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VMMC VMMC Medical Solved Paper-2014

  • question_answer
    Minimum light intensity that can be perceived by normal human eye is about \[{{10}^{-10}}\text{Wb}\,{{\text{m}}^{-2}}\]. What is the minimum number of photons of wavelength 660 nm that must enter the pupil in Is for one to see the object? Area of cross-section of the pupil is \[{{10}^{-4}}\,{{\text{m}}^{2}}.\]

    A) \[3.3\times {{10}^{2}}\]                

    B) \[3.3\times {{10}^{3}}\]

    C) \[3.3\times {{10}^{4}}\]         

    D) \[3.3\times {{10}^{5}}\]

    Correct Answer: C

    Solution :

    Intensity,  $l={{10}^{-10}}Wb{{m}^{-2}}={{10}^{-10}}J{{s}^{-1}}{{m}^{-2}}.$ Let the number of photons required be n. Then,$\frac{nhv}{{{10}^{-4}}}={{10}^{-10}}$ Hence, $n={{10}^{-10}}\times {{10}^{-4}}/hv={{10}^{-14}}\frac{\lambda }{hc}$ $=\frac{{{10}^{-14}}\times 660\times {{10}^{-9}}}{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}=3.3\times {{10}^{4}}$


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