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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    In Young's double slit experiment the 7th maximum with wavelength\[{{\lambda }_{1}}\]is at a distance\[{{d}_{1}}\]and that with wavelength\[{{\lambda }_{2}}\]is at distance\[{{d}_{2}}.\]Then, \[{{d}_{1}}/{{d}_{2}}=\]

    A)  \[{{\lambda }_{2}}/{{\lambda }_{1}}\]

    B)  \[{{\lambda }_{1}}/{{\lambda }_{2}}\]

    C)  \[\lambda _{2}^{2}/\lambda _{1}^{2}\]

    D)  \[\lambda _{1}^{2}/\lambda _{2}^{2}\]

    Correct Answer: B

    Solution :

     The distance from the centre for\[{{7}^{th}}\]maximum with wavelength\[{{\lambda }_{1}},\] \[{{d}_{1}}=6(D{{\lambda }_{1}}/d)\] Similarly, that for wavelength\[{{\lambda }_{2}},\] \[{{d}_{2}}=6(D{{\lambda }_{2}}/d)\] \[\therefore \] \[\frac{{{d}_{1}}}{{{d}_{2}}}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\]


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