A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index 1.33), its focal length changes to

The refractive index of the lens (glass) is $$n_g = 1.5$$ and the refractive index of the surrounding medium is

• for air: $$n_{\text{air}} = 1$$
• for water: $$n_w = 1.33$$

For a thin lens placed in a medium of refractive index $$n_m$$ the lens-maker’s formula is

$$\frac{1}{f_m} = \left(\frac{n_g}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$-(1)$$

Here $$R_1, R_2$$ are the radii of curvature of the two spherical surfaces. The term $$\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ depends only on the shape of the lens and therefore remains the same when the surrounding medium is changed.

Step 1: Focal length in air
Putting $$n_m = n_{\text{air}} = 1$$ in $$(1)$$,

$$\frac{1}{f_{\text{air}}} = (n_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$-(2)$$

Given $$f_{\text{air}} = 24 \text{ cm}$$, so

$$\frac{1}{24} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\Rightarrow$$ $$\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{24 \times 0.5} = \frac{1}{12}$$ $$-(3)$$

Step 2: Focal length in water
Now place the lens in water, i.e. $$n_m = n_w = 1.33$$. Using $$(1)$$ again,

$$\frac{1}{f_w} = \left(\frac{n_g}{n_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$-(4)$$

Substitute $$n_g = 1.5$$, $$n_w = 1.33$$ and the value from $$(3)$$:

$$\frac{1}{f_w} = \left(\frac{1.5}{1.33} - 1\right)\left(\frac{1}{12}\right)$$

Compute the bracket:

$$\frac{1.5}{1.33} = \frac{150}{133} \approx 1.1278$$
$$\frac{1.5}{1.33} - 1 \approx 0.1278$$

Hence

$$\frac{1}{f_w} \approx 0.1278 \times \frac{1}{12} = \frac{0.1278}{12} \approx 0.01065$$

Therefore

$$f_w \approx \frac{1}{0.01065} \approx 93.9 \text{ cm}$$

The nearest value among the options is $$96 \text{ cm}$$.

Final answer: in water the focal length of the lens becomes approximately $$96 \text{ cm}$$ - Option B.