(a) State the two conditions under which total internal reflection occurs.

(b) A transparent container contains layers of three immiscible transparent liquids A, B and C of refractive indices n, $(3/4)n$ and $(2/3)n$, respectively. A laser beam is incident at the interface between A and B at an angle θ as shown in figure. Prove that the beam does not enter region C at all for $\sin \theta \ge 2/3$.

(a) Two conditions for total internal reflection (TIR):

• Light must travel from an optically denser to an optically rarer medium (higher n to lower n).
• The angle of incidence must exceed the critical angle ($i > i_c$) for that pair of media.

(b) Proof that the beam does not enter C for $\sin \theta \ge 2/3$:

Step 1 — refraction at the A–B interface. Applying Snell's law with $n_A = n$, $n_B = (3/4)n$, and refraction angle r in B:

$$n \sin θ = \tfrac{3}{4}n \sin r \;\Rightarrow\; \sin r = \tfrac{4}{3}\sin θ$$

Step 2 — critical angle at the B–C interface. The B–C interface is parallel to A–B, so the ray strikes it at angle r. Critical angle i_c for B → C:

$$\sin i_c = \frac{n_C}{n_B} = \frac{(2/3)n}{(3/4)n} = \frac{2/3}{3/4} = \frac{8}{9}$$

Step 3 — condition for TIR at B–C. The beam fails to enter C when $r \ge i_c$, i.e. $\sin r \ge 8/9$. Substituting $\sin r = (4/3) \sin \theta$:

$$\tfrac{4}{3}\sin θ \ge \tfrac{8}{9} \;\Rightarrow\; \sin θ \ge \tfrac{8}{9}\cdot\tfrac{3}{4} = \tfrac{2}{3}$$

Conclusion: For $\sin \theta \ge 2/3$, the angle in B satisfies $\sin r \ge 8/9 = \sin i_c$, so TIR occurs at the B–C interface and the beam does not enter region C. Hence proved.