February 21, 2023# [Solved] A silica optical fiber with a core diameter large enough to be considered by ray theory analysis has a core ...

## Question:

### A silica optical fiber with a core diameter large enough to be considered by ray theory analysis has a core refractive index of 1.50 and a cladding refractive index of 1.47.

### Determine: (a) the critical angle at the core–cladding interface; (b) the NA for the fiber; (c) the acceptance angle in air for the fiber.

## Solution/Given:

**Here are the given from the question.**

Core refractive index (n_{1}) = 1.50

Cladding refractive index (n_{2}) = 1.47

Critical angle at the core–cladding interface φ_{c} = ?

Numerical aperture (NA for the fiber) = ?

Acceptance angle in air for the fiber (θ_{a}) = ?

**(a) To find the critical angle φ _{c} from the core and cladding refractive index,**

φ_{c} = sin^{−1}(n_{2}/n_{1}) = sin^{−1}(1.47/1.5) = 78.5°

**(b) The NA can be found by using the equation,** NA = (n_{1}^{2} - n_{2}^{2})^{0.5}

= (_{1.5}^{2} - _{1.47}^{2})^{0.5}

= (2.25 − 2.16)^{0.5} = 0.3

**(c) The acceptance angle in air θ _{a}**

= θ_{a} = sin^{−1}(NA) = sin^{−1}(0.30)

= 17.4°