - 1}}} ight)\left( {{1 \over {7/6}}} ight){1 \over {10}} = {2 \over {35}}$$. ..... (1)
Again using lens formula, $${1 \over v} - {1 \over u} = {1 \over {{f_l}'}}$$
$${1 \over v} - {1 \over { - 20}} = {2 \over {35}} \Rightarrow {1 \over v} = {2 \over {35}} - {1 \over {20}} = {1 \over {140}}$$
$$	herefore$$ v = 140 cm
Magnification, $${m_2}' = {v \over u} = {{140} \over { - 20}} =  - 7$$
Magnification produced by the combination,
$${M_2} = {m_1} 	imes {m_2}' = ( - 2) 	imes ( - 7) = 14$$
$$	herefore$$ $$\left| {{{{M_2}} \over {{M_1}}}} ight| = {{14} \over 2} = 7$$

Ray Optics

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