Question

1. The inner core of an optical fiber has a diameter of 0.1 mm and an index of refraction of n₁ = 1.5. The outer core of the fiber has an index of refraction of n₂ = 1.49.

12 = 1.49 ni = 1.5 n2 = 1.49 (a) (7 points) Coming from air, what is the largest angle relative to the aris of the fiber for which light is totally internally reflected in the fiber? (b) (6 points) I want to couple a laser beam that is originally 10 mm in diameter into the fiber. What is the minimum focal length of the lens I should use? (Ignore any effects of diffraction here) (c) (7 points) In the desert, or on a hot road, there can be a thin layer of hot air below cooler air. Hot air has a lower index of refraction than cool air, so it is possible to get total internal reflection. In reality there will be a continuous change in air temperature, but let's imagine there is a 1 cm thick layer of 60C air (n = 1.00023716 = 1 + 2.3716 * 10) below air at 30C (n = 1.00026236 = 1 + 2.6236 * 10-). At certain angles, light from the blue sky will be reflected off the interface between the hot and cold layer and appear as a lake in the distance to a thirsty desert traveler whose eyes are exactly 1.6 meters above the ground. How far away does this lake appear to be? Sky That minge

Answer 1

n2 = 1.49 nị = 1.5 n2 = 1.49

Total internal reflection will take place if angle of incidence at interface of inner core and outer core equals or exceeds critical angle $\theta$ _c

Critical angle $\theta$ _c is given by, sin $\theta$ _c = 1/n₁₂ , where n₁₂ is the relative refractive index of inner core with respect to outer core.

n₁₂ = n₁ / n₂ = 1.5 / 1.49 = 1.0067

hence we get, $\theta$ _c = sin^-1 ( 1/1.0067 ) = sin^-1( 0.9933 ) = 83^o

Hence angle of refraction for light ray entering at at fiber = 7^o

By law of refraction, sin$\theta$ / sin 7 = n₁ = 1.5

sin$\theta$ = 1.5 $\times$ sin 7 = 0.183

hence angle $\theta$ = sin^-1 (0.183) = 10.53^o

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10.530

AS shown in figure, lens is used to couple the laser light so that light rays enters

the fiber at semi-angle 10.53^o

focal length of lens is calculated as follows

tan 10.53 = 5/f or focal length f = 5 / tan10.53 $\approx$ 26.9 mm

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1.00026236 1.00023716 d.

As shown in figure , hot air layer with low refractive index 1.00023716 is below

air layer with refractive index 1.00026236.

This variation of refractive index causes total internal reflection to occur the phenomenon of mirage,

as shown in figure .

relative refractive index of air-layer above hot air-layer = 1.00026236 / 1.00023716 = 1.0000252

Critical angle or angle of incidence $\theta$ is given by, 1/sin$\theta$ = 1.0000252

Hence sin$\theta$ = 1/1.0000252 = 0.999975

Hence angle $\theta$ = sin^-1 (0.999975 ) = 89.6^o

Hence from figure we can see that , tan(90 - 89.6) = 1.6/d

we get the distance d from above equation as, d = 225.4 m