A prism of angle 75° and refractive index √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 ___

A prism of angle 75° and refractive index √3 is coated with thin film of refractive index 1.5 only at the back exit surface

Q. 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 must be ____.

$(\sin 15^\circ = 0.25,\ \sin 25^\circ = 0.43)$

A. between $15^\circ$ and $20^\circ$

B. $15^\circ$

C. $< 15^\circ$

D. $> 25^\circ$

Correct Answer: between 15° and 20°

Explanation

Refractive index of prism:

$$ \mu_1 = \sqrt{3} $$

Refractive index of thin film:

$$ \mu_2 = 1.5 $$

For total internal reflection at the prism–film interface, the angle of incidence at the back surface must be greater than the critical angle.

Critical angle $C$ is given by:

$$ \sin C = \frac{\mu_2}{\mu_1} = \frac{1.5}{\sqrt{3}} = \frac{\sqrt{3}}{2} $$

$$ C = 60^\circ $$

Let the refraction angle inside the prism at first surface be $r$. Using prism geometry:

$$ r + r' = A = 75^\circ $$

For total internal reflection at exit surface:

$$ r' > 60^\circ $$

$$ r < 15^\circ $$

Using Snell’s law at first surface:

$$ \sin i = \mu \sin r = \sqrt{3}\sin r $$

If $r = 15^\circ$,

$$ \sin i = \sqrt{3} \times 0.25 \approx 0.433 $$

$$ i \approx 25^\circ $$

Hence for $r < 15^\circ$, the incident angle must lie slightly below $25^\circ$ and above $15^\circ$.

Final Answer: $$ \boxed{\text{between }15^\circ\text{ and }20^\circ} $$

Explanation

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