RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • 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:\[d=slit-width=2\lambda \] \[\therefore \]  \[2\lambda \sin \theta =n\lambda \] \[\Rightarrow \]               \[2\sin \theta =n\] 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\[5\].


You need to login to perform this action.
You will be redirected in 3 sec spinner