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

Refractive index of the prism: \[ \mu_1 = \sqrt{3} \] Refractive index of the coating: \[ \mu_2 = 1.5 \] Step 1: Critical angle at prism–coating interface
\[ \sin C = \frac{\mu_2}{\mu_1} = \frac{1.5}{\sqrt{3}} = \frac{\sqrt{3}}{2} \] \[ \Rightarrow C = 60^\circ \] For total internal reflection at the back surface: \[ r_2>60^\circ \] Step 2: Refraction inside the prism
For a prism: \[ r_1 + r_2 = A = 75^\circ \] \[ r_1 = 75^\circ - r_2 \] Since $r_2>60^\circ$: \[ r_1<15^\circ \] Step 3: Refraction at the first surface
Using Snell’s law (air to prism): \[ \sin i = \mu_1 \sin r_1 \] \[ \sin i<\sqrt{3} \sin 15^\circ \] Given $\sin 15^\circ = 0.25$: \[ \sin i<1.732 \times 0.25 \approx 0.433 \] \[ \Rightarrow i<25^\circ \] To ensure $r_1<15^\circ$ and hence $r_2>60^\circ$, the angle of incidence must be less than $15^\circ$. \[ \therefore \text{Correct option is (D).} \]