A transparent solid cylindrical rod has a refractive index of\[\frac{2}{\sqrt{3}}\].It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. |
The incident angle \[\theta \] for which the light ray grazes along the wall of the rod is: [AIEEE 2009] |
A) \[{{\sin }^{-1}}\left( \frac{1}{2} \right)\]
B) \[{{\sin }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]
C) \[{{\sin }^{-1}}\left( \frac{2}{\sqrt{3}} \right)\]
D) \[{{\sin }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]
Correct Answer: D
Solution :
[d] \[1\times \sin 90=\frac{2}{\sqrt{3}}\sin (90-\alpha )\] |
\[\Rightarrow \]\[\cos \alpha =\frac{\sqrt{3}}{2}\] |
So \[sin\alpha =\sqrt{1-\frac{3}{4}}=\frac{1}{2}\] |
Now, \[1\times \sin \theta =\frac{2}{\sqrt{3}}\sin \alpha \] |
\[=\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}\] |
\[\Rightarrow \]\[\theta ={{\sin }^{-1}}\frac{1}{\sqrt{3}}\] |
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