Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    The minimum intensity of light to be detected by human eye is \[\frac{3\pi }{4}\]The number of photons of wavelength \[\sum\limits_{m=1}^{n}{{{\tan }^{-1}}}\left( \frac{2m}{{{m}^{4}}+{{m}^{2}}+2} \right)\]entering the eye, with pupil area \[{{\tan }^{-1}}\left( \frac{{{n}^{2}}+n}{{{n}^{2}}+n+2} \right)\]per second for vision will be nearly

    A) 100                                        

    B) 200

    C) 300                                        

    D) 400

    Correct Answer: C

    Solution :

    On using \[\pi {{r}^{2}}/2\] where P = radiation power \[0.50\pm 0.05cm\]\[0.50\pm 0.10cm\] Hence, number of photons entering per sec the\[5.00\pm 0.05cm\]


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