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J & K CET Medical J & K - CET Medical Solved Paper-2009

  • question_answer
    In Youngs double slit experiment, the fringe width with light of wavelength 6000 \[_{A}^{0}\] is 3 mm. The fringe width, when the wavelength of light is changed to 4000 \[_{A}^{0}\]is

    A)  3 mm                                   

    B)  1 mm

    C)  2 mm                                   

    D)  4 mm

    Correct Answer: C

    Solution :

                     Fringe width, \[\beta =\frac{\lambda D}{d}\] \[\frac{\beta }{\lambda }=\frac{D}{d}=\]constant for an experiment \[\therefore \]  \[\frac{\beta }{\lambda }=\frac{\beta }{\lambda }\] \[\Rightarrow \]               \[\frac{3}{6000}=\frac{\beta }{4000}\] \[\Rightarrow \]               \[\beta =2mm\]


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