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Path difference in an interference device

  1. Feb 15, 2016 #1
    1. The problem statement, all variables and given/known data

    In a Young’s two-slit experiment a piece of glass with an index of refraction and a thickness is placed in front of the upper slit. (a) Describe qualitatively what happens to the interference pattern. (b) Derive an expression for the intensity of the light at points on a screen as a function of n, L and θ. Here θ is the usual angle measured from the center of the two slits. (c) From your result in part (b) derive an expression for the values of θ that locate the maxima in the interference pattern.

    2. Relevant equations

    3. The attempt at a solution

    (Sorry my bad English). Let φ be the phase difference between the two waves. When φ is equal to 2π the path difference dsinθ + d2 is equal to one wavelength. I've not found a relationship between L, n and the angles of incidence and refraction. So I've considered simply that the extra distance d2 should be proportional to L and n. When n is equal to zero, the light doesn't bend and the path difference is simply dsinθ. The larger value of n, the greater distance d2. So:
    2π/φ = [dsinθ + L(n - 1)] / λ
    Is it a correct assumption?

    See my sketch below:
    Last edited: Feb 15, 2016
  2. jcsd
  3. Feb 15, 2016 #2
    i wonder when can n=0 if n is something like refractive index of the glass it can not go to zero?
  4. Feb 15, 2016 #3
    I mean when n is equal to 1 :biggrin:

    I think n = 0 means that the atoms will reject all the energy of the wave. Am I right?
  5. Feb 15, 2016 #4
    what is n when expressed as related to speed of light in various medium? is it a number or has some units?
    when we do such experiments what happens when you place a material in one of the beams i.e. after the slit.?
    pl. see a class XI-XII physics text book= its a common example- or see Principles of Physics by Sears,Young and Zemansky

  6. Feb 15, 2016 #5


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    In "front" of the slit might be interpreted as placing the glass plate on the left of the slit as shown below. That way, you don't need to worry about refraction. You can think of the light that passes through the plate as traveling the distance d2 = L through the plate rather than some "slanted" distance.

    dsinθ + d2 is not the effective path difference. The first term is ok, but d2 is not the path difference caused by the plate. You need to compare the number of wavelengths of light that occur over a distance L inside the plate with the number of wavelengths of light that occur over a distance L outside the plate.

    Your expression 2π/φ = [dsinθ + L(n - 1)] / λ is almost correct. Is your fraction on the left sided of the equation upside down? Also, people usually take θ to be positive for angles above the horizontal. If you do that, then you need to check your signs on the right side of the equation.

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  7. Feb 15, 2016 #6
    Okay, but in your sketch with the glass on the left side, the thickness L would not affect the path of the light, because it would not change its direction. Is it correct? If so, we wouldn't need to include the L and n in the result, but it doesn't satisfy the problem.
    Last edited: Feb 15, 2016
  8. Feb 15, 2016 #7


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    The glass will produce a phase difference between the light wave that passes through the top slit and the wave that passes through the bottom slit. You can think of this as due to the shortening of the wavelength when the light is inside the glass.
  9. Feb 15, 2016 #8
    Ok. I'll try to solve considering it soon.

    I thought when d2 + dsinθ = λ, φ should be equal to 2π. Would it be so if d2 were part of the path difference?

    Oh yes, I wrote it wrong. As I mentioned in the quote 2#, I thought 2π is proportional to φ when d2 + dsinθ is equal to λ.
    Last edited: Feb 15, 2016
  10. Feb 16, 2016 #9
    I've tried solving this problem several times and don't get the solution. I know that the light travels more slowly on the glass but don't know how to relate it with the phase angle.
  11. Feb 16, 2016 #10
    first you should try to define the phase angle- relate it to path length and then see how the in turn it can be related to speed of the wave.
  12. Feb 16, 2016 #11
    Wave that passes through upper slit:
    time interval for passing through glass = Ln / c;
    time interval due additional distance on the right side = dsinθ / c;
    time for reaching the screen = t.

    Wave that passes through lower slit:
    time interval for passing through the same distance L = L / c;
    time for reaching the screen = t.

    Time difference between the two waves: (1/c) (Ln + dsinθ - L).
    Now, if this time difference multiplied by the speed of light (additional distance traveled) equals one wavelength, then φ will be equal to 2π.

    Wow... I finally get it. But what if the glass plate were on the right side? Why the angles of incidence and refraction doesn't matter here?
  13. Feb 16, 2016 #12
    try to think over it and do the calculation by taking anguluar deflection /transmission of light- then calculate the time taken to traverse that distance- again do the difference to get path difference and convert it to phase difference- you have found out in doing above calculations so many times that a phase difference due to change in path say dL is equivalent to (2.Pi)/wavelengrth times dL- but due to refracting medium the wavelength has changed . so in new calculation one must use it for the new path inside.then see....
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