Question

Calculate maximum acceptance angle and propogation angle for an optical fiber that has critical angle of 78.5 degree and (mu)clad of 1.52

Answer 1

Given:

An optical fiber that has critical angle of 78.5 degree and (mu)clad of 1.52

To Find:

Calculate maximum acceptance angle and propagation angle?

Solution:

given, $\mu _{2}=1.52$

critical angle , $\theta_{c}=78.5 \ degree$

since, we know that-

$sin\theta_{c}=\dfrac{\mu_{2}}{\mu_{1}}\\\\sin(78.5)=\dfrac{1.52}{\mu_{1}}\\\\0.98=\dfrac{1.52}{\mu_{1}}\\\\\mu_{1}=1.56$

Now, numerical aperture, NA is;

$NA=(\mu_{1}^2-\mu_{2}^2)^{1/2}\\\\NA=(2.4336-2.3104)^{1/2}\\\\NA=0.3509$

so,

$sin\theta_{a}=NA\\\\\theta_{a}=sin^{-1}(0.3509)\\\\\theta_{a}=20.54 \ degree$

thus, acceptance angle is 20.54 degree.