Optics 7 Question 26
27. A ray $O P$ of monochromatic light is incident on the face $A B$ of prism $A B C D$ near vertex $B$ at an incident angle of $60^{\circ}$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$, which of the
following is (are) correct?
(a) The ray gets totally internally reflected at face $C D$
(b) The ray comes out through face $A D$
(c) The angle between the incident ray and the emergent ray is $90^{\circ}$
(d) The angle between the incident ray and the emergent ray is $120^{\circ}$
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Answer:
Correct Answer: 27. (a, b, c) 28. (b, c) $\quad$ 29. (b, d) $\quad$ 30. $\sqrt{\frac{\varepsilon \mu}{\varepsilon _0 \mu _0}}$
$\begin{array}{llll}\text { 31. } \frac{25}{9} & 32 . T & \text { 33. (7) }(2)\end{array}$
$\begin{array}{lll}\text { 35. } 6 & 36.60^{\circ} & \text { 37. } 6.06 m\end{array}$
$\begin{array}{llll}\text { 38. } \frac{\mu _3 R}{\mu _3-\mu _1} & 39.1 .6 & \text { 40. (a) } 4^{\circ} & \text { (b) }-0.04^{\circ}\end{array}$
Solution:
- $\sqrt{3}=\frac{\sin 60^{\circ}}{\sin r}$
$$ \begin{aligned} \therefore \quad r & =30^{\circ} \\ \theta _C & =\sin ^{-1} \frac{1}{\sqrt{3}} \text { or } \sin \theta _C=\frac{1}{\sqrt{3}} \\ & =0.577 \end{aligned} $$
At point $Q$, angle of incidence inside the prism is $i=45^{\circ}$.
Since $\sin i=\frac{1}{\sqrt{2}}$ is greater than $\sin \theta _C=\frac{1}{\sqrt{2}}$, ray gets totally internally reflected at face $C D$. Path of ray of light after point $Q$ is shown in figure.
From the figure, we can see that angle between incident ray $O P$ and emergent ray $R S$ is $90^{\circ}$.
