Question

A Plano convex lens fits exactly into a Plano concave lens. Their plane surfaces are parallel to each other. The lenses are made of different materials of refractive indices $\left({\mu }_1\right)$ of Plano convex lens and $\left({\mu }_2\right)$ of Plano concave lens. $R$ is the radius of curvature of the curved surface of the lenses. Then the focal length of the combination is -

• A. $\frac{R}{({\mu }_1+{\mu }_2)}$
• B. $\frac{R}{2({\mu }_2-{\mu }_1)}$
• 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 effective focal length of the system is,

$\frac{1}{f_{eff}}=\frac{1}{f_1}+\frac{1}{f_2}$

Using, lens maker's formula,

$\frac{1}{f_{eff}}=({\mu }_1-1)(\frac{1}{\infty }-\frac{1}{-R})+({\mu }_2-1)(\frac{1}{-R}-\frac{1}{\infty })$

$\Rightarrow f_{eff}=\frac{R}{{\mu }_1-{\mu }_2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.