Manipal Medical Manipal Medical Solved Paper-2014

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

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

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

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

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

    Correct Answer: B

    Solution :

    \[d\sin \theta =n\lambda \] \[n=1,\]we have \[d\sin \theta =\lambda \] If angle is small, then \[\sin \theta =\theta \] \[d\theta =\lambda \] Half angular width \[\theta =\frac{\lambda }{d}\] Full angular width \[2\theta =\frac{\lambda }{d}\] \[\omega '=\frac{2\lambda }{d}\] \[\frac{\lambda '}{\lambda }=\frac{\omega '}{\omega }\Rightarrow \lambda '=\lambda \frac{\omega '}{\omega }\] \[\lambda '=6000\times 0.7=4200\,\overset{\circ }{\mathop{A}}\,\]


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