Question

A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If the lenses are made of different material of refractive indices ${\mu }_1$ and ${\mu }_2$ and R is the radius of curvature of the curved surface of the lenses, then focal length of the combination is :

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

Answer 1

Answer: (A)

Lens maker formula: $\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$ .......(1)

For Plano-convex lens: $R_1=\infty$ $R_2=-R$

Using (1), $\frac{1}{f_1}=({\mu }_1-1)(\frac{1}{\infty }-\frac{1}{-R})$ $⟹\frac{1}{f_1}=\frac{{\mu }_1-1}{R}$

For Plano-concave lens: $R_1=-R$ $R_2=\infty$

Using (2), $\frac{1}{f_2}=({\mu }_2-1)(\frac{1}{-R}-\frac{1}{\infty })$ $⟹\frac{1}{f_2}=\frac{1-{\mu }_2}{R}$

Thus focal length of combination, $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{{\mu }_1-1}{R}+\frac{1-{\mu }_2}{R}$

$⟹$ $F=\frac{R}{{\mu }_1-{\mu }_2}$