Optics Ques 210
- $A B$ and $C D$ are two slabs. The medium between the slabs has refractive index $2$. Find the minimum angle of incidence of $Q$, so that the ray is totally reflected by both the slabs.
$(2005,2 M)$
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Answer:
Correct Answer: 210.$(60^\circ)$
Solution:
Formula:
Critical Angle and Total Internal Reflection (T. I. R.)
- Critical angles at $1$ and $2$
$ \begin{aligned} & \theta _{C _1}=\sin ^{-1} \left(\frac{\mu _1}{\mu _2}\right)=\sin ^{-1} \left(\frac{1}{\sqrt{2}}\right)=45^{\circ} \\ & \theta _{C _2}=\sin ^{-1}\left( \frac{\mu _3}{\mu _2}\right)=\sin ^{-1} \left(\frac{\sqrt{3}}{2}\right)=60^{\circ} \end{aligned} $
Therefore, minimum angle of incidence for total internal reflection to take place on both slabs should be $60^{\circ}$.
$ i _{\text {min }}=60^{\circ} $
BETA
