Light travels from air into an optical fiber with an index of refraction of 1.35

Question

Light travels from air into an optical fiber with an index of refraction of 1.35.

(A) if the angle of incidence on the end of the fiber is 22degrees, what is the angle of refraction inside the fiber?

ans = 16 degrees

(B) For the above fiber, what would occur for a light ray at the second boundary (fiber/air boundary) if the angle of incidence at boundary 1 is 37 degree?

a. refracted at 29.4 degree
b. reflected at 15.7 degree
c. the ray skims along the surface

d. reflected at 63.5 degree

e. none of these things.

The answer is D, but I do not know how to calculate it myself. Could you please explain to me how to do it?

Explanation / Answer

Light travels from air into an optical fiber with an index of refraction of 1.44. (a) In which direction does the light bend? (b) If the angle of incidence on the end of the fiber is 22 degree, what is the angle of refraction inside the fiber?

Solution:

(a) Since the light is traveling from a rarer region (lower n) to a denser region (higher n), it will bend toward the normal.

(b) We will identify air as medium 1 and the fiber as medium 2. Thus, n1 = 1.00, n2 = 1.44, and θ1 = 22o. Snell's Law then becomes

(1.00) sin 22 = 1.44 sin θ2.

sin θ2 = (1.00/1.44) sin 22 = 0.260

θ2 = sin-1 (0.260) = 15 degree.

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