Question

A light ray in the core (n = 1.5) of a cylindrical optical fiber is incident on the cladding. A ray is transmitted through the cladding (n = 1.21) and into the air. The emerging ray makes an angle 2 = 5.3° with the outside surface of the cladding. What angle 1 did the ray in the core make with the axis?

Answer 1

The light will make angle deg with the surface normal to the core/clad interface. The angle, "a" it enters the cladding at is, from Snell's law;

(1.21)Sin(a) = (1.5)Sin()

When the light is incident on the cladding/air interface it makes the same angle, "a" with that surface normal (draw a diagram and do a little basic geometry to see that). The angle 5.3 degree it enters the air at is, again from Snell's law,

(1)Sin(5.3) = (1.21)Sin(a)

elliminate the Sin(a) from these eqs;

Sin(5.3) = (1.5)Sin()

or, = 3.53 degree

= 3.53 deg (angle 1 did the ray in the core make with the normal)

= 90 - 3.53 = 86.47 deg (angle 1 did the ray in the core make with the axis)