A `60^@` prism has a refractive index of `1.5`. Calculate (a) the angle of incidence for minimum deviation (b) angle of minimum deviation ( c) the angle of emergence of light at maximum deviation (d) angle of maximum deviation.

Explanation

In the position of minimum deviation
`r = (A)/(2) = (60)/(2)`
According to Snell's law
`mu = (sin i)/(sin r) or sin I = mu sin r`
`sin i = 1.5 sin 30^(@) = 0.75`
`I = sin^(-1)(0.75) = 49^(@)`
`delta_(m) = 2i - A = 2 xx 49^(@) - 60^(@) = 38^(@)`
For maximum deviation, `i_(1) = 90^(@)`,
`r_(1) = C = sin^(-1)((1)/(mu)) = sin^(-1)((1)/(3//2))`
`=sin^(-1)0.6667 = 42^(@)`
from `R_(1) + R_(2) = A`,
`r_(2) = A - r_(1) =60^(@) - 42^(@) = 18^(@)`
Again, According to Snell's law ,
`mu sin r_(2) = 1 sin i_(2)`
`(3)/(2)sin 18^(@) = sini_(2)`
`(1.5 xx 0.31) = sin i_(2)`
`:. i_(2) = 28^(@)`
`delta_(max) = i_(1) + i_(2) -A = 90^(@) + 28 - 60 = 58^(@)`