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

Here, `A=60^(@), mu=1.5`
When deviation is minimum , `r=(A)/(2)=(60)/(2)=30^(@)`
As `mu=(sin i)/(sin i)`
`sini=mu sin r=1.5 sin 30^(@)=0.75`
`i=sin^(-1)(0.75)=49^(@)`
(b) `del_(m)=2i-A=2xx49^(@)-60^(@)=38^(@)`
(c) For maximum deviation, `i_(1)=90^(@)`
and `r_(1)=C=sin^(=1)((1)/(mu))=sin^(-1)((1)/(3//2))=sin^(-1)(0.6667)=42^(@)`
As `r_(1)+r_(2)=A`
`r_(2)=A-r_(1)=60-42^(@)=18^(@)`
From `mu=(sini_(2))/(sinr_(2)),`
`sini_(2)=mu sin r_(2)=1.5 sin 18^(@)=1.5xx0.31=0.4650`
`i_(2)=sin^(-1)(0.4650)=28^(@)`
This is the angle of emergent (d) `del_("max")=i_(1)+i_(2)-A=90^(@)+28^(@)-60^(@)=58^(@)`