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Question: In a glass sphere, there is a small bubble \(22 \times {10^{ - 2}}m\) from its centre. If the bubble...

In a glass sphere, there is a small bubble 22×10−2m22 \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 25×10−2m25 \times {10^{ - 2}}m and refractive index of glass is 1.5:
(A) 22.5×10−2m22.5 \times {10^{ - 2}}m
(B) 23.2×10−2m23.2 \times {10^{ - 2}}m
(C) 26.5×10−2m26.5 \times {10^{ - 2}}m
(D) 20.2×10−2m20.2 \times {10^{ - 2}}m

Explanation

Solution

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 RR, uu is the object distance from a pole of the spherical surface, vv is the image distance from a pole of the spherical surface.
μ2v−μ1u=μ2−μ1R\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R} where μ1{\mu _1} is the refractive index of a medium from which rays are incident and μ2{\mu _2} is the refractive index of another medium.

Complete step by step answer
For a spherical surface of radius RR, uu is the object distance from a pole of the spherical surface, vv is the image distance from a pole of the spherical surface.
Let, μ1{\mu _1} is the refractive index of a medium from which rays are incident and μ2{\mu _2} is the refractive index of another medium.
Thus, we know that, μ2v−μ1u=μ2−μ1R\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 22×10−2m22 \times {10^{ - 2}}m from its centre. The radius of the glass sphere is 25×10−2m25 \times {10^{ - 2}}m and the refractive index of glass is 1.5.
Let, μ1{\mu _1} be the refractive index of glass and μ2{\mu _2} be the refractive index of air, RR is the radius of the glass sphere.
Thus, vv is the distance from the surface the bubble appears, if viewed along a diameter of the sphere, and uu is the distance between the periphery of the bubble and that of the sphere.
Given that, the refractive index of glass μ1{\mu _1} is 1.5 and the refractive index of air μ2{\mu _2} is 1. The radius of the glass sphere RR is 25×10−2m25 \times {10^{ - 2}}m.
Assigning the values in the equation, 1v−1.53×10−2=1.5−125×10−2\dfrac{1}{v} - \dfrac{{1.5}}{{3 \times {{10}^{ - 2}}}} = \dfrac{{1.5 - 1}}{{25 \times {{10}^{ - 2}}}}
⇒1v=−0.525×10−2+1.53×10−2\Rightarrow \dfrac{1}{v} = \dfrac{{ - 0.5}}{{25 \times {{10}^{ - 2}}}} + \dfrac{{1.5}}{{3 \times {{10}^{ - 2}}}}
Simplifying the equation,
1v=−1.5+37.575×10−2\dfrac{1}{v} = \dfrac{{ - 1.5 + 37.5}}{{75 \times {{10}^{ - 2}}}}
∴v=75×10−236m=22.5×10−2m\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 22.5×10−2m22.5 \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 μ1{\mu _1} be the refractive index of glass and μ2{\mu _2} be the refractive index of air and not the other way round.