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Interference in thin slits

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    A very thin sheet of brass contains two thin parallel slits. When a laser beam shines on these slits at normal incidence and room temperature (20.0 ), the first interference dark fringes occur at 32.5 from the original direction of the laser beam when viewed from some distance.
    If this sheet is now slowly heated up to 135 , by how many degrees do these dark fringes change position? Coefficient of linear expansion for brass .Ignore any effects that might occur due to change in the thickness of the slits. (Hint: Since thermal expansion normally produces very small changes in length, you can use differentials to find the change in the angle.)
    Express your answer using two significant figures.

    2. Relevant equations



    3. The attempt at a solution

    I first tried to find delta d, the distance between the two paths

    I found it to be equal to dsin(32.5), where d is the distance between the 2 slits

    I think that this is equal to

    dsin(32.5)=λ/2 (because it asks for dark fringes so destructive interference)

    Here is where I start to get lost.

    I think that the addition of heat will cause the distance to change to

    d2=d(1+(α)ΔT) where i used α as coefficient of linear expansion

    I then tried to say that

    dsin(32.5)=d(1+(α)ΔT)sinθ

    However I am stuck because i do not know two variables, d and θ.

    Did I make a mistake or do i have the wrong method?

    Can someone help me out please?

    Thanks
     
  2. jcsd
  3. Jan 16, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    Try following the hint about taking differentials. Unfortunately "d" is the symbol for the differential operator and it's also the symbol for the distance between the slits. Suppose you use "w" for the distance between the slits. Then your equation for the first min is

    ## w sin\theta=\lambda/2##

    As the temperature changes both ##w## and ##\theta## change while ##\lambda## remains constant. What expression do you get if you take the differential of both sides of ## w sin\theta=\lambda/2## ? Can you sove it for ##d\theta##?
     
    Last edited: Jan 16, 2013
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