CET Karnataka Engineering CET - Karnataka Engineering 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)  \[\frac{n}{d(n+\sqrt{2})}\]                         

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

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

    D)  \[\frac{nd}{d+\sqrt{2}n}\]

    Correct Answer: B

    Solution :

    Refractive index \[=\frac{Rea1\text{ }depth\text{ (}d\text{)}}{Apparent\text{ }depth\text{ (}x\text{)}}\] For 1st liquid, \[\sqrt{2}=\frac{d}{{{x}_{1}}}\] \[\Rightarrow \]                               \[{{x}_{1}}=\frac{d}{\sqrt{2}}\] Similarly, for 2nd liquid,                 \[n=\frac{d}{\sqrt{2}}\] \[\Rightarrow \]               \[{{x}_{2}}=\frac{d}{n}\] 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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