JEE Main 2026 Jan 22 Shift 1, Physics Q23: Refraction at Plane and Spherical surfaces

JEE Main2026Jan 22, Shift 1Physics

A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface. The value of x is ____ cm.

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Solution

Refraction at a single spherical surface: $\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$ .

For a parallel beam, $u = \infty$ , so $\frac{μ _{1}}{u} = 0$ . Plug $μ_{1} = 1, μ_{2} = 1.5, R = + 50$ cm (convex surface):

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

So $v = 150$ cm (measured from the pole of the surface).

Distance from the centre of curvature (which is 50 cm beyond the pole, on the same side as the image): $x = v - R = 150 - 50 = 100$ cm.

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JEE Main 2026 Jan 22 Shift 1, Physics Q23: Refraction at Plane and Spherical surfaces

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