If the radii of curvature of the faces of a double convex lens are 10 cm and 15 cm, respectively and its refractive index is 1.5, the focal length of the lens in air is

12 cm

The correct answer is Option (2) → 12 cm

Given:

$R_1 = +10\ \text{cm}, \quad R_2 = -15\ \text{cm}, \quad \mu = 1.5$

Lens maker’s formula:

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

Substitute values:

$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{10} - \frac{1}{-15}\right)$

$\frac{1}{f} = 0.5\left(\frac{1}{10} + \frac{1}{15}\right)$

$\frac{1}{f} = 0.5\left(\frac{5}{30}\right) = \frac{1}{12}$

$f = 12\ \text{cm}$