determine the maximum angle theta for which the light rays incident on the end of the optical fiber of radius 1 mm are subhect to the total internal reflection along the walls of the fiber. assume that the fiber has an index of refraction od 1.2 and that the outside medium is air. note that the rays refracts when entering the fiber g

Explanation:

Let the critical angle be C .

sinC = 1 / μ where μ is index of refraction .

sinC = 1 /1.2

= .833

C = 56°

Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )

sin i / sinr = 1.2 , i is angle of incidence

sini = 1.2 x sinr = 1.2 x sin 34 = .67

i = 42°.  

and 1and 1

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