2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-40

  • question_answer
    Two slits separated by a distance of 1 mm are illuminated with red light of wavelength \[6.5\times {{10}^{-7}}\] m. The interference fringes are observed on a screen placed 1 m from the slits. The distance between third dark fringe & the fifth bright fringe is equal to

    A) 0.65 mm                     

    B) 1.63 mm

    C) 3.25 mm                     

    D) 4.88 mm

    Correct Answer: B

    Solution :

    [b] \[\beta =\frac{\lambda D}{2d}=\frac{6.5\times {{10}^{-7}}}{{{10}^{-3}}}\times 1=0.65\times {{10}^{-3}}=0.65mm\] The distance between fifth bright fringe from third dark fringe \[=2.5\beta =2.5\times .65=1.63mm\].


You need to login to perform this action.
You will be redirected in 3 sec spinner