CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    Light of wavelength \[6000\,\overset{\text{o}}{\mathop{\text{A}}}\,\] falls on a single slit of width 0.1 mm. The second minimum will be formed for the angle of diffraction of

    A)  0.08 rad          

    B)         0.06 rad

    C)  0.12 rad          

    D)         0.15 rad

    E)  0.012 rad

    Correct Answer: E

    Solution :

    Given single slit of width \[d=0.1\text{ }mm\] \[d=0.1\times {{10}^{-3}}m\] or            \[d=1\times {{10}^{-4}}m\] Light of wavelength\[\lambda =6000{\AA}\] or           \[\lambda =6\times {{10}^{-7}}m\] The angle of diffraction                 \[\theta =\frac{n\lambda }{d}\]                 \[\theta =\frac{2\times 6\times {{10}^{-7}}}{1\times {{10}^{-4}}}\]                 \[\theta =12\times {{10}^{-3}}\] \[\theta =0.012rad\]


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