Geometrical OpticsHard
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 must be $\_\_\_\_$ .
( $sin15^{\circ} = 0.25$ and $sin25^{\circ} = 0.43$ )
Options
A.between $15^{\circ}$ and $20^{\circ}$
B.$15^{\circ}$
C.$> 25^{\circ}$
D.$< 15^{\circ}$
Solution
$$r_{1} + r_{2} = 75^{\circ} $$For TIR at back surface
$${\sqrt{3}sinr_{2} = \frac{3}{2}sin90^{\circ} }{r_{2} \geq 60^{\circ} }{r_{1} \leq 15^{\circ} }{1sini = \sqrt{3}sin15^{\circ} }{sini = 1.73 \times .25 }{sini = .433 }{i = 25^{\circ} \Rightarrow i < 25^{\circ}}$$
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