JEE Main 2025 Apr 3 Shift 2, Physics Q21: Refraction at Plane and Spherical surfaces

JEE Main2025Apr 3, Shift 2Physics

Light from a point source in air falls on a spherical glass surface (refractive index, $μ = 1.5$ and radius of curvature $= 50$ 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.

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Solution

Apply the single-surface refraction formula with the standard sign convention (light travels left to right, distances measured from the pole; object on the left has negative coordinate).

$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$ .

Here $μ_{1} = 1$ (air), $μ_{2} = 1.5$ (glass), $R = + 50$ cm (centre on the same side as the outgoing light, i.e. inside the glass), and $v = + 200$ cm:

$\frac{1.5}{200} - \frac{1}{u} = \frac{0.5}{50} = \frac{1}{100}$ .

Solve for $1/ u$ : $\frac{1}{u} = \frac{1.5}{200} - \frac{1}{100} = \frac{3 - 2}{400} = \frac{1}{400}$ .

So $u = - 400$ cm (the negative sign reflects sign convention; the object is 400 cm in front of the surface).

Magnitude $= 400$ cm $= 4$ m.

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JEE Main 2025 Apr 3 Shift 2, Physics Q21: Refraction at Plane and Spherical surfaces

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