A convex lens has 20 cm focal length in air. Its focal length in water would be

(Given: Refractive index of water = 1.33, refractive index for glass = 1.5)

+78.2 cm

The correct answer is Option (1) → +78.2 cm

Using lens maker’s formula:

$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

In air: $\frac{1}{f_{\text{air}}} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$\frac{1}{0.20} = 0.5\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 10$

In water: $\frac{1}{f_{\text{water}}} = \left(\frac{1.5}{1.33} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$\frac{1}{f_{\text{water}}} = \left(1.1278 - 1\right)\times 10 = 0.1278 \times 10 = 1.278$

$f_{\text{water}} = \frac{1}{1.278} \approx 0.782 \ \text{m} = 78.2 \ \text{cm}$

Answer: 78.2 cm