(a) Figure 9.32 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 range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.

(b) What is the answer if there is no outer covering of the pipe?

Explanation

`rArr` Here, `mu_1 = 1.44 mu_2 = 1.68`
`mu_(2) = (mu_2)/(mu_1) = (1)/(sin i._c)`
But sin `i_(c) = (mu_1)/(mu_2) = (1.44)/(1.68) = 0.8571`
`therefore i._(c) ~~ 59^@`
`rArr "If" i. gt i._c`, then total internal refraction
Hence, if `i gt 59^@` of
refraction angle `r lt r_("max")`
where `r_("max") = 90^@ - 59^@ = 31^@`
`therefore` From Snell.s law,
`(sin i_(max))/(sin r_(max)) = mu = 1.68`
`therefore sin i_(max) = 1.68 xx sin r_(max)`
`= 1.68 xx sin 31^@`
`= 1.68 xx 0.5150`
`= 0.8662`
`therefore sin i_(max) ~~ 60^@`
`therefore` Rays enter at `0 lt i lt 60^@` with axis of cylinder perform total internal refraction.
`rArr` For limited length of pipe, minimum value of incidence angle can be decided by ration of diameter and length of pipe.
(b) If there is no layer outside of pipe, then
`sin i.c = (mu_1)/(mu_2)` where `mu_1 =1, mu_2 = 1.68`
`= 1/ 1.68 = 0.5952`
`therefore i_c = 36.5^@`
`therefore` F or `i = 36.5^@, r = 36.5^@ and i. = 90^@ - 36.5^@`
`therefore i = 53.5^@`
Thus , `i gt i_c` hence `0 lt i lt 90^@` , so all the rays perform total internal refraction.