CET Karnataka Medical CET - Karnataka Medical Solved Paper-2013

  • question_answer
    \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] are used to illuminate the slits. \[{{\beta }_{1}}\] and \[{{\beta }_{2}}\] are the corresponding fringe widths. The wavelength \[{{\lambda }_{1}}\] can produce photoelectric effect when incident on a metal. But the wavelength \[{{\lambda }_{2}}\] cannot produce photoelectric effect. The correct relation between \[{{\beta }_{1}}\] and \[{{\beta }_{2}}\] is

    A)  \[{{\beta }_{1}}<{{\beta }_{2}}\]

    B)  \[{{\beta }_{1}}={{\beta }_{2}}\]

    C)  \[{{\beta }_{1}}>{{\beta }_{2}}\]

    D)  \[{{\beta }_{1}}\ge {{\beta }_{2}}\]

    Correct Answer: A

    Solution :

    : Fringe width,  \[\beta =\frac{\lambda D}{d}\] \[\therefore \]  \[B\propto \lambda \] \[\therefore \] \[\frac{{{\beta }_{1}}}{{{\beta }_{2}}}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\] \[{{\lambda }_{1}}\] can produce photoelectric effect and \[{{\lambda }_{2}}\] cannot produce photoelectric effect. \[\Rightarrow \] \[{{\lambda }_{1}}<{{\lambda }_{2}}\]    \[\therefore \]    \[{{\beta }_{1}}<{{\beta }_{2}}\]


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