question_answer

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)        \[r>\frac{R}{n'}\]