Chhattisgarh PMT Chhattisgarh PMT 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:  \[d=slit-width=2\lambda \] \[\therefore \]  \[2\lambda \,\,\sin \,\theta =n\lambda \] \[\Rightarrow \]               \[2\sin \theta =n\] The maximum value of sin 9 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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