Question

An air bubble in a sphere having a 4 cm diameter appears 1cm from the surface nearest to the eye when loaded along the diameter. If ${}_a{\mu }_g$ = 1.5, the distance of the bubble from refracting the surface is?
A) 1.2cm
B) 3.2cm
C) 2.8cm
D) 1.6cm

Answer 1

Hint: Here we used to solve this question with the help ${\displaystyle \frac{{\mu }_2}{v}}-{\displaystyle \frac{{\mu }_1}{u}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}$ where u is the object distance from the pole of spherical and v is the distance from pole of a spherical surface and ${\mu }_2$,${\mu }_1$are the refractive index of air and glass. Substitute the value of v, R, the and other parameters to find the value of the distance of the bubble from refracting the surface that is u.

Formula used:
${\displaystyle \frac{{\mu }_2}{v}}-{\displaystyle \frac{{\mu }_1}{u}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}$
${}_a{\mu }_g={\displaystyle \frac{{\mu }_g}{{\mu }_a}}$

Complete Step by step solution:
First we have to write given values,
There is a given diameter of the sphere i.e 4cm so, its radius is 2cm.
And the air bubbles appear 1cm from the surface nearest to the eye when looked along the diameter.
We have ${}_a{\mu }_g$ = 1.5.
${}_a{\mu }_g={\displaystyle \frac{{\mu }_g}{{\mu }_a}}$
${\mu }_g$and ${\mu }_a$ are the refractive index of air and glass. So, the refractive index of air is 1.

Substitute these values in the formula we have
$\begin{array}{rl}& {}_a{\mu }_g={\displaystyle \frac{{\mu }_g}{{\mu }_a}}\\ & {\Rightarrow }_a{\mu }_g\times {\mu }_a={\mu }_g\\ & \Rightarrow 1.5\times 1={\mu }_g\\ & \therefore {\mu }_g=1.5\end{array}$

For calculating distance of the bubble index of air and glass.
${\displaystyle \frac{{\mu }_2}{v}}-{\displaystyle \frac{{\mu }_1}{u}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}$
where,
u is the object distance from the pole of the sphere.
 v is the distance from the pole of the spherical surface.
${\mu }_2$ ,${\mu }_1$are the refractive index of air and glass.
u is the distance of the bubble from the refracting surface.

We substituting given values we get
$\begin{array}{rl}& {\displaystyle \frac{{\mu }_2}{v}}-{\displaystyle \frac{{\mu }_1}{u}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}\\ & \Rightarrow {\displaystyle \frac{1}{-1}}-{\displaystyle \frac{1.5}{u}}={\displaystyle \frac{1-1.5}{-2}}\\ & \Rightarrow -1-{\displaystyle \frac{1.5}{u}}={\displaystyle \frac{1-1.5}{-2}}\\ & \Rightarrow -{\displaystyle \frac{1.5}{u}}={\displaystyle \frac{0.5}{2}}+1\\ & \Rightarrow -{\displaystyle \frac{1.5}{u}}={\displaystyle \frac{2.5}{2}}\\ & \Rightarrow u=-{\displaystyle \frac{1.5\times 2}{2.5}}\\ & \therefore u=-1.2cm\end{array}$
The distance of the bubble from the refracting surface is 1.2cm.

Hence, the correct option is A.

Note:
In this first, we have to write in the question and then mention formulas to find the distance of the bubble from the refracting surface. A student may forget to keep the values along with their signs in the formula, here the sign of R and v will be negative according to sign convention, and we got the value of u also negative.