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KVPY Sample Paper KVPY Stream-SX Model Paper-3

  • question_answer
    In a YDSE both slits produce equal intensities on the screen. A 100% transparent thin film is placed in front of one of the slits. Now the intensity of the geometrical centre of system on the screen becomes 75% of the previous intensity. The wavelength of the light is \[6000\,\overset{{}^\circ }{\mathop{A}}\,\] and \[{{\mu }_{film}}=1.5.\] The thickness of the film cannot be:

    A) \[0.2\,\mu \,m\]             

    B) \[1.0\,\mu \,m\]

    C) \[1.4\,\mu \,m\]             

    D) \[1.6\,\mu \,m\]

    Correct Answer: D

    Solution :

    \[I={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos f\]
    \[{{I}_{\max }}={{(\sqrt{{{I}_{0}}}+\sqrt{{{I}_{0}}})}^{2}}=4\,{{I}_{0}}\]
    \[I=0.75\,{{I}_{\max }}=3{{I}_{0}}\]
    So \[3{{I}_{0}}={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos f\]
    \[\cos f=\frac{1}{2}\]
    \[f=\frac{\pi }{3},\] \[2\pi -\frac{\pi }{3},\]\[2\pi +\frac{\pi }{3},\]\[4\pi -\frac{\pi }{3}\]
    \[f=\frac{\pi }{3},\] \[\frac{5\pi }{3},\]\[\frac{7\pi }{3},\] \[\frac{11\pi }{3},\] ??.
    Path difference \[Dx=\frac{\lambda }{2\pi }\phi =(m-1)\,t\]
    \[t=\frac{0.6}{\pi }\phi \,\mu m\]= 0.2 mm, 1.0 mm, 1.4 mm, 2.2 mm,


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