Question

A step index fibre has core diameter 40$\mu$m with core refractive index of 1.3 and numerical aperture 0.5, then the refractive index of cladding is

• A. 1.2
• B. 1.1
• C. 0.9
• D. 0.8

Hint:

For total internal reflection to occur, the refractive index of the core ($n_1$) must always be greater than the refractive index of the cladding ($n_2$).
Here, $1.3 > 1.2$, confirming our result is physically plausible.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Given a step-index optical fiber with:
- Core refractive index, $n_1 = 1.3$
- Numerical Aperture, $\text{NA} = 0.5$
We need to calculate the refractive index of the cladding, $n_2$.

Step 2: Key Formula or Approach:
The Numerical Aperture ($\text{NA}$) of a step-index fiber is given by:
\[ \text{NA} = \sqrt{n_1^2 - n_2^2} \] We can rearrange this formula to solve for $n_2$:
\[ \text{NA}^2 = n_1^2 - n_2^2 \implies n_2^2 = n_1^2 - \text{NA}^2 \]

Step 3: Detailed Explanation:

• Identify the given parameters:
\[ n_1 = 1.3 \] \[ \text{NA} = 0.5 \]
• Substitute these values into the rearranged equation:
\[ n_2^2 = (1.3)^2 - (0.5)^2 \] \[ n_2^2 = 1.69 - 0.25 \] \[ n_2^2 = 1.44 \]
• Take the square root of both sides to find $n_2$:
\[ n_2 = \sqrt{1.44} = 1.2 \]

Step 4: Final Answer:
The refractive index of the cladding is 1.2.