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MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • question_answer
    In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is 5 mm. The screen on which the diffraction pattern is displayed is at a distance of 80 cm from the slit. The wavelength is \[6000{AA}\]The slit width (in mm) is about

    A)  0.576                                   

    B)  0.348

    C)  0.192                                   

    D)  0.096

    Correct Answer: C

    Solution :

    Slit width, \[w=\frac{D\lambda }{d}\] Given,\[D=80\text{ }cm=80\times {{10}^{-2}}m,\] \[\lambda =6000{A}\text{ }=6000\times {{10}^{-10}}m,\] \[d=\frac{5}{2}mm=\frac{5\times {{10}^{-3}}}{2}m\] \[\therefore \]  \[w=\frac{80\times {{10}^{-2}}\times 6000\times {{10}^{-10}}\times 2}{5\times {{10}^{-3}}}\] \[w=0.192\times {{10}^{-3}}m=0.192\text{ }mm\]


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