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Derivation of Young's double slit formula problem

  1. Oct 6, 2011 #1
    Hi hi, first post ^^
    Anyway, I encountered this little problem with the derivation of Young's double slit formula. The standard derivation found on the internet and my textbook is shown here: http://schools.matter.org.uk/content/interference/formula.html [Broken]

    When deriving, a few errors had cropped up. I'll make reference to the linked diagram cause my work on paper is kinda messy:
    1) The assumption that triangle S2-Z-S1 is a right-angled triangle. This assumption is made for sinθ=λ/s.
    However, if said triangle is right-angled, then ZY≠S2Y, due to Pythagoras' theorem and some value of S2Z existing.
    Therefore, the path difference cannot be λ for the first fringe, which goes against what we learn.

    2) The assumption that θ reappears again in the triangle Y-(center of S1S2)-X. I don't see any similar triangles after error 1) is corrected. So does doing this base a working on an error which magnifies everything?

    3) Assuming θ is negligible at the final step. This one is slightly more straightforward. If θ is negligible, then there would be no path difference, then there would be no λ, no fringe, and only one central fringe.


    The major problem here is error 1). Unless the idea that sine=opposite/hypotenuses can be applied to NON-right-angled triangles, path difference x=λD/s seems flawed, at least, in terms of this method of derivation.

    So, can someone please enlighten me a little about this?


    Also, I have worked out an alternate equation that I'm pretty sure is wrong. Anyway, here it is:
    Fringe difference x
    =(a/2)+((S2Y)^2-D^2)^(1/2)
    =(a/2)+((S1Y-λ)^2-D^2)^(1/2)
    =(a/2)+Dtanβ
    All 3 are alternate derivations depending on whether you have
    a)distance S2Y
    b)distance S1Y AND λ
    c)angle β, which is the angle Y-S2-horizontal parallel to D

    My derivation comes mostly from a different right angle triangle w.r.t Pythagoras' theorem. If anyone is interested, I can scan my messy drawing to be scrutinized properly.
    I'm just pretty much certain that the derivation approximates too many things. Granted, the results are close enough to warrant the use of ≈, but =?
    Thoughts?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 6, 2011 #2
    I think it is good to think and question things when reading a derivation - as you are doing.
    Here is my contribution:
    re 1) if one takes a compass and draw an arc with centre Y and radius YZ, the arc passes through S2. Now the arc of a circle is always at rt angles with the radius. But since the angle of the sector drawn is very, very small (but it is NOT 0) then the arc is almost a straight line. Hence S1-S2 -Z is very, very close to a rt angled triangle.
    re 2) if my contribution above is accepted then there is not trouble in considering ywo similar traingles.
    re 3) Be careful here. Negligible DOES NOT mean 0 but it means very very small when compared with something else. This idea is used continually in Mathematics and Physics. For example in differentiation we use that delta x and delta y APPROACH 0 and some mathematicians even add (but not attaining the value 0).
     
  4. Oct 6, 2011 #3
    Re the alternative derivations:
    I am assuming your a is identical to s of the web link.
    Your expressions for x ( identical to w of the web link) are correct. But in practice it would be difficult to measure S2Y, or S1Y of angle beta.
     
  5. Oct 6, 2011 #4
    I see. To make life easier, approximations are used. Thanks for your comment :)
    Never considered an arc acting as a straight line.
    And yea my a is s. Neat how you saw the triangle I used.


    Though to be truthful, I hate approximations. Sticks out like a sore thumb.
    I guess I would need a really really precised ruler to get S2Y or S1Y.
     
  6. Oct 6, 2011 #5

    jtbell

    User Avatar

    Staff: Mentor

    You'd better get used to them. :smile: Lots of situations are completely unmanageable without them, unless you use a computer program to solve them numerically instead of algebraically or geometrically.
     
  7. Oct 6, 2011 #6
    In high school, we say the Earth is a sphere. That is an approximation.
    In most problems in dynamics at high school and a little bit beyond high school we usually ignore air resistance. Otherwise problems would be too difficult at that level.
    Physicists and Mathematicians continually use the first few terms of a power series as approximations for functions.Those are approximations too!
    I dare say that without the use of approximations Physics and Mathematics would not make much progress.
     
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