JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 15)
One plano-convex and one plano-concave lens
of same radius of curvature 'R' but of different
materials are joined side by side as shown in
the figure. If the refractive index of the material
of 1 is $$\mu $$1 and that of 2 is $$\mu $$2, then the focal
length of the combination is :
_10th_April_Morning_Slot_en_15_1.png)
$${2R \over { {{\mu _1} - {\mu _2}}}}$$
$${R \over {2 - \left( {{\mu _1} - {\mu _2}} \right)}}$$
$${R \over { {{\mu _1} - {\mu _2}}}}$$
$${R \over {2 \left( {{\mu _1} - {\mu _2}} \right)}}$$
Explanation
For 1st lens $${1 \over {{f_1}}} = \left( {{{{\mu _1} - 1} \over 1}} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right) = {{{\mu _1} - 1} \over R}$$
For 2nd lens $${1 \over {{f_2}}} = \left( {{{{\mu _2} - 1} \over 1}} \right)\left( {{1 \over { - R}}} \right) - 0 = {{{\mu _2} - 1} \over R}$$
$${1 \over {{f_{eq}}}} = {1 \over {{f_1}}} + {1 \over {{f_2}}}$$
$${1 \over {{f_{eq}}}} = {{{\mu _1} - 1} \over R} + {{ - \left( {{\mu _2} - 1} \right)} \over R}$$
$$ \Rightarrow {f_{eq}} = {R \over {{\mu _1} - {\mu _2}}}$$
For 2nd lens $${1 \over {{f_2}}} = \left( {{{{\mu _2} - 1} \over 1}} \right)\left( {{1 \over { - R}}} \right) - 0 = {{{\mu _2} - 1} \over R}$$
$${1 \over {{f_{eq}}}} = {1 \over {{f_1}}} + {1 \over {{f_2}}}$$
$${1 \over {{f_{eq}}}} = {{{\mu _1} - 1} \over R} + {{ - \left( {{\mu _2} - 1} \right)} \over R}$$
$$ \Rightarrow {f_{eq}} = {R \over {{\mu _1} - {\mu _2}}}$$
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