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AMU Medical AMU Solved Paper-2012

  • question_answer
    A parallel beam of monochromatic light is incident normally on narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is

    A)  zero                                     

    B)  \[\pi /2\]

    C)  \[\pi \]                                

    D)  \[2\pi \]

    Correct Answer: C

    Solution :

                     The phase difference \[(\phi )\] between the wavelets from the top edge and bottom edge of the slit is                 \[\phi =\frac{2\pi }{\lambda }(d\sin \theta )\]where d is the width. The first minima of the diffraction pattern occurs at                 \[\sin \theta =\frac{\lambda }{d}\], so,                 \[\phi =\frac{2\pi }{\lambda }\,\left( d\times \frac{\lambda }{d} \right)\]                 \[=2\pi \]


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