Answer to Question #117727 in Optics for Waqas

Question

b) In the step index fiber the relative refractive index difference is=4% and n2 = 1.5 Find a) n1 b)Critical propagation angle c)N.A d)Maximum angle of incidence for guidance

Answer 1

Find $n_1$ from relative refractive index difference $\Delta$:

$$\Delta=\frac{n_1^2-n_2^2}{2n_1^2},\\ n_1=1.56.$$

Calculate the numerical aperture:

$$NA=\sqrt{n_1^2-n_2^2}=0.428.$$

Calculate critical propagation angle:

$$\theta_c=\text{sin}^{-1}\frac{n_1}{n_2}=74.1^\circ.$$

Maximum angle of incidence (or acceptance angle):

$$\theta_{acc}=\text{sin}^{-1}\bigg(\frac{1}{n_0}\sqrt{n_1^2-n_2^2}\bigg)=25.4^\circ.$$

In this equation, $n_0$ is the refractive index of a medium around the fibre. It is close to unity for air.