A capillary tube is made of glass having index of refraction n and is surrounded by air. The outer radius of the tube is R. The tube is filled with a liquid having index of refraction n' (n' < n). For any ray that hits the outer surface of tube from air as shown to also enter the liquid, the minimum internal radius r of the tube is given by: |
A) \[r\,=\,\frac{R}{n}\]
B) \[r\,=\,\frac{R}{n'}\]
C) \[r\,=\,\frac{nR}{n'}\]
D) \[r\,=\,\frac{n'R}{n}\]
Correct Answer: B
Solution :
For all rays to enter liquid maximum value of \[{{\theta }_{3}}<\]critical angle | |
\[\therefore \] \[\sin {{\theta }_{3\,\,\max }}<\frac{n'}{n}\] | ?(1) |
Applying sine rule to \[\Delta \,ABO\] | |
\[\frac{\sin {{\theta }_{2}}}{r}=\frac{\sin \,(\pi -{{\theta }_{3}})}{R}=\frac{\sin {{\theta }_{3}}}{R}\] | |
\[\therefore \] \[{{\theta }_{3}}\] is maximum when \[{{\theta }_{2}}\] is maximum |
\[\Rightarrow \]\[\frac{\sin {{\theta }_{2\,\,\max }}}{r}=\frac{\sin {{\theta }_{3\,\,}}_{\max }}{R}\] | ?(2) |
Also \[\sin {{\theta }_{2}}_{\,\,\max }=\frac{1}{n}\] | ?(3) |
from (1), (2) and (3) \[r>\frac{R}{n'}\] |
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