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VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    In Youngs double slit experiment, the 10th maximum of wavelength \[{{\lambda }_{1}}\]is at a distance of \[{{y}_{1}}\] from the central maximum- When the wavelength of the source is changed to \[{{\lambda }_{2}}\], 5th maximum is at a distance of \[{{y}_{2}}\] from its central maximum. The ratio \[\left( \frac{{{y}_{1}}}{{{y}_{2}}} \right)\] is

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

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

    C)  \[\frac{{{\lambda }_{1}}}{2{{\lambda }_{2}}}\]            

    D)  \[\frac{{{\lambda }_{2}}}{2{{\lambda }_{1}}}\]  

    Correct Answer: A

    Solution :

    Position fringe from central maxima \[{{y}_{1}}=\frac{n{{\lambda }_{1}}D}{d}\] Given,            \[n=10\] \[\therefore \] \[{{y}_{1}}=\frac{10{{\lambda }_{1}}D}{d}\]         ?.(i) For second source \[{{y}_{2}}=\frac{5{{\lambda }_{2}}D}{d}\]               ??(ii) \[\therefore \] \[\frac{{{y}_{1}}}{{{y}_{2}}}=\frac{d}{\frac{5{{\lambda }_{2}}D}{d}}\] \[\Rightarrow \] \[\frac{{{y}_{1}}}{{{y}_{2}}}=\frac{2{{\lambda }_{1}}}{{{\lambda }_{2}}}\]


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