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JEE Main & Advanced JEE Main Paper (Held On 12 May 2012)

  • question_answer
    The maximum number of possible interference maxima for slit separation equal to 1. 8A,, where X is the wavelength of light used, in a Young's double slit experiment is   JEE Main Online Paper (Held On 12 May 2012)

    A) zero                                      

    B)                        3

    C)                        infinite                                 

    D)                        5

    Correct Answer: B

    Solution :

                    As \[\sin \theta =\frac{n\lambda }{d}\]and\[\sin \theta \] cannot be \[1\] \[\therefore \]\[1=\frac{n\lambda }{1.8\lambda }\]or \[n=1.8\] Hence maximum number of possible interference maxim as, \[0,\pm 1\] i.e. 3


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