Haryana PMT Haryana PMT Solved Paper-2001

  • question_answer
    Light of frequency \[5\times {{10}^{14}}Hz\]is travelling in a medium of refractive index 1.5. What is its wavelength? \[(C=3\times {{10}^{8}}m/s)\]

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

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

    C) \[6000\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D) \[9000\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: A

    Solution :

                    Refractive index, \[\mu =\frac{\lambda }{{{\lambda }_{m}}}\] or            \[{{\lambda }_{m}}=\frac{\lambda }{\mu }\] Putting   \[\lambda =\frac{c}{v}\] Than      \[{{\lambda }_{m}}=\frac{c}{v\mu }\] So,          \[{{\lambda }_{m}}=\frac{3\times {{10}^{8}}}{5\times {{10}^{14}}\times 1.5}m=4000\overset{\text{o}}{\mathop{\text{A}}}\,\]


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