A parallel beam of light is incident on a transparent sphere of refractive index $n$ . If the beam finally gets focussed at a point situated at a distance $= 2 \times$ (radius of sphere) from the centre of the sphere, then find $n$ .

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Solution

Given that,

Refractive index of air $= n_{1}$

Refractive index of medium $= n$

Distance of image $= 2 r$

Distance of object $= \infty$

Sphere is considered as a solid transparent with two spherical refractive surface of radius f curvature

$r_{1} = r_{2} = r$

The image is formed at $2 r$ distance from the pole of the 1^st surface and so no refraction takes place in the second surface.

Effectively refraction take place in single surface and image is formed inside the refractive medium.

$\frac{n_{1}}{u} + \frac{n}{v} = \frac{n - n_{1}}{r}$

$\frac{1}{\infty} + \frac{n}{2 r} = \frac{n - 1}{r}$

$n = 2 n - 2$

$n = 2$

Hence, the value of $n$ is $2$

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A parallel beam of light is incident on a transparent sphere of refractive index n. If the beam finally gets focussed at a point situated at a distance =2×(radius of sphere) from the centre of the sphere, then find n.

A parallel beam of light is incident on a transparent sphere of refractive index $n$ . If the beam finally gets focussed at a point situated at a distance $= 2 \times$ (radius of sphere) from the centre of the sphere, then find $n$ .