A parallel beam of light travelling in air is incident on a convex spherical glass surface
Q. A parallel beam of light travelling in air (refractive index $1.0$) is incident on a convex spherical glass surface of radius of curvature $50\ \text{cm}$. 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.
Correct Answer: 100

Explanation

This question is based on refraction at a spherical surface, a very important topic for JEE Main, JEE Advanced and IIT JEE.

Refraction formula at a spherical surface is:

$$ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} $$

Given values:

$$ n_1 = 1.0,\quad n_2 = 1.5 $$

Parallel beam implies:

$$ u = \infty \Rightarrow \frac{n_1}{u} = 0 $$

Radius of curvature:

$$ R = +50\ \text{cm} $$

Substitute values:

$$ \frac{1.5}{v} = \frac{1.5 - 1}{50} $$

$$ \frac{1.5}{v} = \frac{0.5}{50} $$

$$ \frac{1.5}{v} = \frac{1}{100} $$

$$ v = 150\ \text{cm} $$

This distance is measured from the pole of the spherical surface. The centre of curvature is $50$ cm from the pole.

Therefore, distance of the image point from the centre of curvature:

$$ x = v - R = 150 - 50 = 100\ \text{cm} $$

Hence,

$x = 100\ \text{cm}$

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