A thin equiconvex lens of radius of curvature R made up of glass (μ = 1.5) is dipped in water (μ = 1.3). Its focal length in water is

3.25 R

The correct answer is Option (1) → 3.25 R

Given:

Equiconvex lens, radius of curvature $R$, refractive index of lens $\mu_{\text{glass}} = 1.5$, refractive index of water $\mu_{\text{water}} = 1.3$

Lensmaker's formula for lens in a medium:

$\frac{1}{f} = \frac{\mu_{\text{lens}} / \mu_{\text{medium}} - 1}{R_1} - \frac{\mu_{\text{lens}} / \mu_{\text{medium}} - 1}{R_2}$

For equiconvex lens: $R_1 = R$, $R_2 = -R$

$\frac{1}{f} = (\frac{\mu_{\text{glass}}}{\mu_{\text{water}}} - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\frac{1.5}{1.3} - 1) \cdot \frac{2}{R}$

$\frac{1.5}{1.3} = 1.1538 \Rightarrow 1.1538 - 1 = 0.1538$

Therefore:

$\frac{1}{f} = 0.1538 \cdot \frac{2}{R} = \frac{0.3076}{R}$

Focal length:

$f = \frac{R}{0.3076} \approx 3.25 R$

Answer: Focal length in water $f \approx 3.25 R$