A Using geometrical arguments for the lowest propagating mod

A Using geometrical arguments for the lowest propagating mode throught a fiber, derive the minumum angle at which a fiber face must be set in order to ensure that reflected light escapes the given fiber. Let the core index be 1.6, and the cladding index be 1.55. For the geometrical approach, you may assume all rays are in a plane containing the centerline of the fiber and that the fiber is perfect and perfectly straight.

Solution

The critical angle is the minimum angle for the light to escape from the fiber. In other words the critical angle is the maximum angle at which light can reflect inside the fiber and get transmitted further.

The formula for critical angel is sine(crit) = nr/ni

Assuming the geometrical arguments and the fiber is perfectly straight with all the rays in a plane containing the center line of the fiber.

The refractive index is given = 1.55 (That\'s nr)

The reflective index is given = 1.6 (That\'s ni)

sin(thita) = nr/ni

= 1.55/1.6 = 0.96875.

Therefore theta = sin^-1 (0.96875) = 75.6384 Degrees Any angle beyond this would cause the light to escape from the fiber.