JCECE Medical JCECE Medical Solved Paper-2008

  • question_answer
    The maximum number of possible interference maxima for slit-separation equal 'to twice the wavelength in Young?s 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 \[d=\text{slit}-\text{width}=2\lambda \] \[\therefore \] \[2\lambda \sin \theta =n\lambda \] \[\Rightarrow \] \[2\sin \theta =n\] The maximum value of \[\sin \theta \]is 1, hence, \[n=2\times 1=2\] Thus, Eq. (i) must be satisfied by 5 integer values ie, -2, -1, 0,1, 2. Hence, the maximum number of possible interference maxima is 5.


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