Question

Calculate the angle of acceptance of a given optical fiber, if the refractive indices of the core and cladding are 1.563 and 1.498 respectively.​

Answer 1

Answer:

Core index:

1.563

Cladding index:

1.498

Index of input medium:

1

Numerical aperture:

0.446

Acceptance angle:

26.5 °

Answer 2

The angle of acceptance is 26.49°

Step-by-step Explanation:

Given: Refractive index of the core $n_1$ = 1.563

Refractive index of the cladding $n_2$ = 1.498

To Find: The angle of acceptance

Solution:

• Finding the angle of acceptance

For the given optical fiber, the angle of acceptance $i_a$ is given by,

$i_a = sin^{-1} \Big( \frac{\sqrt{n_{1}^{2} - n_{2}^{2}}}{n_{0}} \Big)$

Here, $n_{0}$ is the refractive index of the outer medium; for air, $n_{0} = 1$

Therefore, $i_a = sin^{-1} {\sqrt{n_{1}^{2} - n_{2}^{2}}}$

Substituting the values of $n_1$ and $n_2$ in the above formula, we get,

$\Rightarrow i_a = sin^{-1} {\sqrt{(1.563)^{2} - (1.498)^{2}}}$

$\Rightarrow i_a = sin^{-1} {\sqrt{0.198965}} = 26.49^{o}$

Hence, the angle of acceptance of the given optical fiber is $i_a = 26.49^{o}$