question_answer

A large transparent cube (refractive index\[=1.5\]) has a small air bubble inside it. When a coin (diameter\[2\text{ }cm\]) is placed symmetrically above the bubble on the top surface of the cube, the bubble cannot be seen by looking down into the cube at any angle. However, when a smaller coin (diameter\[1.5cm\]) is placed directly over it, the bubble can be seen by looking down into the cube. What is the range of the possible depths d of the air bubble beneath the top surface?

 

A) \[\frac{3\sqrt{5}}{8}cm<d<\frac{\sqrt{5}}{2}cm\]

B) \[\frac{2\sqrt{3}}{8}cm<d<\frac{3\sqrt{5}}{2}cm\]

C) \[\frac{\sqrt{5}}{7}cm<d<\frac{2\sqrt{5}}{7}cm\]

D) \[\frac{\sqrt{3}}{7}cm<d<\frac{3\sqrt{3}}{7}cm\]

Correct Answer: A

Solution :

[a] The depth of the bubble shall not be more than that shown in the figure, when the coin of diameter \[2\text{ }cm\]is placed above it. A larger depth will mean that that the angle of incidence at points not covered by the coin can be lesser than the critical angle [c].

\[\mu \,\,\sin C=1\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\operatorname{tanC}=\frac{1}{\sqrt{{{\mu }^{2}}-1}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\operatorname{tanC}=\frac{2}{\sqrt{5}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{d}=\frac{2}{\sqrt{5}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d=\frac{\sqrt{5}}{2}\]

The depth at which the bubble will be just visible if the smaller coin is placed can be similarly calculated as

\[\tan C=\frac{2}{\sqrt{5}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3/4}{d}=\frac{2}{\sqrt{5}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,d=\frac{3\sqrt{5}}{8}\]

If the bubble happens to be below this depth, it will be certainly visible.

Therefore, \[\frac{3\sqrt{5}}{8}<d<\frac{\sqrt{5}}{2}cm.\]