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Optics: Converging waves

  1. Apr 24, 2015 #1
    Hi. A spherical wave ##e^{i(kr-\omega t)}## diverging from a single point ##(x=0,y=0,z=-z_0)## can be approximated as a parabolic wave in the paraxial case around the z-axis. I.e., ##k r = k \sqrt{x^2+y^2+z^2} \simeq k (z +\frac{x^2+y^2}{2z})##.

    OK, then let's say a lens is placed such that its optical axis coincides with the ##z## axis and its focus points are at ##-z_0## and ##z_0##. In this case, the outgoing parabolic wave from ##-z_0## will be focused into the point ##z_0##. My question is, how is this to be modeled mathematically? Intuitively I would guess that ##k r \simeq k ( z - \frac{x^2+y^2}{2z})##, but what is ##kr## equal to in the accompanying case of a converging spherical wave? Something ala ##e^{i(kr + \omega t)} e^{i \phi}##, where ##\phi## is some phase factor?

    I would appreciate it if you guys could help me in clearing this stuff up :)

    Thanks
     
    Last edited: Apr 24, 2015
  2. jcsd
  3. Apr 24, 2015 #2

    blue_leaf77

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    If this is converging lens, then the rays will be focused at infinity, that is the outgoing rays are collimated.
     
  4. Apr 25, 2015 #3
    crap, yeah you're right. I was thinking in terms of rays from the object plane being focused into the image plane, but I mixed it up. sorry.

    But anyway, do you know the mathematical form of waves converging to a single point?
     
  5. Apr 25, 2015 #4

    blue_leaf77

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    converging spherical wave must be the inverse of the diverging one, the sign of the wavevector k must change (better change the sign of k rather than the sign of ##\omega t## as you did above, this will mean time reversal).
     
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