VIT Engineering VIT Engineering Solved Paper-2007

  • question_answer
    Following diffraction pattern was obtained using a diffraction grating using two different wavelengths \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\]. With the help of the figure identify which is the longer wavelength and their ratios.

    A)  \[{{\lambda }_{2}}\]is longer than \[{{\lambda }_{1}}\]and the ratio of the longer to the shorter wavelength is 1.5

    B)  \[{{\lambda }_{1}}\] is longer than \[{{\lambda }_{2}}\] and the ratio of the longer to the shorter wavelength is 1.5

    C)  \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] are equal and their ratio is 1.0

    D)  \[{{\lambda }_{2}}\] is longer than \[{{\lambda }_{1}}\] and the ratio of the longer to the shorter wavelength is 2.5

    Correct Answer: C

    Solution :

    The equation of nth principal maxima for wavelength X is given by \[(a+b)\sin \theta =n\lambda \] where a is the width of transparent portion and b is that of opaque portion. The width \[(a+b)\] is called the grating element. The spectral lines will overlap, i.e., they will have the same angle of diffraction if \[{{\lambda }_{1}}={{\lambda }_{2}}\] When a line of wavelength \[{{\lambda }_{1}}\] in order \[{{n}_{1}}\] coincides with a line of unknown wavelength \[{{\lambda }_{2}}\] in order \[{{n}_{2}},\] then \[{{n}_{2}}{{\lambda }_{2}}={{n}_{1}}{{\lambda }_{1}}\] or \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{{{n}_{2}}}{{{n}_{1}}}\]


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