JEE Physics 2023 Question with Solution

Given: refractive index of lens $\mu_{\text{glass}} = 1.5$, refractive index of water $\mu_{\text{water}} = \frac{4}{3}$, and focal length in air $f_{\text{air}} = 18 \, \text{cm}$.

Find: the change in focal length when the lens is immersed in water.

The focal length of a lens in water is given by:

$$\frac{1}{f_{\text{water}}} = \left( \frac{\mu_{\text{glass}}}{\mu_{\text{water}}} - 1 \right) \frac{2}{R}$$

For air:

$$\frac{1}{f_{\text{air}}} = (\mu_{\text{glass}} - 1) \frac{2}{R}$$

The ratio of focal lengths in water and air is:

$$\frac{f_{\text{water}}}{f_{\text{air}}} = \frac{\frac{1}{f_{\text{air}}}}{\frac{1}{f_{\text{water}}}} = \frac{\mu_{\text{glass}} - 1}{\frac{\mu_{\text{glass}}}{\mu_{\text{water}}} - 1}$$

Substitute the values:

$$\frac{f_{\text{water}}}{f_{\text{air}}} = \frac{1.5 - 1}{\frac{1.5}{\frac{4}{3}} - 1}$$

Thus:

$$f_{\text{water}} = 4 f_{\text{air}} = 4 \times 18 = 72 \, \text{cm}$$

The change in focal length is:

$$\Delta f = f_{\text{water}} - f_{\text{air}} = 72 - 18 = 54 \, \text{cm}$$

Therefore, the change in focal length is $54 \, \text{cm}$.

Given: $\mu_{\text{lens}} = 1.5$, $\mu_{\text{medium}} = \frac{4}{3}$, and $f_{\text{air}} = 18 \, \text{cm}$.

Find: change in focal length in water.

Use the effective refractive index:

$$\mu_{\text{eff}} = \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}}$$

So,

$$\mu_{\text{eff}} = \frac{1.5}{\frac{4}{3}} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$$

Hence,

$$\mu_{\text{eff}} - 1 = \frac{9}{8} - 1 = \frac{1}{8}$$

In air,

$$\mu_{\text{lens}} - 1 = 1.5 - 1 = 0.5 = \frac{1}{2}$$

Since focal length is inversely proportional to the refractive factor,

$$\frac{f_{\text{water}}}{f_{\text{air}}} = \frac{\frac{1}{2}}{\frac{1}{8}} = 4$$

Therefore,

$$f_{\text{water}} = 4 \times 18 = 72 \, \text{cm}$$

Now,

$$\Delta f = 72 - 18 = 54 \, \text{cm}$$

So the numerical answer is $54$.