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Interference viewed in the focal plane

  1. Feb 8, 2015 #1
    I'm doing a problem where we have plane monochromatic waves incident on finite double slits (each of width a, separation d), and the transmitted light is observed in the focal plane of a lens of focal length f. The position of the lens isn't specified.

    So I have worked out that the intensity distribution (ignoring the lens at the moment) in the Fraunhofer regime would be
    I(θ)=I(0)cos2(kdsinθ/2)sin2(kasinθ/2)/(kasinθ/2)2

    However I'm finding it hard to understand how the lens fits into the problem exactly. If the lens was right after the screen, I would write sinθ≈tanθ=x/f where x is the position on the screen, and then rewrite my intensity in terms of x instead of θ - this works because the length between slits and screen is clearly f. However, if the lens was anywhere else, which it could well be, the distance between slits and screen wouldn't simply be f anymore. I don't see how the above could hold anymore, and so how would I rewrite the intensity in terms of f now?. So the problem is how does this work out when the lens is in any position

    Thanks for any help!

    Note, I would have this same issue if the aperture was simply two infinitesimal double slits with intensity pattern
    I(θ)=I(0)cos2(kdsinθ/2)
    so feel free to use this if necessary to simplify things.
     
  2. jcsd
  3. Feb 8, 2015 #2
  4. Feb 8, 2015 #3

    Drakkith

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    Be patient. This isn't exactly a question we get often, so it may take some time for someone to answer.
     
  5. Feb 9, 2015 #4
    I appreciate that, it's just it's probably more likely to slip down the list and be forgotten about...
     
  6. Feb 9, 2015 #5

    Andy Resnick

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    It's just a scaling factor that 'converts' the angular extent of the diffraction pattern into a linear extent. Say the lens was placed a distance 'f' from the slits; the interference pattern would be located a distance 'f' from the lens and the linear extent of the angular diffraction pattern given by x'= λf/x.

    http://ocw.mit.edu/courses/mechanic...fer-functions-transforms/MIT2_71S09_lec18.pdf
     
  7. Feb 10, 2015 #6
    Did you mean 'x' when you defined the first 'f'?

    Even so, I still can't quite understand where you obtained x'=λf/x from. I assume you are using the diagram on slide 15/16 with z=x and x''=x', but it's not clear exactly what to do? After this, how would you insert this into the intensity profile exactly (i.e how would you replace θ)?
     
  8. Feb 10, 2015 #7

    Andy Resnick

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    Using text makes the expressions a little clumsy- let me try again.

    In the case for an object placed a distance 'z' in front of a lens with focal length f, the amplitude and phase of the light at image position (xf, yf) in the rear focal plane of the lens is given by the amplitude and phase of the object spectrum at frequencies (xf/λf, yf/λf) with an overall phase factor that goes as 1/λf (1-z/f). Alternatively, the image spectrum (fx, fy) = (xf/λf, yf/λf).

    Now, you have a 1-D intensity pattern: I(0)cos2(kdsinθ/2)sinc2(kasinθ/2) so the situation is somewhat simpler, but in any case you have the spectrum specified in terms of an angular (θ) frequency (call it fx), so in order to figure out 'how big' the pattern is at your image plane simply substitute in the scaling rule above.

    How's that?
     
  9. Feb 10, 2015 #8
    Hmm, θ in my intensity is just the angle subtended at the screen from the centre of the aperture, as here (just obviously we have a lens between aperture and screen):
    http://www.citycollegiate.com/young1.gif
     
  10. Feb 11, 2015 #9

    Andy Resnick

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    Right- adding a lens simply alters the relationship between 'y' and 'θ'.
     
  11. Feb 11, 2015 #10
    Without the lens, I can write sinθ=y/L and so the intensity distribution becomes
    I(θ)=I(0)cos2(kdy/2L)sin2(kay/2L)/(kay/2L)2.

    If the lens is a distance D from the slits, and f from the screen, so that L=D+f, I have sinθ=y/D+f, however I am not given what D is so how can I get rid of it?
     
  12. Feb 12, 2015 #11

    Andy Resnick

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    That's what I wrote up in post #7- what you are calling 'D', I wrote as 'z'.
     
  13. Feb 12, 2015 #12
    I'm not quite sure what you mean by 'object spectrum at frequencies (xf/λf, yf/λf) with an overall phase factor that goes as 1/λf (1-z/f). Alternatively, the image spectrum (fx, fy) = (xf/λf, yf/λf)'. How did you get those expressions and what does 'image spectrum' mean?

    Apart from this, I'm guessing it's impossible to eliminate z - the position of the lens does matter?
     
  14. Feb 12, 2015 #13

    Andy Resnick

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    Ok. The basic idea is that diffraction of light through an aperture can be described in terms of plane waves, and each plane wave is described by it's wavevector. At the screen, all the plane waves add coherently and you get the 'far field diffraction pattern' of your aperture. How much or each possible plane wave there is at the screen is determined by the aperture itself: the far-field diffraction pattern is the (spatial) Fourier transform of the aperture. For example, if the aperture is a circle, the far-field diffraction pattern is an Airy function. For a rectangular aperture, it's a sinc function. etc. etc. The tricky bit is that the argument of the sinc function (or whatever) is the wavevector, not coordinates like (x,y). To convert the argument from wavevectors to coordinates, you have to 'scale' the wavevector, like y = z sinθ (see all the sinθ terms in your post #1?)

    Adding a lens simply changes the scale factor- colloquially, you move the far-field diffraction pattern from infinity to a distance f. The prefactor 1/λf (1-z/f) simply means that if your lens is not a distance 'f' from the object, there is some extra phase component to keep track of - note what happens when the lens is placed a distance z=f from the aperture.

    Does this help?
     
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