A light ray in the core ( n = 1.35) of a cylindrical optical fiber (the figurebe

Question

A light ray in the core (n = 1.35) of a cylindrical optical fiber (the figurebelow) travels at an angle 1 =47.0° with respect to the axis of thefiber. A ray is transmitted through the cladding (n =1.25) and into the air. What angle2 does the exiting ray make with theoutside surface of the cladding?
Picture: http://www.webassign.net/grr/chapter-23/fig-052.gif A light ray in the core (n = 1.35) of a cylindrical optical fiber (the figurebelow) travels at an angle 1 =47.0° with respect to the axis of thefiber. A ray is transmitted through the cladding (n =1.25) and into the air. What angle2 does the exiting ray make with theoutside surface of the cladding?
Picture: http://www.webassign.net/grr/chapter-23/fig-052.gif

Explanation / Answer

The light will make angle 43 (i.e 90-47) deg with the surface normal to the core/clad interface. The angle, "a" it enters the cladding at is, from Snell's law; (1.25)Sin(a) = (1.35)Sin(43) 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 "b" it enters the air at is, again from Snell's law, (1)Sin(b) = (1.25)Sin(a) eliminate the Sin(a) from these eqns; Sin(b) = (1.35)Sin(43) = 0.92069 b = 67.028 deg (angle made with the outside surface normal) = 90 - 67.0283 = 22.97 deg (angle made with the outside cladding surface)

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