Question 158: Ray Optics

Question

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 :

Accepted Answer

For 1
st
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 2
nd
lens $${1 \over {{f_2}}} = \left( {{{{\mu _2} - 1} \over 1}} \right)\left( {{1 \over { - R}}} \right) - 0 = {{{\mu _2} - 1} \over R}$$
$${1 \over {{f_{eq}}}} = {1 \over {{f_1}}} + {1 \over {{f_2}}}$$
$${1 \over {{f_{eq}}}} = {{{\mu _1} - 1} \over R} + {{ - \left( {{\mu _2} - 1} \right)} \over R}$$
$$ \Rightarrow {f_{eq}} = {R \over {{\mu _1} - {\mu _2}}}$$

Answer

$${2R \over {  {{\mu _1} - {\mu _2}}}}$$

Answer

$${R \over {2 - \left( {{\mu _1} - {\mu _2}} \right)}}$$

Answer

$${R \over {  {{\mu _1} - {\mu _2}}}}$$

Answer

$${R \over {2 \left( {{\mu _1} - {\mu _2}} \right)}}$$

Ray Optics

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 :

Choose the correct answer: