JEE Physics 2023 Question with Solution

Given: The bulb is at real depth $8 \, \text{cm}$ in water of refractive index $\mu = \frac{4}{3}$. The plane mirror is at a height of $50 \, \text{cm}$ from the bottom of the tank.

Find: The distance of the image formed by the mirror from the bottom of the tank.

First, find the apparent depth of the bulb as seen from air:

$$\text{Apparent depth} = \frac{\text{Real depth}}{\mu}$$

So,

$$\text{Apparent depth} = \frac{8}{4/3} = 6 \, \text{cm}$$

Thus, the bulb appears $6 \, \text{cm}$ below the water surface.

Since the water surface is $8 \, \text{cm}$ above the bottom, the apparent position of the bulb is $8 - 6 = 2 \, \text{cm}$ above the bottom.

The mirror is $50 \, \text{cm}$ above the bottom, so the distance of this apparent object from the mirror is

$$50 - 2 = 48 \, \text{cm}$$

A plane mirror forms the image at the same distance behind the mirror. Therefore, the image is $48 \, \text{cm}$ above the mirror.

Hence, the distance of the image from the bottom of the tank is

$$50 + 48 = 98 \, \text{cm}$$

Therefore, the required distance is $98 \, \text{cm}$.

Given: Real depth of bulb $= 8 \, \text{cm}$, refractive index of water $= \frac{4}{3}$, mirror height from bottom $= 50 \, \text{cm}$.

Find: Position of the final image from the tank bottom.

For an observer in air, the bulb inside water appears raised. Its apparent depth below the water surface is

$$\frac{8}{4/3} = 6 \, \text{cm}$$

So the apparent object for the mirror lies at a height

$$8 - 6 = 2 \, \text{cm}$$

from the bottom.

Now treat this apparent point as the object for the plane mirror. Object distance from mirror:

$$50 - 2 = 48 \, \text{cm}$$

For a plane mirror, image distance equals object distance. Therefore, image lies $48 \, \text{cm}$ behind the mirror.

So image position from bottom is

$$50 + 48 = 98 \, \text{cm}$$

Therefore, the answer is $98$.