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Violation in diffraction? Lagrange (Optical) invariant

  1. Feb 22, 2015 #1
    It says you can not change with lenses the value L - radiance. Below I have an example where it proves that you can or where am I wrong? (I made L for 2D case, in 3D case everything the same - L2>L1)


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  3. Feb 22, 2015 #2

    mfb

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    You cannot change it from left to right with a lense, but different setups can have different values of course.
     
  4. Feb 22, 2015 #3
    well, lagrange invariant says you can not.
     
  5. Feb 23, 2015 #4

    Andy Resnick

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    I don't understand what you are trying to show. The optical invariant (or etendue) can be written a few different ways, but it is the product of aperture stop and numerical aperture of the system. The etendue can always decrease (for example, by stopping down a lens), but never increase- one way to show this is to calculate the determinant of the ABCD matrix of the optical system: for a thin lens, it equals 1, showing that etendue is conserved in this case. Are you trying to demonstrate something by connecting the two lenses into a beam concentrator/expander?

    Perhaps you are confusing 'radiance' and 'etendue'- they are not the same thing. Etendue/optical throughput/optical invariant is a geometrical property of the system, while radiance is a property of spatially extended emitters.
     
  6. Feb 23, 2015 #5

    mfb

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    "Something is conserved/invariant" does not mean "this has the same value everywhere, for all setups".
    Energy is conserved, but a truck on a highway still has a larger energy than a snail moving across a road.

    And I agree with Andy Resnick, I don't understand what your setups are supposed to show.
     
  7. Feb 23, 2015 #6
    you focus some light (Watt) in a specific area (m2) and get a angle of radiance(sr - steradian). Then if you want at least the same amout of light into a smaller area the angle of radiance must increase. That's the Etendue/optical throughput/optical invariant. However, in the examples I have provided, I focus more light in the same area with the same angle thus breaking the L rule or what am I missing?
     
  8. Feb 23, 2015 #7

    mfb

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    If the light source is the same - it is not in your setups. And only if your focussing is ideal.
    Also, where is the area?
     
  9. Feb 23, 2015 #8

    Andy Resnick

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    I think you are mixing concepts- radiometry and ray optics, specifically. To be sure, a single 'ray' is associated with radiance, but if you are talking about collections of rays you have to be careful. For example, if a single ray enters through the entrance pupil and exits through the exit pupil, then the radiance is conserved (for that ray)- if there is vignetting, then the integrated radiance (luminous flux) is no longer conserved.

    Ok- you have some luminous flux that passes through a lens: we should be talking in terms of irradiance (W/m^2) and intensity (W/sr). Then there's a second surface that the light is incident upon. Certainly, the irradiance incident on that second surface can be much higher than the irradiance of the initial optical field. However, say the second surface is underfilled by the incident light. You don't magically increase anything; in fact, if the second surface is a lens, you don't access the full optical power of the lens and can't focus the light down any further than you did previously.

    Think of it this way- I have a microscope objective and fully illuminate the back aperture. Then, the light is focused down as normal- the spot size is (approximately) given by Abbe's formula. However, if I underfill the back aperture, the spot size is larger than Abbe's formula specifies because I don't access the full numerical aperture of the lens. In the limit that I illuminate the back aperture with a single ray, a single ray emerges from the lens and is not focused at all.

    Does that help?
     
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