JEE Physics 2025 Question with Solution

Given: Height of the drone camera is $18 \, \text{km}$, size of the camera film is $2 \, \text{cm} \times 2 \, \text{cm}$, and area of the landscape photographed is $400 \, \text{km}^2$.

Find: The focal length $f$ of the lens.

Using similar triangles, the areas are related by

$$\frac{A_{\text{image}}}{A_{\text{landscape}}} = \left( \frac{f}{h} \right)^2$$

Here,

$$A_{\text{image}} = 2 \, \text{cm} \times 2 \, \text{cm} = 4 \, \text{cm}^2$$

and

$$A_{\text{landscape}} = 400 \, \text{km}^2 = 400 \times (10^5)^2 = 400 \times 10^{10} \, \text{cm}^2$$

Also,

$$h = 18 \, \text{km} = 18 \times 10^5 \, \text{cm}$$

Substitute into the relation:

$$\frac{4}{400 \times 10^{10}} = \left( \frac{f}{18 \times 10^5} \right)^2$$

So,

$$\frac{4}{400 \times 10^{10}} = \frac{f^2}{(18 \times 10^5)^2}$$

Hence,

$$f^2 = \frac{4 \times (18 \times 10^5)^2}{400 \times 10^{10}}$$

On calculating,

$$f = 0.9 \, \text{cm}$$

Therefore, the focal length of the lens is $0.9 \, \text{cm}$. The correct option is A.

Given: The image is formed on a film of size $2 \, \text{cm} \times 2 \, \text{cm}$, the photographed landscape has area $400 \, \text{km}^2$, and the camera is at height $18 \, \text{km}$.

Find: The focal length $f$.

The solution uses the fact that the image formed on the film is a scaled-down version of the landscape. Therefore, the ratio of corresponding lengths is

$$\frac{\text{image side}}{\text{landscape side}} = \frac{f}{h}$$

and hence the ratio of areas is

$$\frac{A_{\text{image}}}{A_{\text{landscape}}} = \left( \frac{f}{h} \right)^2$$

Now evaluate each quantity from the given data:

$$A_{\text{image}} = 2 \times 2 = 4 \, \text{cm}^2$$

Convert the landscape area into square centimetres:

$$1 \, \text{km} = 10^5 \, \text{cm}$$

Therefore,

$$400 \, \text{km}^2 = 400 \times (10^5)^2 = 400 \times 10^{10} \, \text{cm}^2$$

Convert the height also into centimetres:

$$18 \, \text{km} = 18 \times 10^5 \, \text{cm}$$

Now substitute:

$$\frac{4}{400 \times 10^{10}} = \left( \frac{f}{18 \times 10^5} \right)^2$$

This gives

$$f = 0.9 \, \text{cm}$$

Therefore, the focal length of the lens is $0.9 \, \text{cm}$.