JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

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

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