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AMU Medical AMU Solved Paper-2014

  • question_answer
    In a double slit experiment, the distance between slits is 5.0 mm and the slits are 1.0 m from the screen. Two interference patterns can be seen on the screen; one due to light with wavelength 480 nm and the other due to light with wavelength 600 nm. Find the separation on the screen between the third order bright fringes of the two interference patterns.

    A)                  0.072 mm          

    B)                  0.063 mm

    C)  0.037 mm          

    D)  0.019 mm

    Correct Answer: A

    Solution :

                    \[{{x}_{1}}=\frac{(2m-1)\,D{{\lambda }_{1}}}{2d}\]                 \[=\frac{(2\times 2-1)\times 1\times 480\times {{10}^{-9}}}{2\times 5\times {{10}^{-3}}}\]                 \[=3\times 48\times {{10}^{-6}}=0.144\,mm\]                 \[{{x}_{2}}=\frac{mD{{\lambda }_{1}}}{d}\]                 \[=\frac{3\times 1\times 480\times {{10}^{-9}}}{5\times {{10}^{-3}}}=288\times {{10}^{-6}}\]                 = 0.288 mm So, the third order bright fringes \[=\frac{{{x}_{2}}-{{x}_{1}}}{2}\] \[=\frac{0.288-0.144}{2}=\frac{0.144}{2}=0.072\,mm\]


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