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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • 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 at a distance of 80 cm from the slit The length is \[6000\overset{\text{o}}{\mathop{\text{A}}}\,\].The slit width in (mm) is about:

    A)  \[0.576\]            

    B)         \[0.348\]            

    C)         \[0.192\]            

    D)         \[0.096\]

    E)  \[0.048\]

    Correct Answer: C

    Solution :

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


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