Correct Answer - A
Taking refraction at`AC`
`mu sin r_(2)=1 sin 90^(@)`
`sqrt(2) sinr_(2)=1`
`r_(2)=45^(@)`
Since,`r_(1)+r_(2)=A`
`r_(1)+45^(2)=60^(@)`
`r_(1)=15^(@)`
Taking refraction at`AB`
`1 sin i=sqrt(2)sin 15^(@)`
`sin i=sqrt(2) sin(45^(@)-30^(@))`
`=sqrt(2)[sin45^(@)cos30^(@)-cos45^(@)-sin30^(@)]`
`=sqrt(2)[(1)/(sqrt(2)).(sqrt(3))/(2)-(1)/(sqrt(2)).(1)/(2)]`
`sin^(-1)=((sqrt(3)-1)/(2))`
`i=sin^(-1)((sqrt(3)-1)/(2))`