What will be the minimum angle of incidence i | vedclass

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(C) Let $i$ be the angle of incidence at the first interface (between $\mu_2$ and $\mu_1$). For total internal reflection $(TIR)$ to occur at the firs

Vedclass · Accepted Answer

$60$

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(C) Let $i$ be the angle of incidence at the first interface (between $\mu_2$ and $\mu_1$). For total internal reflection $(TIR)$ to occur at the first interface,the angle of incidence must be greater than or equal to the critical angle $c_1$.$\sin c_1 = \frac{\mu_1}{\mu_2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow c_1 = 45^{\circ}$. So,$i \ge 45^{\circ}$.Now,let $r$ be the angle of refraction at the first interface. By Snell's law,$\mu_2 \sin i = \mu_1 \sin r$. However,for $TIR$ to occur at the second interface (between $\mu_2$ and $\mu_3$),the angle of incidence at that interface must be greater than or equal to the critical angle $c_2$.$\sin c_2 = \frac{\mu_3}{\mu_2} = \frac{\sqrt{3}}{2} \Rightarrow c_2 = 60^{\circ}$.Since the light ray is traveling from $\mu_2$ to $\mu_3$,the angle of incidence at the second interface is $i$ (due to alternate interior angles). Thus,$i \ge 60^{\circ}$.To satisfy both conditions,$i$ must be at least $60^{\circ}$.

What will be the minimum angle of incidence $i$ such that total internal reflection takes place at both interfaces (in $^{\circ}$)? (Given: $\mu_1 = \sqrt{2}, \mu_2 = 2, \mu_3 = \sqrt{3}$)

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