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A double-slit (Young experiment) problem

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    In Young's double-slit interference experiment with light of wavelength [tex]\lambda[/tex], two light waves come from the slits and are superposed at the point P on the screen. These waves have the following forms:

    E1 = E0 sin(wt + π/4)
    E2 = E0 sin(wt + Φ)

    Find the possible values of Φ if we have destructive interference at the point P.

    2. Relevant equations


    3. The attempt at a solution

    I attempted the problem two different ways and I had two different results.

    Φ = 2kπ - π/4
    Φ = 2kπ + 5π/4

    Help me out please.
  2. jcsd
  3. Aug 4, 2010 #2
    Can you show us how you got that?
  4. Aug 4, 2010 #3
    Phase difference is Φ - π/4.

    So d/λ = (Φ - π/4)/2π

    d = (2k+1)/2 λ for destructive interference

    So, (2k+1)/2 λ = (Φ - π/4)/2π λ

    Φ = 2kπ + 5π/4


    The second way, I added 2 wave functions together and I equated the expression to zero.

    I found Φ = 2kπ - π/4
  5. Aug 4, 2010 #4
    How did you arrive at Φ = 2kπ - π/4?
  6. Aug 4, 2010 #5
    Let's see. The sum of the two waves is:

    E = 2E0 sin(wt + π/8 + Φ/2) cos(π/8 - Φ/2) = 0

    That is satisfied when

    wt + π/8 + Φ/2 = kπ

    wt=2πft=2π, so we can drop it.

    Φ = 2kπ - π/4
  7. Aug 4, 2010 #6
    The condition E=0 must be satisfied at ALL TIME for destructive pattern, and the bold line is not true at all time.
    Find another condition :wink:
  8. Aug 4, 2010 #7
    Hmm, thanks hikaru. How can I proceed? I have no idea.
  9. Aug 4, 2010 #8
    What about cos(π/8 - Φ/2)?
  10. Aug 5, 2010 #9
    Oh yes, I found it now. I must equate the cosine term (cos(π/8 - Φ/2)) to zero. But why exactly?
  11. Aug 5, 2010 #10
    Because of this: "The condition E=0 must be satisfied at ALL TIME for destructive pattern" :wink:
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