JEE MAIN - Physics (2025 - 3rd April Evening Shift - No. 22)
Light from a point source in air falls on a spherical glass surface (refractive index, $\mu=1.5$ and radius of curvature $=50 \mathrm{~cm}$ ). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ___________m.
Answer
4
Explanation
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$$\begin{aligned} & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{200}-\frac{1}{-\mathrm{x}}=\frac{1.5-1}{50} \\ & \frac{1}{\mathrm{x}}=\frac{1}{100}-\frac{3}{400} \\ & \mathrm{x}=400 \mathrm{~cm} \\ & \mathrm{x}=4 \mathrm{~m} \end{aligned}$$
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