When r2=C,∠N2RC=90∘ Where C= critical angle As sinC=
1
µ
=sinr2 Applying Snell's law at ' R ' µsinr2=1sin90∘ Applying Snell's law at ' Q ' 1×sinθ=µsinr1 But r1=A−r2 So, sinθ=µsin(A−r2) sinθ=µsinAcosr2−cosA [using (i)] From (i) cosr2=√1−sin2r2=√1−
1
µ2
By eq. (iii) and (iv) sinθ=µsinA√1−
1
µ2
−cosA on further solving we can show for ray not to transmitted through face AC θ=sin−1[µsin(A−sin−1(
1
µ
)].
So, for transmission through face AC θ>sin−1[µsin(A−sin−1(