Question

A plano-convex lens fits exactly into a plano-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 the combination is

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

Answer 1

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

From the lensmaker formula for plano convex lens,

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

For plano concave lens,

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

For the combination,

$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}=({\mu }_1-{\mu }_2)\left(\frac{1}{R}\right)$

This gives us $f_{eq}=\frac{R}{{\mu }_1-{\mu }_2}$