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.

Answer

A, B, C, D

Explanation

(i) The ray incident on AB at M makes an angle of incidence i. It gets refracted at M. The angle of refraction is r. Applying Snell's law at M
`n=(sini)/(sinr)`…(i)
from fig `angleAPM=180^circ-(45^circ+90^circ-r)=45^circ+r`
and `C=90^circ-(45^circ+r)=45^circ-r`
The ray after refraction at M enter the prism and strikes its diagonal face AC making an angle C with the normal at P. Here C is the critical angle, therefore, the ray after refraction at P makes angle of refraction `90^circ`
Applying Snell's law at P
`(n)/(n_1)=(sin90^circ)/(sinC)impliessinC=(n_1)/(n)` ...(ii)
From (i), `sini=nsinr=nsin(45^circ-C)`
`=n[sin45^circcosC-cos45^circsinC]`
(n)/(sqrt2)[sqrt(1-sin^(2)C)-sinC]`
`sini=(n)/(sqrt2)[sqrt(1-(n_1^2)/(n_1^2)-(n_1)/(n_2)]]` [from (ii)]
`implies``i=sin^(-1)[(1)/(sqrt2){sqrt(n^2-n_1^2)-n_1}]`
(ii) Angle of incidence at Ab for which the refracted ray passes through the diagonal face undeviated. For this to happen, the angle of incidence of ray MP on diogonal face should be zero. It means that the ray should strike normal to AC. Applying Snell's law at M, we gen `n=(sini^`)/(sinr)`
Since `angleAP^`M=90^circ angle AMP=45impliesr=45^circ`
`:'``sini^`=nsinr=nsin45^circ=(1.352)/(sqrt2)=0.956`
`:'` `implies``i^`=72.94^circ`