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JEE Main & Advanced AIEEE Solved Paper-2007

  • question_answer
    In a Young's double-slit experiment, the intensity at a point where the path difference is\[=\frac{T}{4}=\frac{2}{4}=0.5s\] (\[\pi /2\]being the wavelength of the light used), is\[{{v}_{a}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{({{10}^{-3}})}{OA}\]. If\[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{({{10}^{-3}})}{\sqrt{{{(\sqrt{2})}^{2}}+{{(\sqrt{2})}^{2}}}}\]denotes the maximum intensity, \[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{OB}\]is equal to       AIEEE  Solved  Paper-2007

    A)  \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{2}\]                     

    B)         \[{{V}_{A}}-{{V}_{B}}=0\]                           

    C)         \[=\frac{\frac{1}{2}qV}{qV}=\frac{1}{2}\]                            

    D)         \[I={{I}_{0}}(1-{{e}^{-t/\tau }})\]

    Correct Answer: D

    Solution :

    As phase difference\[I\]path difference i.e.,        \[{{M}_{O}}\] As          \[l={{l}_{\max }}\,{{\cos }^{2}}(\phi /2)\] or           \[\frac{l}{{{l}_{0}}}\,={{\cos }^{2}}\,(\phi /2)\] or            \[\frac{l}{{{l}_{0}}}\,={{\cos }^{2}}\,(\pi /6)\,\]     (where,\[{{l}_{\max }}\,={{l}_{0}}\])                                 \[{{M}_{O}}{{c}^{2}}\]


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