(a) Conditions for Total Internal Reflection (TIR)
- Condition 1: The light must travel from a denser medium to a rarer medium.
- Condition 2: The angle of incidence in the denser medium must be greater than the critical angle for the given pair of media.
(b) Mathematical Proof
Step 1 (Governing Law & Given Data): [1 Mark]
According to Snell's Law, for any interface between two media with refractive indices $n_1$ and $n_2$ and angles of incidence $i$ and refraction $r$:
$$n_1 \sin i = n_2 \sin r$$Given refractive indices: $n_A = n$, $n_B = \frac{3}{4}n$, $n_C = \frac{2}{3}n$.
Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]
Let $\theta_1$ be the angle of incidence at the A-B interface and $\theta_2$ be the angle of refraction into medium B. Applying Snell's Law at the A-B interface:
$$n_A \sin \theta = n_B \sin \theta_2$$ $$n \sin \theta = \left( \frac{3}{4}n \right) \sin \theta_2 \implies \sin \theta_2 = \frac{4}{3} \sin \theta$$Now, consider the interface between B and C. Let $\theta_3$ be the angle of refraction into medium C. Applying Snell's Law at the B-C interface:
$$n_B \sin \theta_2 = n_C \sin \theta_3$$ $$\left( \frac{3}{4}n \right) \sin \theta_2 = \left( \frac{2}{3}n \right) \sin \theta_3$$Substituting $\sin \theta_2$ from the first interface:
$$\left( \frac{3}{4}n \right) \left( \frac{4}{3} \sin \theta \right) = \left( \frac{2}{3}n \right) \sin \theta_3$$ $$n \sin \theta = \frac{2}{3}n \sin \theta_3 \implies \sin \theta_3 = \frac{3}{2} \sin \theta$$Step 3 (Final Concluded Answer): [0.5 Mark]
For the beam to not enter region C, it must undergo TIR at the B-C interface, which requires $\sin \theta_3 \ge 1$.
Given $\sin \theta \ge \frac{2}{3}$, then:
$$\sin \theta_3 = \frac{3}{2} \sin \theta \ge \frac{3}{2} \left( \frac{2}{3} \right) = 1$$Since $\sin \theta_3 \ge 1$ is impossible for a real angle, the light undergoes Total Internal Reflection at the B-C interface and does not enter region C. Hence Proved.
💡 EL10 Concept Insight: Think of light like a ball rolling across different surfaces; when it hits a boundary where it's forced to bend too sharply to stay in the next layer, it simply "bounces" back off the surface instead of crossing over.