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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    At two points P and Q on a screen in Young's double slit experiment, waves from slits \[{{S}_{1}}\] and \[{{S}_{2}}\] have a path difference of 0 and \[\frac{\lambda }{4}\]respectively. The ratio of intensities at P and Q will be:     AIEEE  Solved  Paper (Held On 11 May  2011)

    A)  \[2:1\]

    B) \[\sqrt{2}:1\]

    C) \[4:1\]

    D) \[3:2\]

    Correct Answer: A

    Solution :

                                    \[\Delta {{x}_{1}}=0\]                 \[\Delta \phi ={{0}^{o}}\]                 \[{{I}_{1}}={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos {{0}^{o}}=4{{I}_{0}}\]                 \[\Delta {{x}_{2}}=\frac{\lambda }{4}\]                 \[\Delta \theta =\frac{2\pi }{\lambda }.\frac{\lambda }{4}=\left( \frac{\pi }{2} \right)\]                 \[{{I}_{2}}={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos \frac{\pi }{2}=2{{I}_{0}}\]                 \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{4{{I}_{0}}}{2{{I}_{0}}}=\frac{2}{1}.\]


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