A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)

The focal length of a thin lens in a surrounding medium is obtained from the Lens-Maker’s formula:

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

For a given lens the radii $$R_1$$ and $$R_2$$ do not change; only the refractive-index term varies when the medium is changed.

Case 1: Lens in air
Here $$\mu_{\text{medium}} = 1$$, $$\mu_{\text{lens}} = 1.6$$ and the focal length is given as $$f_1 = 12\ \text{cm}$$.

Substituting in $$(1)$$:

$$\frac{1}{12} = \left(\frac{1.6}{1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\Rightarrow \frac{1}{12} = (1.6 - 1)K$$ where $$K = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$.

Hence $$K = \frac{1}{12 \times 0.6} = \frac{1}{7.2}\quad -(2)$$

Case 2: Lens in water
Now $$\mu_{\text{medium}} = 1.28$$, while $$\mu_{\text{lens}} = 1.6$$ remains the same. Let the new focal length be $$f_2$$.

Using $$(1)$$ again:

$$\frac{1}{f_2} = \left(\frac{1.6}{1.28} - 1\right)K$$ First evaluate the refractive-index factor: $$\frac{1.6}{1.28} = 1.25 \; \Rightarrow \; 1.25 - 1 = 0.25$$.

Therefore $$\frac{1}{f_2} = 0.25 \times K$$ Insert $$K$$ from $$(2)$$: $$\frac{1}{f_2} = 0.25 \times \frac{1}{7.2} = \frac{0.25}{7.2}$$ $$\frac{1}{f_2} = 0.034722\dots$$ Hence $$f_2 = \frac{1}{0.034722\dots} = 28.8\ \text{cm}$$.

Convert to millimetres: $$28.8\ \text{cm} = 28.8 \times 10 = 288\ \text{mm}$$.

Thus, the focal length of the lens in water is $$288\ \text{mm}$$.

Option B is correct.