Question

A right angled prism is to be made by selecting a proper material and the angles A and B $(B\le A),$ as shown in figure. It is desired that a ray of light incident on the face AB emerges parallel to the incident direction after two internal reflections.
(a) What should be the minimum refractive index n for this to be possible ?
(b) For n = 5/3 is it possible to achieve this with the angle B equal to 30 degrees ?

Answer 1

(a) At P, angle of incidence $i_A$ = A
and at Q, angle of incidence $i_B$ = B
If T1R satisfies for the smaller angle of incidence than for
larger angle of incidence is automatically satisfied.
$B\le A\therefore i_B\le i_A$
Maximum value of B can be $45^{\circ }$. Therefore, if condition of
TIR is satisfied, then condition of TIR will be satisfied for all
value of $i_{A}and{i}_B$
Thus, $45^{\circ }\ge {\theta }_c$
$orsin45^{\circ }\ge sin{\theta }_c$
$or\frac{1}{\sqrt{2}}\ge \frac{1}{\mu }or\mu \ge \sqrt{2}$
$\therefore$ $\mu ornis\sqrt{2}.$
$(b)Forn=\frac{5}{3},sin{\theta }_c=\frac{1}{n}=si{n}^{-1}\left(\frac{3}{5}\right)\approx 37^{\circ }$
$If=30^{\circ },then{i}_B=30^{\circ }$
$then=60^{\circ }or{i}_A=60^{\circ }$
$i_A>{\theta }_{c}but{i}_B<{\theta }_c$
i.e. TIR will take place at A but not at B.
or we write : sin $i_B<sin{\theta }_c<sin{i}_A$
$orsin30^{\circ }<\frac{3}{5}<sin60^{\circ }$
$or0.5<0.6<0.86$