A biconvex lens has radii of curvature of 10 cm each. If the refractive index of the material of the lens is 1.5, what will be its focal length in air?

10 cm

The correct answer is Option (2) → 10 cm

Using lens maker’s formula:

$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Here, $n = 1.5$, $R_1 = +10 \text{ cm}$, $R_2 = -10 \text{ cm}$

$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{10} - \frac{-1}{10}\right)$

$\frac{1}{f} = 0.5\left(\frac{1}{10} + \frac{1}{10}\right)$

$\frac{1}{f} = 0.5 \times \frac{2}{10} = 0.1$

$f = 10 \ \text{cm}$

Focal length = 10 cm