JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 8)
A plano convex lens of refractive index $$\mu $$1 and focal length ƒ1 is kept in contact with another plano concave lens of refractive index $$\mu $$2 and focal length ƒ2. If the radius of curvature of their spherical faces is R each and ƒ1 = 2ƒ2, then $$\mu $$1 and $$\mu $$2 are related as -
$$3{\mu _2} - 2{\mu _1}$$ = 1
$${\mu _1} + {\mu _2}$$ = 3
$$2{\mu _1} - {\mu _2}$$ = 1
$$2{\mu _2} - {\mu _1}$$ = 1
Explanation
$${1 \over {2{f_2}}} = {1 \over {{f_1}}} = \left( {{\mu _1} - 1} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right)$$
$${1 \over {{f_2}}} = \left( {{\mu _2} - 1} \right)\left( {{1 \over { - R}} - {1 \over \infty }} \right)$$
$${{\left( {{\mu _1} - 1} \right)} \over R} = {{\left( {{\mu _2} - 1} \right)} \over {2R}}$$
$$2{\mu _1} - {\mu _2} = 1$$
$${1 \over {{f_2}}} = \left( {{\mu _2} - 1} \right)\left( {{1 \over { - R}} - {1 \over \infty }} \right)$$
$${{\left( {{\mu _1} - 1} \right)} \over R} = {{\left( {{\mu _2} - 1} \right)} \over {2R}}$$
$$2{\mu _1} - {\mu _2} = 1$$
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