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Phase shift of light incident on 2 slits

  1. Apr 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Light of wavelength λ = 450 nm is incident upon two thin slits that are separated by a distance d = 25 μm. The light hits a screen L = 2.5 m from the screen. It is observed that at a point y = 2.8 mm from the central maximum the intensity of the light is I = 55 W/m2.

    a) Write an equation for the phase shift between the light from the two slits at the observation point in terms of the given variables.
    ANSWER: Φ = 2πdy/(λL)

    b) For the given data, what is the phase shift, in radians?
    ANSWER: 0.3909 radians

    c) What is the intensity of the light at the two slits Io in W/m2?

    2. Relevant equations
    Small angle approximation: y = mλ/d (for bright fringes)
    Young's equation: Φ = kdsinθ
    Path Difference Equation: ΔX = λΔΦ/2π
    3. The attempt at a solution
    I've worked out both of the first two parts, and the answers are correct. I'm not sure how to get the original intensity though. I tried dividing the intensity I at y by the phase shift Φ, which yielded 140.82 W/m2, but for some reason that's wrong.
     
  2. jcsd
  3. Apr 30, 2016 #2
    There exists a formula giving the intensity of the interference pattern as a function of the phase shift. It goes like this (for 2 slits) :

    I=4Iocos2(Φ/2)
     
  4. May 2, 2016 #3
    Is there a way to derive that?
     
  5. May 2, 2016 #4
    Well, you know that light is an oscillating wave made of an electric and magnetic field. When you add two waves, like in Young's experiment, you could look at it as if you add up the electric fields of the waves. So add :

    E1sint(wt+Φ1) and E1sin(wt+Φ2)

    With a little work, you end up with :

    E2=2E1cos(ΔΦ/2)sin(wt+ ((Φ12)/2)

    So you know that the amplitude of this electric field is 2E1cos(ΔΦ/2).
    We almost have it. You also know, maybe from your courses on mechanical waves, that power is proportional to the square of the amplitude. We have Io=k*E12

    Therefore, I=k*E22=k*4E12cos2(ΔΦ/2)

    Finally, because Io=k*E12, I=4Iocos2(ΔΦ/2)
     
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