Answer to Question #193445 in Electrical Engineering for Ganesh P

Question

Briefly indicate with the aid of suitable diagrams the difference between meridional and skew ray paths in step index fibers. Derive an expression for the acceptance angle for a skew ray which changes direction by an angle 2y at each reflection in a step index fiber in terms of the fiber NA and y. It may be assumed that ray theory holds for the fiber. A step index fiber with a suitably large core diameter for ray theory considerations has core and cladding refractive indices of 1.44 and 1.42 respectively. Calculate the acceptance angle in air for skew rays which change direction by 150° at each reflection

Answer 1

The light rays during the journey inside the optical filter through the core across the core axis sunlight rays are known as meridional rays. The passage of such rays in a step-index file is similar, the rays which never cross the axis of the core are known as the skew rays. Skew rays describe the angular helical as they progress along with the filter. The skew rays will not utilize the full area of the core and they travel farther than meridional rays and undergo higher attenuation

Meridional rays

Skew ray

The acceptance angle for a skew ray

Considering the ray traveling from the air of reflective index (no) to core of reflective index (ni). The ray AB resolves the direction of the ray path AB to the core ratio at point B as the incident and reflection Ray's at point B are in the same place as cos theta. r is the angle between the core radius and the projection of the ray onto a plane BRS normal to the core axis and theta is the angle between The ray and a line AT down parallel to the core axis, this to resolve the ray path AB relative to the radius BR in these two perpendicular planes requires multiply by cos r and $sin \theta.$

Hence the reflection at point Bat an angle theta may be given by 

Cos r sin theta=0= cos theta

Using the trigonometric relationship sin theta + cos theta = 1 becomes $Cos r sin \theta= \sqrt{1-sin^2 \phi}.....2$

Of the limiting case of total internal reflection is now considered then theta becomes equal to the critical angle theta for the core-cladding interface

We know that $sin \phi_0=\frac{n_1}{n_2}$

Hence equation 2 may be written as $cos r sin \theta <cos \theta_c=\sqrt{1-\frac{n_2^2}{n_1^2}}......3$

Using snell law at the point A is given by $n_osin \theta=n_1sin \theta......4$

So, $sin \theta_{as}=\frac{n_1 cos \phi_c}{n_0 cos r}=\frac{n_1}{n_0 cos r}[1-\frac{n_2^2}{n_1^2}]^{0.5}...5$

Given that refractive index of core (n1) =1.44 refractive index of cladding (n2) =1.42

Skew Ray's which charge direction by 150 degrees 

The skew rays change direction by 150 degrees at each reflection, therefore, r=75 degrees. 

The acceptance angle for skew rays is given by $sin \theta cos r =NA \implies sin \theta_o=\frac{NA}{cos r}$

But $NA=\sqrt{n_1^2-n_2^2}=\sqrt{1.44^2-1.42^2}=0.2392$

Therefore, the acceptance angle , $\theta_o=sin^{-1}[\frac{NA}{cos r}]=sin^{-1}[\frac{0.2392}{cos 75}]=67.52^o$