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 :
![](https://imagex.cdn.examgoal.net/3rsa0w2w9jwysrb5x/f8da4f76-fd09-4020-92cb-233d15d084f5/13f81750-9020-11e9-86c1-73f7f991b766/file-3rsa0w2w9jwysrb5y.png)

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: