Question

A plano convex lens fits exactly into a plano concave lens. Their plane surfaces 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

Answer: (D)

$\frac{R}{(\mu_1 - \mu_2)}$

Answer 2

Equivalent focal length is given by $\frac{1}{f_{eq}} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\frac{1}{f_{eq}} = \left(\mu_{1}- 1\right)\left(\frac{1}{\infty} - \frac{1}{-R}\right)+ \left(\mu_{2} -1\right)\left(\frac{1}{-R} - \frac{1}{\infty}\right)$
$\Rightarrow f_{eq} = \frac{R}{\mu_{1} - \mu_{2}}$