CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    In Youngs experiment, the third bright band for light of wavelength coincides with the fourth bright band for another source of light in the same arrangement. Then the wavelength of second source is

    A) \[3600\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B)        \[4000\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C)        \[5000\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D)        \[4500\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    E)                 \[5500\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: D

    Solution :

    \[x=\frac{mD{{\lambda }_{1}}}{d}=\frac{(m+1)D{{\lambda }_{2}}}{d}\] \[\Rightarrow \]\[3\times 6000=4{{\lambda }_{2}}\] Or           \[{{\lambda }_{2}}=\frac{3\times 6000}{4}=4500{\AA}\]


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