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If you put an aperture in a laser beam to block some of it, I would imagine that the spot it can be focused to becomes larger due to diffraction. The numerical aperture of the system is limited by that truncation of the beam. For a regular laser focusing without an aperture, the beam diameter determines the numerical aperture, not the lens diameter.

But if the aperture becomes an infinitesimal pinhole, then it's the same as a point source emitting light. And the equation for how well resolved that point can be is defined by the airy disk using the aperture size of the lens diameter.

So as the aperture is shrunk down, do the two relationships describe different things? For example, does the NA associated with the aperture of the laser beam describe how small of a Gaussian spot can be formed, and the NA associated with the lens itself describe the size of the airy disk that all the light coming through the aperture can be condensed into?

I guess fundamentally it's just confusing because a coherent beam is usually implied to only have a NA related to the beam diameter, but after passing through an aperture it diffracts so the whole diameter of the lens is relevant in picking up and focusing the higher diffraction orders.

Or perhaps my understanding is totally messed up?

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Aside from aberrations, the spot size that can be formed by a laser beam going through a lens system is determined by the last aperture that restricts the beam. Note that if an aperture is wider than the beam, it is not restricting the beam -- so it is necessary to look upstream for the "last" aperture to restrict the beam.

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  • $\begingroup$ What if I want to form an image of the beam instead of focusing it? It also seems to be counterintuative that the restricting aperture controls the beam, because all points on the wavefront diffract in a hemisphere, so the lens size should matter in capturing more diffraction orders from the aperture itself. (photolithography does this) $\endgroup$ – Yifan Kong Apr 16 at 14:41
  • $\begingroup$ If the beam as it's emitted from the laser is collimated, then it is effectively a point source at infinity, passing through an effective aperture corresponding to the width of the beam. You can form an image of that effective aperture. The easiest way to understand diffraction optics, I think, is to learn the math. $\endgroup$ – S. McGrew Apr 16 at 14:49
  • $\begingroup$ I understand the math behind formation of an airy disk, but I suppose I need to learn the math to understand power distribution in the wavefront as it evolves. Is this image I drew correct however? imgur.com/a/yjoxOKE If it is, I can see why a condenser is used in a microscope to provide a matching NA to the lens - but another question would be how the power distribution in the wavefront changes between focusing and image forming? If you can give me a hint before I learn that math that would be awesome! $\endgroup$ – Yifan Kong Apr 16 at 15:08
  • $\begingroup$ The image you drew is not correct. If you trace the three rays from the top edge of aperture "A", they should come to a focus somewhere on the right hand side of the lens. Instead, you have drawn the top two rays parallel after the lens. Same mistake at the bottom edge of the aperture. $\endgroup$ – S. McGrew Apr 16 at 15:51
  • $\begingroup$ Doh! Thanks for the catch. I fixed it imgur.com/rb9zDXz But the conclusion is correct? $\endgroup$ – Yifan Kong Apr 16 at 15:55
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I would look at it like this. Aperture cleans up the mode of the light. Provided the intensity of light is unifrom over the aperture, I cannot see the problem with putting a tube lens and an objective after it and focusing it into the diffraction limited spot.

Basically, your argument is that we are messing with the numerical aperture of the light beam, and limit the ammount of k-vectors available to us. I would say that a microscope arrangement can effectively shift the energy between different k-vectors (same frequency of course), so it really does not matter. What does matter however is how clean your mode is, since odd modes tend to occupy more space (cannot be focused into such small spots)

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  • $\begingroup$ From my reading of Gaussian optics, truncation of the beam to the limit will result in a final spot size dependent on the Numerical Aperture of the truncation aperture, not of the lens - even at aperture ratios which would be leave the aperture effectively uniformly illuminated. Page 9: cvilaseroptics.com/file/general/… d = K*wavelength * focal length / aperture diameter (of beamstop) $\endgroup$ – Yifan Kong Apr 16 at 5:15
  • $\begingroup$ I think we are assuming different things here. You seem to be talking about a single truncation followed by the lens. I talked about the following problem, given a beam (of unknown quality) will passing it through the pin-hole, followed by arbitrary arrangement of lenses allow you to focus the light into a smaller spot? The answer to my question is 'yes'. You truncate your beam, recollumate it (see 'Spatial Filtering' on page 10 of your document), then pass it through a beam expander, then focus it. $\endgroup$ – Cryo Apr 16 at 21:25
  • $\begingroup$ ...You focus it with the objective. Now this will lead to a smaller focal spot, because (a) the expanded beam will illuminate high proportion of the objective back-aperture, (b) the spatial filtering provided by the pin-hole will reduce the high-order mode content in the beam, leaving only the fundamental mode, which is easier to focus. $\endgroup$ – Cryo Apr 16 at 21:29
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If you had a perfect laser (which is impossible) all beams are parallel and with an aspheric lens the spot is very small, limited by Airy disk of lens. Adding any aperture increases diffraction at the aperture, thus spot size increases because the rays are no longer parallel, there is no perfect focus any more. The aperture of the lens, usually much larger that an aperture in the beam, does not have as great as an affect. i.e. it's effect is << than the beam aperture.

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  • $\begingroup$ But at the limit of an infinitesimally small aperture for the beam, doesn't the beam just become a point source of light? (the beam becomes a single point source of light propagating out by diffraction only) A point source of light is governed by the aperture of the lens when considering how small of a spot it can be focused to. $\endgroup$ – Yifan Kong Apr 16 at 5:05
  • $\begingroup$ You can make a point source of light with the aperture but in doing so you will cause significant diffraction (almost infinite) so the point source is radiating in all directions. If you try and focus this (well first of all you have lost a lot of light unless your lens is huge) the best you can output is a beam with all rays parallel, kind of the reverse of what you started with. There is an important optics concept called entendue, basically it says angle times area is conserved as you go thru each aperture or lens or other optic. $\endgroup$ – PhysicsDave Apr 16 at 13:30
  • $\begingroup$ For etendue see: physics.stackexchange.com/questions/234996/… $\endgroup$ – PhysicsDave Apr 16 at 13:31
  • $\begingroup$ Some of the confusion here (at least for me) is using a lens to focus light to point or using the lens to focus as in make an image. $\endgroup$ – PhysicsDave Apr 16 at 13:41
  • $\begingroup$ Changing a point source into parallel rays only applies if the point source is at the front focal plane. Otherwise it is refocused into an image (if the point is beyond the focal plane) Focusing light to a point vs forming an images is really confusing me. Empirically, it is possible to demagnify and project light passing through very small apertures onto a plane, because that's how photolithography systems work. Their resolution is not limited by the NA of the mask aperture but the NA of the lens and how much of the diffracted light it can capture. Thoughts? $\endgroup$ – Yifan Kong Apr 16 at 14:37
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For a well corrected lens, one with a small amount of aberration, putting an aperture will increase the size of the focused spot. The basic formula for the diameter of the first dark ring is (from Smith, Modern Optical Engineering, page 453) $$B = \frac{(2.44 \lambda f)}{D}$$ where B is the diameter of the dark ring, $\lambda$ is the wavelength, $f$ is the focal length, and $D$ is the aperture diameter. If you put an aperture that decreases D, you will increase the diameter of the dark ring. With Gaussian beams, the formulas are different, but the effect is the same: putting an aperture on something causes the light to diffract more, and that spreads out the light at the focal plane.

The one practical disclaimer is: If the lens is not well corrected, but has significant aberrations, then decreasing the aperture can actually decrease the spot size. The reason is that light passing through the edge of the lens, which forms aberrations that increase the spot size, is now blocked by the smaller aperture. The resulting spot might have lower overall power, but also have a lower spot size. To know if this is true you must know about the details of the lens.

"decrease" changed to "increase", in sentence 1, per PhysicsDave. Gotta keep 'em straight!

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  • $\begingroup$ In you first sentence I think you meant to type increase not decrease? $\endgroup$ – PhysicsDave Apr 16 at 13:33

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