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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    In Youngs double-slit experiment, the angular width of fringe formed on a distance screen is \[\text{0}\text{.}{{\text{1}}^{\text{o}}}\text{.}\] If wavelength of light is \[\text{6000}\,\overset{\text{o}}{\mathop{\text{A}}}\,,\] then spacing between the slits is :

    A)  \[3.4\times {{10}^{-4}}\,m\]                      

    B)  \[4.3\times {{10}^{-4}}m\]

    C)  \[5.4\times {{10}^{-4}}m\]                         

    D)  \[6.3\times {{10}^{-4}}m\]

    Correct Answer: A

    Solution :

                     Spacing between the slits \[d=\frac{\lambda }{\theta }\] \[=\frac{6000\times {{10}^{-10}}}{0.1\times (\pi /180)}\] \[=3.4\times {{10}^{-4}}m\]


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