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Homework Help: Distance between each double-slit and the light band.

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data
    http://img225.imageshack.us/img225/3118/photo1587hx4.jpg [Broken]
    This #74 is on the SAT Practice Subject Test in Physics, which is also the Physics test administered in January 2003.


    2. Relevant equations
    I don't even know how to begin this problem.


    3. The attempt at a solution
    I know Young's double-slit equation: nλL=dx
    But I don't know any equation for the 'x' and 'y' in the diagram, the distance from the actual slits to the band other than the Pythagorean theorem.
    I tried solving with those, and I get a difference, but it's not one of the provided answers.
    In my attempt I called the slits A (top) and B (bottom) (since 'x' is the fringe distance), and the distances (x and y in the diagram) Za and Zb.
    The system of equations I tried to manipulate was:
    Za2=L2 + (d/2 - x)2
    Zb2=L2 + (d/2 + x)2
    nλL=dx
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 30, 2008 #2

    alphysicist

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    Homework Helper

    That is the double slit equation for small angles; the original double slit equation is:

    [tex]
    d \sin\theta= n \lambda
    [/tex]

    Look at how this equation was derived in your textbook; what does [itex]d\sin\theta[/itex] represent?
     
  4. Oct 30, 2008 #3
    I don't have a textbook, I'm just going over the questions I missed online.
    I've gotten all of them except this one. If you could just point me in the right direction, I would be most grateful.
     
  5. Oct 30, 2008 #4

    alphysicist

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    Homework Helper

    Look at the diagram at the top of this page (just ignore the lens in the middle of the diagram):

    http://www.physics.hmc.edu/courses/Ph51/two_slit.html

    Right next to the slits, you can see a part labeled [itex]d\sin\theta[/itex]. If you consider that part, you can see that [itex]d\sin\theta[/itex] is how much more that one light ray has to travel than the other; it's that path length difference between the two paths.

    For constructive interference (maximum intensity), the equation that describes it is the one in my last post: [itex]d\sin\theta = n \lambda[/itex]. Putting both of these things together, do you see what the answer to the problem is now?
     
  6. Oct 30, 2008 #5
    Thank you, this helps a lot.
     
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