A right angles prism `(45^(@),90^(@), 45^(@))` of refractive index n has a plate of refractive index `(n_(1)ltn)` cemented to its diagonal face. The assembley is in air. A ray is incident on AB.
a. Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle.
b. Assuming `n=1.351`, calculate the angle of incidence at AB for which the refracted rey passes through the diagonal face undeviated.

Explanation

a. `sinC=(n_(1))/(n)`
From Figure , `(90^(@)-r_(1))+45^(@)+(90^(@)-C)=180^(@) `
`rArr r_(1)=45^(@)-C`
From Snell's law,
`(sin i)/(sin r_(1))=n`
`:. sin i=n sin r_(1)`
`=n sin (45^@-C)`
`=n(sin 45^@ cos C-cos 45^@ sinC)`
`=(n)/sqrt(2) (cosC-sinC)`
`(n)/(sqrt(2))[sqrt(1-sin^(2)C)-sinC]`
`[sqrt(1-((n_(1))/(n))^(2))-(n_(1))/(n)]`
`(1)/(sqrt (2))[sqrt((n^(2)-n_(1)^(2)))-n_(1)]`
`rArr i=sin^(-1) {(1)/(sqrt(2))(sqrt(n^(2)-n_(1)^(2)))-n_(1)}`
b. When refracted ray passes through the diagonal face undeflected, the incidence at diagonal face is perpendicular.
`r_(2)=0` so `r_(1)+r_(2)=45^@rArr r_(1)=45^@`
Again, `(sini)/(sinr_(1))=nrArrsini=nsinr_(1)=1.352sin45^@`
or `sin i =1.3520020 (1)/(sqrt (2))=0.956`
or `i=sin^(-1)(0.956)=72^@58^(')`.