Coherent light with wavelength 590 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance

Question

Coherent light with wavelength 590 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe (the first dark fringe next to a central maximum) be observed at this same point on the screen?
Express your answer in micrometers (not in nanometers).

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Thái Dương 3 years 2021-08-21T19:20:09+00:00 1 Answers 31 views 0

Answers ( )

  1. giabao
    0
    2021-08-21T19:22:00+00:00
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    Answer:

    λ’ = 392nm

    Explanation:

    In order to find the wavelength of the second light, you first calculate the separation between the slits, by using the following formula, for the distance to the central peak of the bright fringes:

    y_m=\frac{m\lambda D}{d}      (1)

    m: order of the bright fringe = 1

    λ: wavelength of the light = 590nm = 590*10^-9m

    D: distance from the slits to the scree = 3.00m

    d: separation between the slits =

    y = distance to the central peak of the first fringe = 4.84mm = 4.84*10-3 m

    You solve the equation (1) for d:

    d=\frac{m\lambda D}{y_m}=\frac{(1)(590*10^{-9}m)(3.00m)}{4.84*10^{-3}m}\\\\d=3.65*10^{-4}m

    Next, you have for the first dark fringe, the following formula:

    y_1=(1+\frac{1}{2})\frac{\lambda' D}{d}     (2)

    λ’: wavelength of the second light

    You solve the previous equation for λ’ and replace the values of the other parameters:

    y_1=\frac{3}{2}\frac{\lambda'D}{d}\\\\\lambda'=\frac{2dy_1}{3D}=\frac{2(3.65*10^{-4}m)(4.84*10^{-3}m)}{3(3.00m)}\\\\\lambda'=3.92*10^{-7}m=392nm

    The, a light with a wavelength of 392nm has its first dark fringe in the same position of the first bright fringe of the light of 590nm

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