JEE Main & Advanced Physics Ray Optics Sample Paper Topic Test - Refraction of Light Through Curved Surfaces

  • question_answer
    A thin convergent glass lens \[\mu =1.5\] has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index \[{{\mu }_{1}}\] it acts as a divergence lens of focal length 100 cm. The value of \[{{\mu }_{1}}\] should be

    A) 3/2

    B) 4/3

    C) 5/3

    D) 2

    Correct Answer: C

    Solution :

    [c] When the lens in air, we have
    \[{{P}_{a}}=\frac{1}{{{f}_{a}}}=\frac{{{\mu }_{g}}-{{\mu }_{a}}}{{{\mu }_{a}}}\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\]
    When the lens is in liquid, we have
    \[{{P}_{l}}=\frac{1}{{{f}_{I}}}=\frac{{{\mu }_{g}}-{{\mu }_{a}}}{{{\mu }_{l}}}\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\]
    Here,     \[{{P}_{a}}=5,{{P}_{l}}=-1,{{\mu }_{a}}=1,{{\mu }_{g}}=1.5\]
    On solving, we get, \[{{\mu }_{l}}=\frac{5}{3}\]

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