JIPMER Jipmer Medical Solved Paper-2009

  • question_answer
    The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Youngs double-slit experiment, is

    A)  infinite                

    B)         five

    C)  three                   

    D)         zero

    Correct Answer: B

    Solution :

    For possible interference maxima on the screen, the condition is \[d\sin \theta =n\lambda \]                                        ?(i) Given: \[d\sin \theta =n\lambda \] \[\therefore \]  \[2\lambda \sin \theta =n\lambda \] \[\Rightarrow \]                  \[2\sin \theta =n\] The maximum value of sin \[\theta \] is 1, hence, \[n=2\times 1=2\] Thus, Eq. (i) must be satisfied by 5 integer values i.e.,\[-2,\,-1,\,0,\,1,\,2.\] Hence, the maximum number of possible interference maxima is 5.


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