For a prism, angle of prism is `60^@` and it.s refractive index is 1.5, find (1) angle of incidence corresponding to the angle of minimum deviation and (2) angle of emergence for angle of maximum deviation.

Explanation

For minimum deviation,
`r_1 = r_2 and A = r_1 + r_2`
`therefore A = 2r_1`
`r_1 = A/2 = 60 /2 = 30^@`
Now, n = 1.5 and `n = (sin i)/(sin r_1)`
`therefore n sin r_1 = sin i`
`therefore 1. 5 xx sin 30^@ = sin i`
`therefore 1.5 xx 0.5 = sin i `
`therefore i = 48^@ 35`
(2) For maximum deviation , `i = 90^@`
` therefore 1.5 = (sin 90^@)/(sin r_1) therefore r_1 = 41^@ 48.`
`therefore r_2 = A - r_1 = 60 - 41^@ 48. = 18^@ 12.`
`( because r_1 + r_2 =A)`
`therefore 1.5 sin r_2 sin e (because n sin r_2 = sin e)`
`therefore sin e = 0.4685`
`therefore e = 27.9^@`