Question

A plano convex lens fits exterctly into a phone concave lens . Their plane surface are parallel to each other . If lenses are made of different materials of refractive indices ${\mu }_{1}and{\mu }_2$ and R is the radius of curvature of the curved surface of the lenses , then the focal length of combination is

• A. $\frac{2R}{{(\mu }_2-{\mu }_1)}$
• B. $\frac{R}{2{(\mu }_1+{\mu }_2)}$
• C. $\frac{R}{2{(\mu }_1-{\mu }_2)}$
• D. $\frac{R}{{(\mu }_1-{\mu }_2)}$

Answer 1

The correct option is D $\frac{R}{{(\mu }_1-{\mu }_2)}$

The combined focal length of the system of both lens is given as

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$ …… (1)

Now applying lens maker formula for Plano-concave lens

$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Hence,

$\frac{1}{f_1}=\left({\mu }_1-1\right)\left(\frac{1}{\infty }-\frac{1}{-R}\right)\Rightarrow \frac{1}{f_1}=\frac{\left({\mu }_1-1\right)}{R}$ …… (2)

Now for the Plano-convex lens,

Hence,

$\frac{1}{f_2}=\left({\mu }_2-1\right)\left(-\frac{1}{R}-\frac{1}{\infty }\right)\Rightarrow \frac{1}{f_2}=\frac{-\left({\mu }_2-1\right)}{R}$ …… (3)

Substitute the value of equation (2) and (3) in equation (1), hence

$\frac{1}{f}=\frac{\left({\mu }_1-1\right)}{R}+\frac{-\left({\mu }_2-1\right)}{R}\Rightarrow \frac{1}{f}=\frac{\left({\mu }_1-{\mu }_2\right)}{R}$

$\therefore f=\frac{R}{\left({\mu }_1-{\mu }_2\right)}$