Two parallel slits are illuminated with monochromatic light of wavelength 567 nm. An interference pattern is formed on a screen some distanc

Question

Two parallel slits are illuminated with monochromatic light of wavelength 567 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.83 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (Hint: The angle to the fourth dark band and the angle to the first bright band are small enough that tan θ ≈ sin θ.)

in progress 0
Nick 6 months 2021-08-09T20:11:17+00:00 1 Answers 12 views 0

Answers ( )

    0
    2021-08-09T20:12:18+00:00

    Answer:

    a)1984.5nm

    b)523mm

    Explanation:

    A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference

    θ=(m+0.5)λ where m= 1,2.3….

    Where given from the question the 4th dark Fringe which will take place at m= 3

    θ=7/2y

    Where y= 567nm

    = 7/2(567)=1984.5nm

    But

    B)tan θ ≈ y/d

    And sinθ = mλ/d

    y=mλd when m= 1 which is the first bright we have

    Then y=(1× 567.D)/d

    But the distance from Central to the 4th dark Fringe is 1.83cm then

    y= 7λD/2d= 1.83cm

    D/d=(2)×(1.83×10^-2)/(7×567×10^-9)

    =92221.5

    y= (567×10^-9)× (92221.5)

    =0.00523m

    Therefore, the distance between the first and center is y1-y0= 523mm

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )