A prism of angle $$75^{o}$$ 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 must be______.
($$\sin 15^{o}$$ = 0.25 and $$\sin 25^{o}$$ = 0.43)

Step 1: Find the critical angle ($$C$$) for the back surface

The critical angle at the interface between the prism and the thin film is given by:

$$ \sin C = \frac{\mu_{\text{film}}}{\mu_{\text{prism}}} $$

$$ \sin C = \frac{1.5}{\sqrt{3}} = \frac{3/2}{\sqrt{3}} = \frac{\sqrt{3}}{2} $$

$$ C = 60^\circ $$

Step 2: Condition for Total Internal Reflection (TIR)

For TIR to occur at the back exit surface, the angle of incidence at that internal surface ($$r_2$$) must be greater than or equal to the critical angle:

$$ r_2 \ge 60^\circ $$

Step 3: Relate $$r_2$$ to the first refraction angle ($$r_1$$)

In a prism, the internal angles are related to the prism angle ($$A$$) by the formula $$r_1 + r_2 = A$$. Given $$A = 75^\circ$$:

$$ r_1 = 75^\circ - r_2 $$

Substitute the inequality condition for $$r_2$$:

$$ r_1 \le 75^\circ - 60^\circ $$

$$ r_1 \le 15^\circ $$

Step 4: Find the condition for the initial incident angle ($$i$$)

Apply Snell's Law at the first interface (air to prism):

$$ 1 \cdot \sin i = \mu_{\text{prism}} \cdot \sin r_1 $$

$$ \sin i = \sqrt{3} \sin r_1 $$

Since $$r_1 \le 15^\circ$$, and the sine function increases with angle in this range:

$$ \sin i \le \sqrt{3} \sin 15^\circ $$

Substitute the given value $$\sin 15^\circ = 0.25$$:

$$ \sin i \le \sqrt{3} \times 0.25 $$

$$ \sin i \le 1.732 \times 0.25 $$

$$ \sin i \le 0.433 $$

Step 5: Conclusion

We are given that $$\sin 25^\circ = 0.43$$.

Since our condition is $$\sin i \le 0.433$$, the maximum possible value for the incident angle $$i$$ is roughly $$25.6^\circ$$.

Therefore, for Total Internal Reflection to occur, the incident angle must satisfy:

$$ i \le 25.6^\circ $$
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