Physics - Geometrical Optics Question with Solution | TestHub

For a plano-convex lens, the focal length $f$ is given by the lens maker's formula:

$\frac{1}{f}=(\mu -1)\left(\frac{1}{R}-\frac{1}{\infty }\right)=\frac{\mu -1}{R}$

The power of a lens is $P=\frac{1}{f}$.

For the first lens, $P_1=\frac{\mu -1}{R_1}$

For the second lens, $P_2=\frac{\mu -1}{R_2}$

When two thin lenses are kept coaxially together, the equivalent power is the sum of their individual powers:

$P=P_1+P_2=\frac{\mu -1}{R_1}+\frac{\mu -1}{R_2}$

$P=(\mu -1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=(\mu -1)\left(\frac{R_1+R_2}{R_1{R}_2}\right)$

Since both lenses are made of the same material, $\mu$ is constant. Thus, the equivalent power is proportional to $\frac{R_1+R_2}{R_1{R}_2}$.

Answer: $\frac{{\mathrm{R}}_1+{\mathrm{R}}_2}{{\mathrm{R}}_1{\mathrm{R}}_2}$