As shown in figure, a plano-concave lens is placed in such a way that it becomes completely fit with plano-convex lens. Their plane surfaces are parallel. If their refractive indices are 1.6 and 1.5 respectively and radius of curvature is R, then focal length of combination is .....

Options

A. `(R)/(6.2)`
B. `(R)/(3.1)`
C. `(R)/(0.2)`
D. `(R)/(0.1)`

Answer

A, D

Explanation

`f=(R)/(mu_1-mu_2)=(R)/(1.6-1.5)=(R)/(0.1)`
For first lens,` (1)/(f_1)=(n-1)((1)/(R_1)-(1)/(R_2))`
`R_1= infty,R_2=-R,n=mu_1`
`(1)/(f_1)=(mu_1-1)((1)/(infty)+(1)/(R))`
`therefore (1)/(f_1)=(mu-1)/(R)` … (1)
For second lens, `(1)/(f_2)=(n-1)((1)/(R_1)-(1)/(R_2))`
`R_1=-R,R_2=infty,n-mu_2`
`(1)/(f_2)=(mu_2-1)(-(1)/(R)-(1)/(infty))`
`infty (1)/(f_2)=(1-mu_2)/(R)` ... (2)
Focal length of combination, `1/f=(1)/(f_1)+(1)/(f_2)=(mu_1-1)/(R)+(1-mu_2)/(R)`
`therefore 1/f=(mu_1-mu_2)/(R)`
`therefore f=(R)/(mu_1-mu_2)=(R)/(1.6-1.5)=(R)/(0-1)`