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Absorbing Material Over a Slit in Deriving Double Slit Fraunhofer Interfernce Pattern

  1. Apr 5, 2010 #1
    In class we have been deriving the Fraunhofer interference pattern for a double slit experiment. I now have a question of having to derive it if one of the slits is covered with an absorbing material. I have got the equation out but I am missing a constant K...

    1. The problem statement, all variables and given/known data

    We have two identical slits of width b whose centres are separated by a distance a. Now suppose that one slit is covered with an absorbing material, so that it transmits less light than the other slit.Assume that the absorption only affects theamplitude of the light transmitted by the affected slit, multiplying the electric field amplitude by a factor c, but does not affect the phase. It can be shown that the Fraunhofer intensity pattern is given by:

    I([tex]\theta[/tex]) = [[tex]\frac{I(0)}{K}[/tex]][([tex]\frac{Sin(\beta}{\beta}[/tex])]2[ 1 + c2 + 2cCos(2[tex]\alpha[/tex])] That alpha is not ment to be superscript but I cant get it out.

    (a) What is the value of K?
    (d) Derive the above equation. Hint: Assume that the unaffected slit is centred at the origin, and that the other slit is centred at x = a. Stay in complex notation as long as possible.

    2. Relevant equations

    Where K is a constant
    [tex]\beta[/tex] = ub/2 = (1/2)kbsin([tex]\theta[/tex])

    [tex]\alpha[/tex] = ua/2 = (1/2)kasin([tex]\theta[/tex])


    3. The attempt at a solution

    (a) I think this is relatively simple. I let [tex]\theta[/tex] = 0 into the equation above. Then used Le Hopitals rule/ looked at the sinc function and found K = (1 + c)2

    (b) Now the problem here is that I can derive the above equation with no problems but I am missing the constant K!! Do I introduce it as a normalising factor? I dont know where it comes in :(

    Here is the summary of my derivation

    E(u) = A[tex]\int[/tex] eiux dx + Ac[tex]\int[/tex] eiux dx

    Where the first integrand has goes from -b/2 to b/2 and the second intergrand goes from a-(b/2) to a+(b/2)

    Integrating and factorising I get

    (A/iu)(eiu(b/2) - e-iu(b/2))(1+ceiua)

    Now using Euler's Formula and introducing the idenity [tex]\beta[/tex]

    Ab([tex]\frac{Sin\beta}{\beta}[/tex](1 + Ceiua)

    Now squaring to get intenisty/irradiance, making sure to muliply by the complex conjugate and using Eulers formula to introduce Cosine and Alpha

    I = A2b2([tex]\frac{Sin\beta}{\beta}[/tex])2(1 + c2 + 2cCos(2[tex]\alpha[/tex]) Where alpha is not in superscipt.

    So assuming I(0) = A2b2 I have the above equation without K!!

    So I dont know how I am going to get that (1 + c)2 factor in there??!

    Maybe I(0) = (1 + c)2A2b2 ?? But then the absorbing material wont affect the amplitude would it?

    Thanks for any help!! :)
     
  2. jcsd
  3. Apr 6, 2010 #2
    Re: Absorbing Material Over a Slit in Deriving Double Slit Fraunhofer Interfernce Pat

    I guess no one can help :( I am going to run with I(0) = A2b2(1+c)2 I find this when I put [tex]\theta[/tex]=0. Thus I have to introduce K as a denomiator to cancel the (1+c)2.

    The only problem I have with that is that I swear it looks like it will cancel c's effect on the amplitude...
     
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