Q66 / 75 · JEE Main 2025 (3 Apr Shift 2) · Physics

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.

Correct answer: 4

Solution

Figure
$\begin{aligned} & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{200}-\frac{1}{-\mathrm{x}}=\frac{1.5-1}{50}\end{aligned}$
$\begin{aligned} & \frac{1}{x}=\frac{1}{100}-\frac{3}{400} \\ & x=400 \mathrm{~cm} \\ & x=4 m\end{aligned}$

Hence, the correct answer is 4.

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