JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 18)
A photograph of a landscape is captured by a drone camera at a height of 18 km . The size of the camera film is $2 \mathrm{~cm} \times 2 \mathrm{~cm}$ and the area of the landscape photographed is $400 \mathrm{~km}^2$. The focal length of the lens in the drone camera is :
2.8 cm
0.9 cm
2.5 cm
1.8 cm
Explanation
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$$\begin{aligned} &\mathrm{H}=18 \mathrm{~km}\\ &\text { Size of camera film }=2 \mathrm{~cm} \times 2 \mathrm{~cm}\\ &\begin{aligned} & \mathrm{A}_{\text {image }}=400 \mathrm{~km}^2 \\ & \mathrm{x}=20 \times 10^3 \mathrm{~m}=2 \times 10^4 \mathrm{~m} \\ & \mathrm{y}=2 \times 10^{-2} \mathrm{~m} \\ & \frac{\mathrm{x}}{\mathrm{y}}=10^6=\frac{18 \mathrm{Km}}{\mathrm{f}} \\ & \mathrm{f}=18 \times 10^{-3} \mathrm{~m}=18 \mathrm{~mm} \\ & \mathrm{f}=1.8 \mathrm{~cm} \end{aligned} \end{aligned}$$
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