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KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • question_answer
    When a YDSE is performed using a monochromatic source of light, fringe width is found to be 1.2 mm The minimum distance above central maxima at which intensity is \[{{(3/4)}^{\operatorname{th}}}\]of the maximum value is:

    A) 0.1mm

    B) 0.2mm

    C) 0.3mm

    D) 0.4mm

    Correct Answer: B

    Solution :

    given, \[\beta =1.22\operatorname{mm}\]
    For \[{{\operatorname{I}}_{R}}=\frac{3}{4}{{I}_{0}}\]
    \[\therefore \]      \[\frac{3}{4}{{I}_{0}}=2I+21\cos \phi \]
    Or \[\frac{3}{4}(4I)=4I{{\cos }^{2}}\frac{\phi }{2}\] Or \[\cos \frac{\phi }{2}=\frac{\sqrt{3}}{2}\] Or \[\phi ={{60}^{0}}\]
    Now, \[\Delta x=\frac{\phi \times  \lambda }{2\pi }=\frac{\left( \pi /3 \right)\times \lambda }{2\pi }=\frac{\lambda }{6}\]
    or \[d\sin \theta =\frac{\lambda }{6}\] or \[d\times \frac{y}{D}=\frac{\lambda }{6}\] or \[y=\left( \frac{D\lambda }{d} \right)\times \frac{1}{6}\]\[=\frac{1.2}{6}=0.2\operatorname{mm}.\]


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