Cross section of a light pipe is shown in the following figure :

Inner core is material of refractive index 1.68 and its outer covering is a material of refractive index 1.44. Find the maximum angle at which the ray can be incident with the axis of pipe for which total internal reflection occurs inside the pipe.

Explanation

First of all let us find critical angle for this pair of materials .
`sin theta_(c) = (1.44)/(1.68)`
`rArr" "sin theta_(c) = 0.8571`
`rArr" " theta_(c) = 59^(@)`
For total internal reflection to occur inside we can write the followings :
` " "i ge 59^(@)`
`rArr" " 90- r ge 59^(@)`
`rArr" " r le 31^(@)`
Hence `31^(@)` is maximum value of r. Using Snell,s law we canf in correspoding maximum value of i .
`mu sin theta = ` constant .
`rArr" " 1 xx sin i_("max") = 1.68 xx sin r_("max")`
`rArr " "sin i_("max") = 1.68 xx sin 31`
`rArr" "sin i_("max") = 1.68 xx 0.515`
`rArr" " sin i _("max") = 0.8652`
`rArr " " i_("max") = 60^(@)`