Question 425: Ray Optics

Question

What will be the minimum angle of incidence such that the total internal reflection occurs on both the surfaces?

Accepted Answer

If $$c_{1}$$ be the critical angle for the interface AB in figure.
$$\begin{array}{rlr}
n_{2} \sin c_{1} & =n_{1} \sin 90^{\circ} & \left(n_{1}=\sqrt{2}ight. \
2 \sin c_{1} & =\sqrt{2} 	imes 1 & \left.n_{2}=2ight) \
\sin c_{1} & =\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=\sin 45^{\circ} \
c_{1} & =45^{\circ} \
n_{1} & =\sqrt{2} \
n_{2} & =2 \
\hline n_{3} & =\sqrt{3}
\end{array}$$
Since, AB and CD are parallel, the angle of 
incidence for both will be equal.
If $$c_{2}$$ be the critical angle for the interface CD.
$$\begin{aligned}
n_{2} \sin c_{2} & =n_{3} \sin 90^{\circ} \quad\left(n_{3}=\sqrt{3}ight) \
2 \sin c_{2} & =\sqrt{3} 	imes 1 \
\sin c_{2} & =\frac{\sqrt{3}}{2}=\sin 60^{\circ} \
c_{2} & =60^{\circ}
\end{aligned}$$
The minimum value of the angle of incidence must be equal to the greater of the values of $$c_{1}$$ and $$c_{2}$$.
$$	herefore$$ Minimum angle of incidence is $$60^{\circ}$$.

Answer

30$$^\circ$$

Answer

45$$^\circ$$

Answer

50$$^\circ$$

Answer

60$$^\circ$$

Ray Optics

What will be the minimum angle of incidence such that the total internal reflection occurs on both the surfaces?

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