The refracting angle of a glass prism is `30^@.` A ray is incident onto one of the faces perpendicular to it. Find the angle `delta` between the incident ray and the ray that leaves the prism. The refractive index of glass is `mu=1.5.`

Explanation

Given, `A=30^(@),mu=1.5and i_(1)=0^(@)`
Since, `i_(1)=0^(@),`therefore,`r_(1)` is also equal to `0^(@)`.
Further, since, `r_(1)+r_(2)=A`
`therefore r_(2)=A=30^(@)`
Using, `mu=(sini_(2))/(sinr_(2))`
we have `1.5=(sin i_(2))/(sin 30^(@))`
or `sini_(2)=1.5sin30^(@)=1.5xx1/2=0.75`
`therefore i_(2)=sin^(-1)(0.75)=48.6^(@)`
Now, the deviation, `sigma=(i_(1)+i_(2))-A=(0^(@)+48.6^(@))-30^(@)`
`delta=18.6^(@)`