CET Karnataka Medical CET - Karnataka Medical Solved Paper-2012

  • question_answer
    A point source of light is kept below the surface of water \[\left( {{n}_{w}}=\frac{4}{3} \right)\] at a depth of \[\sqrt{7}m\]. The radius of the circular bright patch of light noticed on the surface of water is ... m.

    A)  \[\frac{3}{\sqrt{7}}\]    

    B)  \[3\]      

    C)  \[\frac{\sqrt{7}}{3}\]    

    D)  \[\sqrt{7}\]

    Correct Answer: B

    Solution :

    : Here, Refractive index of water, \[{{n}_{w}}=\frac{4}{3}\] \[\sin C=\frac{1}{{{n}_{w}}}=\frac{1}{\left( \frac{4}{3} \right)}=\frac{3}{4}\] \[\therefore \] \[\tan C=\frac{3}{\sqrt{7}}\] From figure, \[\tan C=\frac{R}{\sqrt{7}}\] Equating (i) and (ii), we get \[\frac{3}{\sqrt{7}}=\frac{R}{\sqrt{7}}\] or \[R=3cm\]


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