Question

Consider an optical fiber with an index of refraction of 1.26. You are positioning a laser beam to enter the flat end of the optical fiber. As the laser light approaches the air to optical fiber boundary, what is the maximum angle of incidence if the beam is not to escape from the fiber?

Answer 1

Index of refraction of air (n1) =1

Index of refraction of fiber(n2) =1.26

Now we have to find out maximum angle of incidence so that beam is not to escape from the fiber i. e. satisfies total internal reflection.

Now from snell's law at first refraction

(n1) (sini) =(n2) (sinr)

Put these Value in this equation

(1)(sini)=(1.26)(sinr)

sini=(1. 26)root(1-cos*2)

Now for second refration

(n1) sin(90)=(n2)sin(90-r)

(n1) /n2=cosr

So sini=(1. 26)root(1-(n1)*2/(n2)*2)

sini=(1.26)root(1-1/(1.26)*2)

sini=root((1.26)*2-1)

sini=root(1.587-1)

sini=root(0.587)=0.776

i=sin-1(0.776)