1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Exact wave interference distances

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Finding exact distances / angles for some maximum constructive / destructive interference 'bands' for a two-source interference
    - fixed source-source distance, d
    - screen parallel with source-source line segment
    - screen fixed distance away from sources, L
    - wavelength difference for the antinode, λ (let us consider for now only whole wavelength differences i.e. antinodal lines) - the antinode considered lies on the screen (where we see the interference pattern).
    - aim is to find the distance between maximum interference and central antinodal line - on the screen.

    2. Relevant equations
    Note: please do not try to bring in the good old interference equations (which ARE really approximations, albeit good ones when used for the SUPPOSED small angles and large distance L). It defeats the purpose.

    Let the antinode's distances to each source be (x) and (x + λ) ...considering only a single wavelength difference for now...
    Let the distance across the screen from the antinode to the central antinodal line be p

    So we can form equations using Pythagoras:
    x2 = L2 + (0.5d - p)2
    (x + λ)2 = L2 + (0.5d + p)2

    so that
    [L2 + (0.5d - p)2]0.5 = [L2 + (0.5d + p)2]0.5 - λ

    3. The attempt at a solution
    I manipulated tediously getting:

    p2 = λ2(4L2 + d2 - λ2)/(4d2 - 4λ2)

    so p is found taking the square-root of this beast(???)

    This gives p (for varying L, d and some wavelength difference) with distances to which
    - the 'approximations' (i.e. the answers given using the good old equations) are decently close ... though not very satisfyingly much
    - If numbers are plugged in, for example, to check, you may confirm the relationship between the wavelength difference between sources, λ, and the distance p from the central antinodal IS NOT REALLY DIRECTLY PROPORTIONAL. This is to mean ... that in the gaps / widths (or however you consider the interference pattern) do not have uniform distances, i.e. not quite evenly spread!!! How can that be? did I do something wrong?
  2. jcsd
  3. Nov 10, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    As you realize by now, the interference pattern close to the slits is much more complicated than the simple far field approximation (Fraunhofer limit) would suggest. (Even more so when you add single slit diffraction to the mix.) I haven't checked your expression, but what should be true is that when L is large enough (and thus d/L & λ/L are both << 1) you should get that p = λL/d. I don't see your expression giving that, so I suspect an error.
  4. Nov 11, 2009 #3
    How do you check...? using limit? If so how?
    I havn't yet found errors... but then I am bad at finding errors usually, anyway...
    by the way, I simplified it to this... which won't be useful if there is an error of course...

    p = 0.5λ[(4L2)/(d2 - λ2) + 1]0.5

    this was a quadratic, I omitted the -ve answer... since it was just the -ve of this one.

    I did check on the (graphic) calculator...
    - plugged in formula with variables
    - set variables to different numbers...
    The answers found using approximation were decently close to the one the formula gave, especially when L was large (say 10000) and d and wavelength small... if I divided the answer (equivalent to our 'p') by # wavelengths, then answer got SLIGHTLY BIGGER each time I used the successive # wavelenths (i.e. say 1, 2, 3 ... wavelengths: the p divided by # wavelengths would give, say, approx 10.1, 10.2, 10.4 etc. respectively, illustrating the non-linear relationship between 'p' and wavelength difference...

    This seemed to work as I expected, so that I didn't suspect errors as much as I do normally... but if you used limits and found it wrong, I've probably done some mistake... :oops:
    Last edited: Nov 11, 2009
  5. Nov 11, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    The reason why I suspect an error is as follows. Rewrite the expression, factoring out L2 from the numerator and d2 from the denominator:
    p2 = (λ2L2/d2)(1 + d2/4L2 - λ2/4L2)/(1 - λ2/d2)

    For large L, where d/L and λ/L are both << 1, the highlighted fraction should go to 1. The numerator is well-behaved, but not the denominator.
  6. Nov 23, 2009 #5
    Are you sure you are using the formula with no omissions?
    I see you used the good old estimation, like this:

    x = nλL/d

    so that if you make (nλL/d) a multiplier of the rest of the expression, then:
    - when the values are respectively large (for L) or small (λ, d) as we said before, this part of the expression is supposed to be a good estimation (of course).
    - So the rest of the expression (0.5 square-root) should not be affecting it much i.e. should be close to multiplication identity = 1 :-)

    I had developed that formula (now has the order number, n, too):

    x = 0.5nλ[4L2/(d2 - n2λ2) + 1]0.5

    So I'd manipulate it like so (more or less what you did...)

    x = (nλL/d){[4 + (d2 - n2λ2)/L2]/[4 - n2λ2/d2]}0.5

    so since

    x = (nλL/d)

    the whole square-root following that part in the formula should go to 1 as you said... (with conditions applied...) that is if

    n = whatever (though preferably small values, whole numbers: n ≥ 0)


    {[4 + (d2 - n2λ2)/L2]/[4 - n2λ2/d2]}0.5 → 1

    and it does! (I think, for me anyway)

    I checked:
    λ = 10-6
    d = 10-7
    L = 1000
    n = 1


    {[4 + (d2 - n2λ2)/L2]/[4 - n2λ2/d2]}0.5 = 1.0050378...

    Looks okay to me... :-)

    The whole point about this formula:

    x = 0.5nλ[4L2/(d2 - n2λ2) + 1]0.5

    is that as the order number, n, increases, the estimation is less and less accurate :-) that's why even when using small number to check, n will affect result more and more as it increases.

  7. Nov 23, 2009 #6

    Doc Al

    User Avatar

    Staff: Mentor

    The problem is the behavior of λ/d in the denominator. You happened to choose values that make λ/d = 1/10, but what if you chose λ/d = 10?
  8. Nov 25, 2009 #7

    Maybe the wavelength needs to also be smaller than gap between sources(???) for estimation to apply? Is that possible?

    I don't feel as if I truly know anything now...

    Would you like to try deriving this thing using Pythagoras too? Maybe you'll get something different as I'd've done a mistake...

    cheers :-)
  9. Nov 25, 2009 #8
    Just to make sure:
    I changed the variables to their normal use in this context
    ...so that distances for any antinodal on the screen are such as that
    - if to one source one distance is p
    - the other would be longer by path difference of a whole number of wavelengths i.e. nλ
    (where n is an integer larger than zero)
    and x is the distance along the screen from our supposed antinodal to the central...

    p2 = L2 + (0.5d - x)2
    (p + nλ)2 = L2 + (0.5d + x)2

    Isolating p...

    p = [L2 + (0.5d - x)2]0.5
    p = [L2 + (0.5d + x)2]0.5 - nλ


    [L2 + (0.5d - x)2]0.5 = [L2 + (0.5d + x)2]0.5 - nλ
    Square both sides​
    L2 + (0.5d - x)2 = L2 + (0.5d + x)2 - 2nλ[L2 + (0.5d + x)2]0.5 + n2λ2
    Cancel out and simplify​
    2dx + n2λ2 = 2nλ[L2 + (0.5d + x)2]0.5
    Square both sides again​
    [dx/(nλ) + (nλ)/2]2 = L2 + (0.5d + x)2
    Expand, cancel out and simplify​
    [d2/(n2λ2)]x2 + dx + (n2λ2)/4 = L2 + dx + d2/4
    [d2/(n2λ2) - 1]x2 = L2 + d2/4 - n2λ2/4
    [d2/(n2λ2) - 1]x2 = (4L2 + d2 - n2λ2)/4
    Dividing by x-squared coefficient​
    x2 = (4L2 + d2 - n2λ2)/[4d2/(n2λ2) - 4]
    So you can get now​
    x2 = (n2λ2/4)[(4L2 + d2 - n2λ2)/(d2 - n2λ2)]
    x2 = (n2λ2/4)[4L2/(d2 - n2λ2) + 1]

    Finally... (taking square-root)

    x = (nλ/2)[4L2/(d2 - n2λ2) + 1]0.5
  10. Nov 26, 2009 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Your calculation looks good to me.
    Yes, I believe that is the case. (Although I didn't realize it until after I went through your calculation.)

    Good work! :approve:
  11. Nov 27, 2009 #10
    So this is the new two-source-interference 'fringe-distance' formula to be used and taught for all educational bodies, systems etc. from now on :biggrin:

    Well, more seriously... it occurred to me when I realised - we don't really use wavelengths even close (to the gap) in size - the visible spectrum is... what, like 300 - 800 nanometres...? who'd making gaps that size for this experiment...? I don't know for what ranges nor contexts this approximation would be used... maybe longer-wavelength-electromagnetic-radiation-interference-experimenting is done too...? So then it would seem that the estimation (which is after all an... ummm... estimation) is even less effective for such cases, where wavelength is larger than source-source distance.

    More conditions for application of the x = nλL/d approximation than one would like to mention...

    I realise also that the larger the order (n) the approximation becomes worse too - really, effectively it is exactly the same as using a larger wavelength. So If you calculate for the, say, 42nd fringe, the error would become just as large as for the second fringe using 42 times larger wavelength so even for those normal visible spectrum calculations, I want to keep n small (like, 1, 2, 3, 4...) but not ... 42 (!)

    You also mentioned sometime early single source diffraction...?
    I looked here... barrrghh! :eek:
    http://en.wikipedia.org/wiki/Diffraction_formalism" [Broken]

    how does that effect...?

    Cheers :-)
    Last edited by a moderator: May 4, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook