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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 1

  • question_answer
    In Young's double slit experiment, when the slit plane is illuminated with light of wavelength\[{{\lambda }_{1}}\], it was observed that point P is closest point from central maximum O, where intensity was 75% the intensity at O. When the light of wavelength \[{{\lambda }_{2}}\] is used, point P happens to be the nearest point from O where intensity is 50% of that at O. Find the ratio \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\] .

    A) \[\frac{1}{3}\]                                  

    B) \[\frac{1}{2}\]

    C) \[\frac{3}{2}\]          

    D) \[\frac{4}{5}\]

    Correct Answer: C

    Solution :

    [c] \[{{\left( \frac{{{I}_{P}}}{{{I}_{\max }}} \right)}_{{{\lambda }_{1}}}}={{\cos }^{2}}\frac{\phi }{2}=0.75\] [\[\phi =\]phase difference between two waves arriving at P] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( \frac{\phi }{2} \right)=\frac{\sqrt{3}}{2}\]             \[\phi =\frac{\pi }{3}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{2\pi }{{{\lambda }_{1}}}{{(\Delta x)}_{1}}=\frac{\pi }{3}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{(\Delta x)}_{1}}=\frac{{{\lambda }_{1}}}{6}\]              ?. (i) Similarly \[{{\left( \frac{{{I}_{P}}}{{{I}_{\max }}} \right)}_{{{\lambda }_{2}}}}={{\cos }^{2}}\frac{\phi '}{2}=0.5\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\cos \frac{\phi '}{2}=\frac{1}{\sqrt{2}}\]             \[\phi '=\frac{\pi }{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{(\Delta x)}_{2}}=\frac{{{\lambda }_{2}}}{4}\]                ? (ii) Because \[{{(\Delta x)}_{1}}={{(\Delta x)}_{2}}\]             \[\frac{{{\lambda }_{1}}}{6}=\frac{{{\lambda }_{2}}}{4}\Rightarrow \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{3}{2}\] 


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