At what angle should a ray of light be incident on the face of a prism of refracting angle `60^(@)` so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Explanation

Ray PQ incidents at AB and after refration QR incidents at AC with critical angle `(i_c)` .
`therefore r_1 = i_c`
Now, sin `i_c = (1)/(mu)= (1)/(1.524) = 0.6542`
`therefore i_c = 41^@`
But `r_1 + r_2 = A`
`therefore r_1 = A-r_2 =60^@ -I_c " " [because A = 60^@]`
`therefore r_1 = 60^@ - 41^@ = 19^@`
`rArr` From Snell.s law,
`mu = (sin i)/(sin r_1)`
`therefore = (sin i)/(sin r_1)`
`therefore sin i = mu sin r_1`
`= 1.524 xx sin 19^@`
`= 1.524 xx 0.3256`
`therefore sin i = 0.4962`
`therefore sin i = 0.4962`
` therefore i = 29.75^@`
` therefore i ~~ 30^@`