The figure bellow shows a ray of light entering one end of an optical fiber at a

Question

The figure bellow shows a ray of light entering one end of an optical fiber at an angle of incidence ?i = 50.0 degrees. The index of refraction of the fiber is 1.62.

a) Find the angle ? the rays makes with the normal when it reaches the curved surface of the fiber

b) Show that the internal reflection from the curved surface is total

c) Show that the internal reflection from the curved surface of the fiber is always total for any incident angle ?i, provided the index of refraction of the fiber exceeds 1.41.

Explanation / Answer

a) n air sin(thita)i = n sin (90-thita)

90-(thita) = sin inv n air sin(thita)i / n

thuta = 90 - sin inv (1.00)sin50.0 / 1.62

= 61.8degree

b) (thita) > (thita) , so the internal reflection is total.

c) sin (thita)c = n2/n1

(thuta)c = sin inv n2/n1

sin inv 1.00/1.41

= 45.17degree   

45.17degree < 61.8degree

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