JEE Physics 2025 Question with Solution

Given: The first liquid has refractive index $\mu_1 = 1.2$ and height $h_1 = 60 \, \text{cm}$. The second liquid has refractive index $\mu_2 = 1.6$ and height $h_2 = H$. The total apparent shift of the bottom is $40 \, \text{cm}$.

Find: The value of $H$.

For a liquid layer of height $h$ and refractive index $\mu$, the apparent shift is:

$$\text{Shift} = h\left(1 - \frac{1}{\mu}\right)$$

For the first liquid:

$$\text{Shift}_1 = 60\left(1 - \frac{1}{1.2}\right)$$ $$= 60\left(1 - \frac{5}{6}\right)$$ $$= 60 \times \frac{1}{6} = 10 \, \text{cm}$$

For the second liquid:

$$\text{Shift}_2 = H\left(1 - \frac{1}{1.6}\right)$$ $$= H\left(1 - \frac{5}{8}\right)$$ $$= H \times \frac{3}{8} = \frac{3H}{8} \, \text{cm}$$

The total apparent shift is given as $40 \, \text{cm}$, so:

$$\text{Shift}_1 + \text{Shift}_2 = 40$$ $$10 + \frac{3H}{8} = 40$$ $$\frac{3H}{8} = 30$$ $$3H = 240$$ $$H = 80 \, \text{cm}$$

Verification:

$$10 + \frac{3 \times 80}{8} = 10 + 30 = 40 \, \text{cm}$$

This matches the given apparent shift.

Therefore, the value of $H$ is $80$.

Given: The upper liquid has height $H$ and refractive index $1.6$. The lower liquid has height $60 \, \text{cm}$ and refractive index $1.2$. The apparent shift is $40 \, \text{cm}$.

Find: The value of $H$.

The apparent depth of the bottom when viewed from above is the sum of apparent depths of both layers:

$$y = \frac{H}{1.6} + \frac{60}{1.2}$$

The real depth is:

$$H + 60$$

So the apparent shift is:

$$(H + 60) - y = 40$$

Substitute $y$:

$$H + 60 - \left(\frac{H}{1.6} + \frac{60}{1.2}\right) = 40$$ $$H + 60 - \frac{H}{1.6} - \frac{60}{1.2} = 40$$

Since

$$\frac{60}{1.2} = 50$$

we get:

$$H + 60 - \frac{H}{1.6} - 50 = 40$$ $$H - \frac{H}{1.6} = 30$$ $$H\left(1 - \frac{1}{1.6}\right) = 30$$ $$H\left(1 - \frac{5}{8}\right) = 30$$ $$\frac{3H}{8} = 30$$ $$H = 80 \, \text{cm}$$

Therefore, the value of $H$ is $80$.