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Interference in Soap films

  1. May 9, 2008 #1
    [SOLVED] Interference in Soap films

    1. The problem statement, all variables and given/known data
    The sketch View Figure shows a thin film of soapy water of uniform thickness t and index of refraction n = 1.33 suspended in air. Consider rays 1 and 2 emerging from the film: Ray 1 represents the reflected wave at the air-film interface and ray 2 represents the wave that reflects off the lower surface of the film.

    Which of the following statements correctly describes rays 1 and 2 and their path difference?
    Check all that apply.

    a.Ray 1 undergoes a half-cycle phase shift. Ray 2 does not undergo any phase shift.
    b.Both rays 1 and 2 undergo a half-cycle phase shift.
    c.Ray 2 undergoes a half-cycle phase shift. Ray 1 does not undergo any phase shift.
    d.The path difference between rays 1 and 2 is about t.
    e.There is no path difference between the two rays.
    f.The path difference between rays 1 and 2 is about 2t.

    http://session.masteringphysics.com/problemAsset/1020409/16/1020409.jpg

    2. Relevant equations

    2nd = (m + 0.5) (lambda)

    3. The attempt at a solution

    I know there is a 180 degree phase change at the first interface and no change at the second. Using my physics textbook it says with thickness d, for case of normal incidence, extra length is 2d.

    I am still confused at what parts are true.
     
  2. jcsd
  3. May 10, 2008 #2
    i think the answers are b, and f

    could someone please verify?
     
  4. May 10, 2008 #3
    ok i solved the first part, but having difficulty with the 2nd part.

    When red light is directed normal to the surface of the soap film, the soap film appears black in reflected light. What is thinnest the film can be?

    Let lambda_0 be the wavelength of red light in air.

    a. {lambda_0}/{2}
    b. 3{lambda_0}/{5.32}
    c. {lambda_0}/{2.66}
    d. {lambda_0}/{1.33}
    e. {lambda_0}/{4}

    Ok so I know we have to find destructive interference which occurs according to 2nd = (m+ 0.5) lambda. So I have to solve for the wavelength of red light in this situation and then find the appropriate thickness it can be. I am having trouble with this one actually.
     
  5. May 11, 2008 #4
    just a hunch but I thought about plugging in some numbers given in the problem:

    2(1.33)d = (m + 0.5)(lamda)

    d = ((m + 0.5)lamda)/(2.66)

    if we move the 0.5 down, and assign m = 3 I get choice B = (3*lambda)/(5.32)

    does this logic make sense? it seems to fit the equation?
     
  6. May 11, 2008 #5

    alphysicist

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    The condition for this case will be

    [tex]
    2 n d = m \lambda_0
    [/tex]
     
  7. May 11, 2008 #6
    ok so using that... the thinnest it can be is when m = 1, and therefore

    d = (lambda) / (2.66) ?
     
  8. May 11, 2008 #7

    alphysicist

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    That looks right to me; does it make sense?
     
  9. May 11, 2008 #8
    just curious then, the formula for constructive interference in a thin film is given by:
    2nd = (m+0.5)lambda

    and the destructive interference in a thin film is then given by:
    2nd = m(lambda)

    ?
     
  10. May 11, 2008 #9

    alphysicist

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    It depends on the number of phase reversals that take place due to reflection. (When light reflects from a material with a higher n than it has it undergoes a phase reversal, or a 180 degree phase shift.)

    If there is one phase reversal, as there is here, then 2nd=m(lambda) is the destructive condition. If there is zero or two phase reversals, then 2nd=m(lambda) is the constructive condition.

    (An example of two reversals would be a soap film on glass; the layers (air/soap/glass) would have indices (1/1.3/1.5). So the light that reflects off the soap undergoes a phase reversal, and the light that reflects off the glass also has a phase reversal.)
     
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