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Manipal Medical Manipal Medical Solved Paper-2011

  • question_answer
    Youngs double slit experimental arrangement is shown in figure. If\[\lambda \]is the wavelength of light used and\[<{{S}_{1}}C{{S}_{2}}=\theta ,\]then the fringe width will be

    A)  \[\frac{\lambda }{\theta }\]

    B)  \[\frac{\lambda }{2\theta }\]

    C)  \[\lambda \theta \]

    D)  \[\frac{2\lambda }{\theta }\]

    Correct Answer: A

    Solution :

     Fringe width, \[\beta =\frac{D\lambda }{d}\] But here,    \[\theta =\frac{d}{D}\Rightarrow d=D\theta \] \[\therefore \] \[\beta =\frac{D\lambda }{D\theta }=\frac{\lambda }{\theta }\]


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