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 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

The correct option is A $\frac{R}{{\mu }_1-{\mu }_2}$

Using the lens maker's formula, the focal length of the combination will be:
$\frac{1}{f}=({\mu }_1-1)(\frac{1}{\infty }+\frac{1}{R})+({\mu }_2-1)(\frac{1}{\infty }-\frac{1}{R})$
So, $\frac{1}{f}=\frac{{\mu }_1-{\mu }_2}{R}$
or, $f=$$\frac{R}{{\mu }_1-{\mu }_2}$