JEE Physics 2023 Question with Solution

Given: The object is initially at $12 \, \text{cm}$ in front of a plane mirror, and the mirror is moved $4 \, \text{cm}$ towards the stationary object.

Find: The shift in the position of the image.

For a plane mirror, the image is formed at the same distance behind the mirror as the object is in front of it.

Initially, the object is placed $12 \, \text{cm}$ away from the mirror. Therefore, the distance between the object and the image is:

$$\text{Distance} = 2 \times 12 = 24 \, \text{cm}$$

The mirror is shifted towards the object by $4 \, \text{cm}$, so the new distance between the object and the mirror becomes:

$$12 - 4 = 8 \, \text{cm}$$

Therefore, the distance between the object and the image is:

$$\text{Distance} = 2 \times 8 = 16 \, \text{cm}$$

The shift in the image position is the difference between the initial and final distances between the object and the image:

$$\text{Shift} = 24 - 16 = 8 \, \text{cm}$$

Therefore, the image shifts by $8 \, \text{cm}$ towards the mirror. The correct option is D.

Using the diagram, the initial object-image separation is $24 \, \text{cm}$ and the final object-image separation is $16 \, \text{cm}$. Hence,

$$I_1 I_2 = 24 \, \text{cm} - 16 \, \text{cm} = 8 \, \text{cm}$$

So the image position shifts by $8 \, \text{cm}$ towards the mirror.