(a) Fig. shows a cross-section of a 'light pipe' made of a glass fibre of refractive index `1.68`. The outer covering of the pipe is made of a material of refractive index `1.44`. What is the axis of the pipe for which total reflection inside the pipe take place as shwon.
(b) What is the answer if there is no outer covering if the pipe ?

Explanation

Here, `mu_2 = 1.68, mu_1 = 1.44`
As `mu = (mu_2)/(mu_1) = (1)/(sin C) :. sin C = (mu_1)/(mu_2) = (1.44)/(1.68) = 0.8571 :. C = sin^-1 (0.0871) = 59^@`
Total internal reflection would take place when `I gt C` i.e., `I gt 50^@` or when `r lt r_(max)`, where
`r_(max) = 90^@ - C = 90^@ - 59^@ = 31^@`
As `(sin i_(max))/(sin r_(max)) = 1.68`
`:. sin i_(max) = 1.68 sin r_(max) = 1.68 sin 31^@ = 1.68 xx 0.5156 = 0.8662 :. i_(max) = 60^@`
Hence all rays which are incident in the range `/_ i lt 60^@` will suffer total internal reflection in the pipe.
It should be noted that lower limit of `i` is not `0^@`. It will be determined by the ratio of diameter to the length of pipe.
(b) If there is no outer coating of pipe, `mu_2 = 1.68, mu_1 = 1`
As `sin C' = (1)/(mu)=(mu_1)/(mu_2)=(1)/(1.68) = 0.5952 :. C' = sin^-1(0.5952)=36.5^@`
Now, `i = 90^@` will have `r = 36.5^@` `("from"(sin i)/(sin r) = mu = 1.68)`
`:. i' = 90^@ - 36.5^@ = 53.5^@`, which is greater than `C'`. Thus all rays incident at angles in the rage zero to `906@` will suffer total internal reflection.