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Waves - Diffraction - Young

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A monochromatic light source is incident normally on a diffraction grating with 600 lines per mm. The beam passes through the grating and the fringes are observed on a screen 5m away. The apparatus is then immersed in a liquid whose refractive index is 1.3. In consequence the distance between the adjacent bright fringes is reduced by 0.45m. Find the wavelength of this light source in air.

    2. Relevant equations

    3. The attempt at a solution

    I derived the formula: x = n[tex]\lambda[/tex]D / d , where D = 5m and d = 1 / (600 x 10^3) = 1.67 x 10^-6m

    then I said: x - 0.45 = (1.3[tex]\lambda[/tex] x D x n)/ d

    but there are two unknowns, and this is where I'm stuck. I have the feeling that simultaneous equations will be needed.

    Thanks for any help!
    Last edited: Jan 16, 2010
  2. jcsd
  3. Jan 16, 2010 #2
    How can I use the refractive index to solve the question?
  4. Jan 16, 2010 #3

    Doc Al

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    Staff: Mentor

    Two problems:
    (1) The distance between adjacent fringes is not reduced by 0.45 m, it is reduced to 0.45 m. In other words: x = 0.45 m.
    (2) In that formula, n is the order of the fringe, not the index of refraction.

    How does the presence of the liquid affect the wavelength of the light?
  5. Jan 16, 2010 #4
    The question clearly says BY 0.45 m.

    Yes I'm aware of this: I used n to be 1 (first order maximum). Hence I am considering the angle between n=1 and n=0.

    The liquid decreases the wavelength. does this mean: 1.3[tex]\lambda[/tex] = 1.0[tex]\lambda[/tex]' (assuming that the refractive index for air is n=1)

    Can you please help me find a method of how to tackle this question? I still don't see it unfortunately! Thanks!
    Last edited: Jan 16, 2010
  6. Jan 16, 2010 #5

    Doc Al

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    Staff: Mentor

    Hmmm... Did you edit your post? I don't recall reading that. In any case, OK.

    Yes. λair = 1.3λliquid

    Your initial idea of simultaneous equations was correct. Set up two diffraction equations, one for liquid and one for air.
  7. Jan 16, 2010 #6
    ok but which equation should I use?


    x - 0.45 = (1.3[tex]\lambda[/tex] x D x n)/ d

    and x = ([tex]\lambda[/tex] x D x n) / d ?

    Thanks for your help!
  8. Jan 16, 2010 #7

    Doc Al

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    Staff: Mentor

    Why 1.3λ? You want xliquid = λliquid (D/d). (Expressed in terms of x and λ, of course.)
    OK. (That's the equation for air.)
  9. Jan 17, 2010 #8
    But I assume you want to express [tex]\lambda[/tex]liquid in terms of λair.

    So i came to the conclusion:

    In water: x - 0.45 = (Dλair/1.3) / d
    In air: x = (λairD) / d
    (I have not included n in these to equations, as I have taken n = 1)

    Solving this gives: λair= 6.5 x 10^-7 m
    Is this correct? I have a feeling that it isn't (because the question is worth a lot of marks- which hints that it takes more than a few steps!) :S Thanks!
    Last edited: Jan 17, 2010
  10. Jan 17, 2010 #9

    Doc Al

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    Staff: Mentor

    Looks good to me.
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