Question

A step index fiber has a relative refractive index of 0.88%. What is the critical angle at the core cladding interface?
A. ${{60}^{o}}$
B. ${{75}^{o}}$
C. ${{45}^{o}}$
D. None of these

Answer 1

Hint: We are given the relative refractive index of a step index fiber. We know the equation to find the relative refractive index. Using this we can find the ratio of the refractive index. Substituting this in the equation to find the critical angle we get the solution.

Formula used: Relative refractive index,
$\dfrac{{{n}_{1}}-{{n}_{2}}}{{{n}_{1}}}$
Critical angle,
\[\theta ={{\sin }^{-1}}\left( \mu \right)\]

Complete step by step answer:
In the question we have a step index fiber.
We are given the relative refractive index of the fiber,
Relative refractive index = 0.88%
We know the formula for relative refractive index,
$\dfrac{{{n}_{1}}-{{n}_{2}}}{{{n}_{1}}}$ , were ‘${{n}_{1}}$’ is the refractive index of the first medium and ‘${{n}_{2}}$’ is the refractive index of the second medium.
Therefore, in this case we have relative refractive index,
$\dfrac{{{n}_{1}}-{{n}_{2}}}{{{n}_{1}}}=\dfrac{0.88}{100}$
Simplifying this we get,
$\begin{align}
  & \Rightarrow 100\left( {{n}_{1}}-{{n}_{2}} \right)=0.88{{n}_{1}} \\
 & \Rightarrow 100{{n}_{1}}-100{{n}_{2}}=0.88{{n}_{1}} \\
 & \Rightarrow 100{{n}_{1}}-0.88{{n}_{1}}=100{{n}_{2}} \\
 & \Rightarrow \left( 100-0.88 \right){{n}_{1}}=100{{n}_{2}} \\
\end{align}$
$\Rightarrow \dfrac{{{n}_{2}}}{{{n}_{1}}}=\dfrac{\left( 100-0.88 \right)}{100}$
$\Rightarrow \dfrac{{{n}_{2}}}{{{n}_{1}}}=0.9912$
Now we have the ratio of ${{n}_{2}}$ and ${{n}_{1}}$ as 0.9912
In the question we are asked to find the critical angle at the core cladding interface.
We know the equation to find the critical angle,
\[\theta ={{\sin }^{-1}}\left( \mu \right)\], were ‘$\theta $’ is the critical angle and ‘$\mu $’ is the ratio of the refractive indices, i.e.
$\mu =\dfrac{{{n}_{2}}}{{{n}_{1}}}$
We have found the value of ratio of ${{n}_{2}}$ and ${{n}_{1}}$ earlier to be 0.9912, i.e.
$\dfrac{{{n}_{2}}}{{{n}_{1}}}=0.9912$
We can substitute this value in the equation of critical angle.
Therefore we get the critical angle as,
$\begin{align}
  & \theta ={{\sin }^{-1}}\left( 0.9912 \right) \\
 & \theta ={{82.39}^{o}} \\
\end{align}$
Therefore we got the critical angle at the core cladding interface as ${{82.39}^{o}}$
This value is not in the given options.

So, the correct answer is “Option D”.

Note: Consider a situation where a ray passes from one medium to another. In this situation at a particular angle of incidence the angle of refraction becomes right angled, this angle of incidence that makes the angle of refraction ${{90}^{0}}$ is the critical angle.
In case of a step index fiber the refractive index of the cladding is smaller than the refractive index of the core.