Question

Optical engineers need to know the cone of acceptance of an optical fiber. This is the maximum angle that an entering light ray can make with the axis of the fiber if it is to be guided down the fiber. What is the cone of acceptance of an optical fiber for which the index of refraction of the core is 1.55 while that of the cladding is 1.45? You can model the fiber as a cylinder with a flat entrance face.

Short answer

The cone of acceptance of this optical fiber is 33.2^o

Step-by-step solution

Step 1. Given information is :Refractive index of air n1 = 1.00Refractive index of core n2 = 1.55Refractive index of cladding n3 = 1.45

We need to find the cone of acceptance of this optical fiber

Step 2. Finding critical angle and using Snell's Law

The critical angle for total internal reflection is expressed as:

$$\begin{gathered} {\theta }_c={\sin }^{-1}\left(\frac{n_3}{n_2}\right) \\ {\theta }_c={\sin }^{-1}\left(\frac{1.45}{1.55}\right)=69.3° \end{gathered}$$

Angle of incidence of Light ${\theta }_2$can be expressed as,

$$\begin{gathered} {\theta }_2=90°-69.3° \\ {\theta }_2=20.7° \end{gathered}$$

Using Snell's Law for finding cone of acceptance ${\theta }_{max}$,

$$\begin{gathered} n_1\sin {\theta }_{max}=n_2\sin {\theta }_2 \\ \sin {\theta }_{max}=\frac{n_2}{n_1}\sin {\theta }_2 \\ {\theta }_{max}={\sin }^{-1}\left(\frac{n_2}{n_1}\sin {\theta }_2\right) \\ {\theta }_{max}={\sin }^{-1}\left(\frac{1.55}{1.00}\sin (20.7)°\right) \\ {\theta }_{max}=33.2° \end{gathered}$$