At what angle should a ray of light be incident on the face of a prism of refracting angle `60^(@)` so that it just suffers total internal reflection at the other face ? The refractive index of the prism is 1.524.

Explanation

Let C=critical angle for the material of the prism.
`therefore sinC=1/(mu)=1/1.524=0.6562`
`therefore C=sin^(-1)(0.6562)=41^(@)`
We know that for a prism,A=`r_(1)+r_(2)`
Here, `r_(2)=C=41^(@),A=60^(@)`
`therefore60^(@)=r_(1)+41^(@)orr_(1)=60^(@)-41^(@)=19^(@)`
For the face AB, `r_(1)=19^(@),mu=1.524`

`therefore` According to Snell's law
`mu=(sini)/(sinr_(1))`
or `sini=musinr_(1)=1.524sin19^(@)`
`=1.524xx0.3256=0.4962`
`i=sin^(-1)(0.4962)=29.75^(@)`
`therefore` Angle of incidence,i=`29.75^(@)=30^(@)`