Question

An initially parallel cylindrical beam travels in a medium of refractive index $\mu \left( I \right)={{\mu }_{0}}+{{\mu }_{2}}I$, where ${{\mu }_{0}}$​ and ${{\mu }_{2}}$ are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. The speed of light in the medium is,
A. Minimum on the axis of the beam.
B. The same everywhere in the beam.
C. Directly proportional to the intensity I.
D. maximum on the axis of the beam.

Answer 1

Hint: To solve this question, we must know that the speed of light is inversely proportional to the refractive index of the medium in which the beam travels. Also, intensity of a light beam is directly proportional to the power and inversely proportional to the area of the beam. With the help of these relations, we will find the speed of the wave.
Formula Used:
$\begin{align}
  & \text{Intensity=}\dfrac{\text{Power}}{\text{Area}}=\dfrac{P}{4\pi {{r}^{2}}} \\
 & \mu =\dfrac{c}{v} \\
\end{align}$

Complete answer:
For calculating the speed of the light in the given medium, we will find the relation between the refractive index and the distance from the axis of the light, which in this case will be the radius of the cylindrical beam.
Now, it is given that the refractive index of the medium is,
$\mu \left( I \right)={{\mu }_{0}}+{{\mu }_{2}}I$ --- (1)
I in this equation is intensity. Intensity of a spherical wave is given as,
$\text{Intensity=}\dfrac{\text{Power}}{\text{Area}}=\dfrac{P}{4\pi {{r}^{2}}}$
But, here we have a spherical beam, so we take all other parameters in the expression except the radius as k in this equation.
$\Rightarrow \text{Intensity=I}=\dfrac{k}{{{r}^{2}}}$ -- (2)
Where, r is the radius of the cylindrical beam.
Now, we will substitute this expression for intensity in (1),
$\Rightarrow \mu \left( I \right)={{\mu }_{0}}+{{\mu }_{2}}\left( \dfrac{k}{{{r}^{2}}} \right)$ --- (3)
We know that refractive index of a medium is given by,
$\mu =\dfrac{c}{v}$
Where, c is the speed of light in vacuum and v is the speed of light in that particular medium.
From (3), we have the expression for the refractive index of the given medium.
$\begin{align}
  & \Rightarrow \mu \left( I \right)=\dfrac{c}{v} \\
 & \Rightarrow {{\mu }_{0}}+{{\mu }_{2}}\left( \dfrac{k}{{{r}^{2}}} \right)=\dfrac{c}{v} \\
 & \Rightarrow v=\dfrac{c}{{{\mu }_{0}}+{{\mu }_{2}}\left( \dfrac{k}{{{r}^{2}}} \right)} \\
 & \therefore v=\dfrac{c\cdot {{r}^{2}}}{{{\mu }_{0}}{{r}^{2}}+{{\mu }_{2}}k} \\
\end{align}$
From this equation for velocity of light in the medium, we can understand that by increasing the value of r, i.e. radius of the cylinder or the distance from the axis, velocity increases and by decreasing r, the velocity decreases.
So, it is clear that the speed of light increases when we move away from the axis and decreases when we approach the axis. Thus, speed of light is minimum on the axis of the beam.

Therefore, option A is correct.

Note:
While solving this type of problems, it is important to determine the correct relation between velocity of the light and the distance from the axis of the beam. Then, by analyzing that expression, we can find the dependence of a particular parameter on other parameters and reach the solution.