VMMC VMMC Medical Solved Paper-2010

  • question_answer
    Angular width of central maxima in the Fraunhofers diffraction pattern is measured. Slit is illuminated by the light of wavelength 6000\[\overset{0}{\mathop{A}}\,\]. If slit is illuminated by light of another wavelength, angular width decreases by 30%. Wavelength of light used is

    A)  3500\[\overset{0}{\mathop{A}}\,\]       

    B)         4200\[\overset{0}{\mathop{A}}\,\]       

    C)         4700\[\overset{0}{\mathop{A}}\,\]       

    D)         6000\[\overset{0}{\mathop{A}}\,\]

    Correct Answer: B

    Solution :

    The condition for minima is given by \[d\sin \theta =n\lambda \] For n = 1, we have \[d\,\sin \theta =\lambda \] If angle is small, then \[\sin \theta =\theta \] \[\Rightarrow \]               \[d\theta =\lambda \] Half angular width \[\theta =\frac{\lambda }{d}\] Full angular width \[2\theta =2\frac{\lambda }{d}\] Also       \[\omega =\frac{2\lambda }{d}\] \[\therefore \]  \[\frac{\lambda }{\lambda }=\frac{\omega }{\omega }\]or \[\lambda =\lambda \frac{\omega }{\omega }\] or            \[\lambda =6000\times 0.7\]                 \[=4200\overset{\text{o}}{\mathop{\text{A}}}\,\]


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