JEE Physics 2023 Question with Solution

Given:

• Apparent distance from one side, $d_1' = 12 \, \text{cm}$
• Apparent distance from the opposite side, $d_2' = 4 \, \text{cm}$
• Side of the ice cube, $t = 24 \, \text{cm}$

Find: The refractive index $\mu$ of the ice cube.

Using the refraction relation,

$$\mu = \frac{\text{real depth}}{\text{apparent depth}}$$

From the first side,

$$\mu = \frac{d_1}{d_1'} = \frac{d_1}{12}$$

So,

$$d_1 = 12\mu$$

From the opposite side,

$$\mu = \frac{d_2}{d_2'} = \frac{d_2}{4}$$

So,

$$d_2 = 4\mu$$

The sum of the real distances equals the side of the cube:

$$d_1 + d_2 = 24$$

Substituting,

$$12\mu + 4\mu = 24$$ $$16\mu = 24$$ $$\mu = \frac{24}{16} = \frac{3}{2}$$

Therefore, the refractive index of the ice cube is $\frac{3}{2}$. The correct option is B.

Given: The bubble is inside a cube of side $24 \, \text{cm}$ and appears at $12 \, \text{cm}$ from one side and $4 \, \text{cm}$ from the opposite side.

Find: $\mu$.

Let the real distances of the bubble from the two opposite faces be $d_1$ and $d_2$. Then,

$$d_1 + d_2 = 24$$

For viewing through a plane refracting surface,

$$d_1' = \frac{d_1}{\mu}, \qquad d_2' = \frac{d_2}{\mu}$$

Hence,

$$d_1 = \mu d_1' = 12\mu$$ $$d_2 = \mu d_2' = 4\mu$$

Adding these,

$$12\mu + 4\mu = 24$$ $$16\mu = 24$$ $$\mu = \frac{3}{2}$$

Thus, the refractive index is $\frac{3}{2}$, so the correct option is B.