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KVPY Sample Paper KVPY Stream-SX Model Paper-26

  • 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 76 % of the previous intensity. The wavelength of the light is 6000\[\overset{\text{o}}{\mathop{\text{A}}}\,\] and \[{{\mu }_{fjlm}}=1.5.\] The thickness of the film cannot be -

    A) \[0.2\text{ }\mu \text{m}\]

    B) \[1.0\text{ }\mu \text{m}\]

    C) \[1.4\text{ }\mu \text{m}\]

    D) \[1.6\text{ }\mu \text{m}\]

    Correct Answer: D

    Solution :

    \[I={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos f\]
    \[{{I}_{\max }}={{(\sqrt{{{I}_{0}}}+\sqrt{{{I}_{0}}})}^{2}}=4\,\,{{I}_{0}}\]
    \[{{I}_{0}}=0.75\,\,{{I}_{\max }}=3\,\,{{I}_{0}}\] 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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