JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    In a Young's double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is: [JEE MAIN Held on 07-01-2020 Evening]

    A) 3.9 mm

    B) 6.9 mm

    C) 5.9 mm

    D) 4.9 mm

    Correct Answer: C

    Solution :

    [c] Fringe-width, \[\omega =\frac{D}{d}\,\lambda =\frac{1.5}{0.15\times {{10}^{-3}}}\times 589\times {{10}^{-9}}\,m\]       \[=589\times {{10}^{\,2}}\text{ }mm=5.89\text{ }mm\] \[~\approx 5.9\text{ }mm\]    

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