Question

In a glass sphere, there is a small bubble $22 \times {10^{ - 2}}m$ from its centre. If the bubble is viewed along a diameter of the sphere, from the side on which it lies, how far from the surface will it appear? The radius of glass sphere is $25 \times {10^{ - 2}}m$ and refractive index of glass is 1.5:
(A) $22.5 \times {10^{ - 2}}m$
(B) $23.2 \times {10^{ - 2}}m$
(C) $26.5 \times {10^{ - 2}}m$
(D) $20.2 \times {10^{ - 2}}m$

Answer 1

Hint : A glass sphere acts as a spherical surface, and the light ray incident behaves exactly like one incident on a spherical surface. The image distance is found with respect to the distance between the periphery of the bubble and that of the sphere.

Formula Used: The formulae used in the solution are given here.
For a spherical surface of radius $R$, $u$ is the object distance from a pole of the spherical surface, $v$ is the image distance from a pole of the spherical surface.
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$ where ${\mu _1}$ is the refractive index of a medium from which rays are incident and ${\mu _2}$ is the refractive index of another medium.

Complete step by step answer
For a spherical surface of radius $R$, $u$ is the object distance from a pole of the spherical surface, $v$ is the image distance from a pole of the spherical surface.
Let, ${\mu _1}$ is the refractive index of a medium from which rays are incident and ${\mu _2}$ is the refractive index of another medium.
Thus, we know that, $\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$.
Given that, in a glass sphere, there is a small bubble $22 \times {10^{ - 2}}m$ from its centre. The radius of the glass sphere is $25 \times {10^{ - 2}}m$ and the refractive index of glass is 1.5.
Let, ${\mu _1}$ be the refractive index of glass and ${\mu _2}$ be the refractive index of air, $R$ is the radius of the glass sphere.
Thus, $v$ is the distance from the surface the bubble appears, if viewed along a diameter of the sphere, and $u$ is the distance between the periphery of the bubble and that of the sphere.
Given that, the refractive index of glass ${\mu _1}$ is 1.5 and the refractive index of air ${\mu _2}$ is 1. The radius of the glass sphere $R$ is $25 \times {10^{ - 2}}m$.
Assigning the values in the equation, $\dfrac{1}{v} - \dfrac{{1.5}}{{3 \times {{10}^{ - 2}}}} = \dfrac{{1.5 - 1}}{{25 \times {{10}^{ - 2}}}}$
$ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 0.5}}{{25 \times {{10}^{ - 2}}}} + \dfrac{{1.5}}{{3 \times {{10}^{ - 2}}}}$
Simplifying the equation,
$\dfrac{1}{v} = \dfrac{{ - 1.5 + 37.5}}{{75 \times {{10}^{ - 2}}}}$
$\therefore v = \dfrac{{75 \times {{10}^{ - 2}}}}{{36}}m = 22.5 \times {10^{ - 2}}m$.
If the bubble is viewed along a diameter of the sphere, from the side on which it lies, it appears $22.5 \times {10^{ - 2}}m$ from the surface.
Hence the correct answer is Option A.

Note
The light ray passes from the glass sphere to the air bubble contained inside. Thus ${\mu _1}$ be the refractive index of glass and ${\mu _2}$ be the refractive index of air and not the other way round.