Critical Angle A common trick in fiber optics is to send several colors of light

Question

Critical Angle A common trick in fiber optics is to send several colors of light down the same fiber (this is a form of what's called multiplexing). Each color of light can carry a separate signal, letting you squeeze more information into your beam. This, of course, leaves you with the problem of how to separate the colors back out once your beam gets to where its going. One way to do this involves total internal reflection. The index of refraction of glass isn't fixed; it's actually a function of frequency. Specifically, the index of the glass n and the angular frequency of the light are related by n2=1+C/(w0^2w^2C where C = 443 1030 rad2/s2 and w = 29.5 1015 rad/s. Since different colors of light see different indices of refraction, different colors of light will see different critical angles. Suppose we have a beam with yellow light (frequency = 3.20 1015 rad/s) and red light (frequency = 2.90 1015 rad/s) traveling through this glass. Eventually it's going to hit a glass-air boundary. At what angle of incidence should the light hit the boundary if we want the yellow light to stay in the glass and the red light to leave? Choose the smallest angle of incidence that works, and enter it in degrees (unit "deg").

Explanation / Answer

The propogation of light along optical fiber is based on the principle of total internal reflection.

let n1 and n2 be the refractive indices of two media such that n1>n2.when a ray of light travelling in a medium of higher refractive index n1 strikes a second medium of lower refractive index n2 making an angle of incidence i with the normal, this ray is refracted into the second medium with the angle of refraction r and moves away from the normal.According to snells law,

n1 sin i = n2 sin r

when the angle of incidence i increases ,the angle of refraction r also increases.For a particular angle of incidence ,i = crital angle

i = sin-1 (n2/n1)

n1 = 1 for air as medium

n2 = 1 + 4431030 / (wo2 -w2 -c)

wo for yellow beam= 2* pie * frequency

= 2* 3.14* 3.201015

= 20.1023742 rad/sec

n2 = 1 + (4431030/ 404.1054485 - 870.848953 - 4431030)

= 1 - 4431030/4431496.744

= 1 - 0.999894675

= 0.01

angle of incidence i = sin-1 (0.01)

= 0.572967 degrees

for red light ,

n2 = 1 + 4431030 / 18.2183742 - 870.25 - 4431030

= 1 - 4431030 / 4431882.032

= 1 - 0.99

=0.01

the angle of incidence = 0.572 degrees

to make the red light to leave and yellow to stay we need to have the angle of incidence = 0.572 degrees

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