A diverging lens of glass (μ = 1.5) has both surfaces of the same radius of curvature (R). On immersion in a medium of refractive index (μ = 1.8), it will behave as

Converging lens with focal length f = 3R

The correct answer is Option (3) → Converging lens with focal length f = 3R

Given:

Refractive index of lens, μ_lens = 1.5

Refractive index of medium, μ_medium = 1.8

Both surfaces have radius R

Lens maker's formula in a medium:

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

For a diverging lens in air, R1 > 0, R2 < 0, both of magnitude R:

$\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{R} - \frac{-1}{R} = \frac{2}{R}$

Substitute values:

$\frac{1}{f} = \frac{1.5 - 1.8}{1.8} \cdot \frac{2}{R} = -\frac{0.3}{1.8} \cdot \frac{2}{R} = -\frac{1}{3R}$

The focal length is negative, so it behaves as a converging lens in the medium with magnitude:

$f = 3R$

Converging lens with focal length f = 3R