Question

A prism of angle $75^\circ$ and refractive index $\sqrt{3}$ is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle angle must be ___. ($\sin 15^\circ = 0.25$ and $\sin 25^\circ = 0.43$)}

• A. $> 25^\circ$
• B. $15^\circ$
• C. between $15^\circ$ and $20^\circ$
• D. $< 15^\circ$

Hint:

Condition for TIR at emergence: $r_2>C \implies r_1<A-C$. Use Snell's law to find the corresponding incident angle limit.

Answer 1

The Correct Option is D

To solve this problem, we need to ensure total internal reflection (TIR) at the back exit surface of the prism. For TIR to occur, the angle of incidence inside the prism must be greater than the critical angle. 

1. First, find the critical angle. The critical angle \( \theta_c \) is given by the formula: \(\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)\), where \( n_2 \) is the refractive index of the thin film coating (1.5) and \( n_1 \) is the refractive index of the prism ( \(\sqrt{3}\)).
2. Calculate the critical angle: \(\theta_c = \sin^{-1}\left(\frac{1.5}{\sqrt{3}}\right)\)
3. Simplify the fraction: \(\sin \theta_c = \frac{1.5}{1.732} \approx 0.866\), leading to \( \theta_c = \sin^{-1}(0.866) \approx 60^\circ\).
4. For TIR, the angle of incidence must be greater than \( 60^\circ \). However, the question is about the incident angle on the prism. This involves understanding how the refractive angle affects this condition.
5. We know the angle of refraction \( r \) inside the prism should satisfy: \(r = 75^\circ - i\), where \( i \) is the angle of incidence on the prism.
6. For TIR at the back surface: \(75^\circ - i > 60^\circ\), which gives \(i < 15^\circ\).

Hence, the incident angle must be less than \( 15^\circ \) for total internal reflection to occur at the back exit surface of the prism.

Therefore, the correct answer is \(< 15^\circ\).