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JCECE Medical JCECE Medical Solved Paper-2004

  • question_answer
    In double slit experiment, the angular width of interference fringes with sodium light \[(\lambda =5890\,\overset{\text{o}}{\mathop{\text{A}}}\,)\]is \[\text{0}\text{.2}{{\text{0}}^{\text{o}}}\text{.}\] The change in wavelength required to increase the angular width by 10% will be:

    A)  zero

    B)  increased by \[6479\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C)  increased by \[589\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D)  decreased by \[589\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

     Angular fringe width \[\theta =\frac{\beta }{D}=\frac{\lambda }{2d}\] \[\therefore \] \[\frac{\theta }{\theta }=\frac{\lambda }{\lambda }\] or \[\frac{\theta -\theta }{\theta }=\frac{\lambda -\lambda }{\lambda }\] \[\therefore \] \[\Delta \lambda =\frac{\Delta \theta }{\theta }\lambda \] Given, \[\frac{\Delta \theta }{\theta }=10%,\lambda =5890\,\overset{\text{o}}{\mathop{\text{A}}}\,\] \[\Delta \lambda =\frac{10}{100}\times 5890=589\,\overset{\text{o}}{\mathop{\text{A}}}\,\] Hence, fringe width increases by \[589\overset{\text{o}}{\mathop{\text{A}}}\,.\]


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