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J & K CET Medical J & K - CET Medical Solved Paper-2004

  • question_answer
    In Young?s double slit experiment the amplitudes of two sources are 3a and a respectively. The ratio of intensities of bright and dark fringes will be :

    A)  3 : 1                                      

    B)  4 : 1

    C)  2 : 1                                      

    D)  9:1

    Correct Answer: B

    Solution :

                    For bright fringe, the resultant amplitude is given by \[=3a+a=4a\] For dark fringe the resultant amplitude is given by \[=3a-a=2a\] \[\therefore \] \[Since,\text{ }intensity=K\text{ }{{(amplitude)}^{2}}\] \[\therefore \]\[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(4a)}^{2}}}{{{(2a)}^{2}}}=\frac{4}{1}\]


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