Ray Optics

JEE Main level, numerical answer, medium

132 of 295

Question 132 of 295

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 cmx \mathrm{~cm} from the centre of the curvature of the spherical surface. The value of xx is ____\_\_\_\_ cm.

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Answer: 100

For refraction at a single spherical surface, we use the formula: μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}
Given data:
Refractive index of air μ1=1.0\mu_1 = 1.0
Refractive index of glass μ2=1.5\mu_2 = 1.5
Radius of curvature R=+50R = +50 cm (convex surface)
Object is at infinity for a parallel beam, so u=−∞u = -\infty
Substituting the values:
1.5v−1.0−∞=1.5−1.050\frac{1.5}{v} - \frac{1.0}{-\infty} = \frac{1.5 - 1.0}{50}
1.5v−0=0.550\frac{1.5}{v} - 0 = \frac{0.5}{50}
1.5v=1100\frac{1.5}{v} = \frac{1}{100}
v=1.5×100=150v = 1.5 \times 100 = 150 cm
The image is formed at a distance of 150150 cm from the pole (vertex) of the spherical surface.
The question asks for the distance xx from the centre of curvature.
Distance from centre of curvature x=∣v−R∣x = |v - R|
x=150−50=100x = 150 - 50 = 100 cm

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