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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2007

  • question_answer
      A vessel of height 2d is half-filled with a liquid of refractive index \[\sqrt{2}\] and the other half with a liquid of refractive index n (the given liquids are immiscible). Then the apparent depth of the inner surface of the bottom- of the vessel (neglecting the thickness of the bottom of the vessel) will be

    A)  \[\frac{n}{d\left( n+\sqrt{2} \right)}\]

    B)  \[\frac{d\left( n+\sqrt{2} \right)}{n\sqrt{2}}\]

    C)  \[\frac{\sqrt{2}n}{d\left( n+\sqrt{2} \right)}\]

    D)  \[\frac{nd}{d+\sqrt{2n}}\]

    Correct Answer: B

    Solution :

    Refractive index \[\mu =\frac{\text{Real}\,\text{depth(d)}}{\text{Apparent}\,\text{depth(x)}}\] For 1st liquid, \[\sqrt{2}=\frac{d}{{{x}_{1}}}\] \[\Rightarrow \] \[{{x}_{1}}=\frac{d}{\sqrt{2}}\] Similarly, for 2nd liquid, \[n=\frac{d}{{{x}_{2}}}\] \[\Rightarrow \] \[{{x}_{2}}=\frac{d}{n}\] Total apparent depth \[={{x}_{1}}+{{x}_{2}}\] \[=\frac{d}{\sqrt{2}}+\frac{d}{n}\] \[=\frac{d(n+\sqrt{2})}{n\sqrt{2}}\]


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