Imaging Fresnel Lens Theory

Tom.G
If one were to put a little "smoke" in the air,
That reminds me of an old joke that sorta strikes home here.

Three optical engineers were in the lab one day where a technician had just completed a prototype.

The engineers were arguing about the light path and location of focal points in the apparatus, each had a different idea of what was going on.

The technician, after listening to this for a while, decided he was fed up with the discussion. He light up a cigarette, blew some smoke into the light path, and walked away.

Cheers,
Tom

sophiecentaur and hutchphd
sophiecentaur
Gold Member
2020 Award
What screen? I made no mention of a screen.
I know you didn't but it's the only way you will get your 'point sources' that are not at minus infinity from the laser sourced beams he's talking about. Where would you find point sources along those beams until they hit something? In smoke, of course you will see many point sources but each smoke particle (tiny screen) takes some light from the beam. If you focus the beams to a point then that is a point image but where is that in this setup?

If you are correct then how is his Image focus at an image plane that is at F? F is where an object at infinity is focussed and not where his light source is in the diagram? If I have got it wrong then can you help me by explaining that away? Because of the coherence, each of the beams from his coherent source has a source that's the same as the laser's. (Near enough)

If you look at a laser beam it doesn't spread out as if from a point at the end of the laser. The beam will be parallel (as we know) and not a wide cone, which it would be from a point on the exit. The spread of the beam from a laser depends on the effective number of reflections within the laser and the physical length. One short tube becomes a much longer tube and a narrower beam. (As in my idea of an infinity mirror.)

PS the gizmo on the end of the laser is not a diverging lens.

hutchphd
Homework Helper
If you look at a laser beam it doesn't spread out as if from a point at the end of the laser. The beam will be parallel (as we know) and not a wide cone, which it would be from a point on the exit.

Laser beam through diffraction grating:

Do you now understand what I am talking about? I am happy to discuss.

Tom.G
sophiecentaur
Gold Member
2020 Award
Do you now understand what I am talking about? I am happy to discuss.
I'm happy too; it is sorting out my mind usefully.
Yes. We both see the same thing but I think your interpretation is 'not right' or perhaps 'limited'. Looking closer, you can see that there is dispersion which forms, say four separate orders (just looking at a single colour).

Imagine that the grating (representing what the OP has) behind a screen so that all you can examine is those beams. They all emanate from, apparently, a single location - the grating. However, a close up of one beam will show a beam with parallel sides (same angle all the way along - ignore the diffraction due to the limited aperture). If you had to use that beam to estimate where the source is, you would say that it's at 'infinity'. So you would have conflicting observations; beams spreading out from a nearby 'point' but single beams with apparent sources at infinity. You are seeing one thing and I am seeing another.

Now replace the grating with a box with a halogen bulb plus simple filter and an array of holes in it and look at one of those mimic beams. The beam will be diverging as if from a source that's inside the box; that's conventional non-coherent optics. You could definitely say that the source is a point. No conflict here and we will both interpret it in the same (conventional) way.

If you use a convergent lens on both of those bundles of light, you will get different results. The individual split laser beams will form sharp spots at F and there will be a scaled image, consisting of defocussed spots, formed in the plane at v according to the f and the u.
For the halogen bulb arrangement, you will get a sharp image of the array of holes at v. Point sources to point images.

There are a couple of caveats, concerning the information we have been supplied with. Was the "Image Plane" described in the diagram, actually identified? Probably the least worst result of focussing using the fresnel. My reasoning is based on a good lens and it fits with what has been reported. It could be that all that's being seen is due to the aberrations of the fresnel lens so a good conventional lens should be used as a reference.

However, the whole situation as described is, itself, very defocussed and the description (including that strange diagram of the odd shaped figure) is in very unfamiliar terms.

Cheers

hutchphd
Homework Helper
Very clear explanation...much appreciated. I think I still disagree with the following (color enhanced):
Iimagine that the grating (representing what the OP has) behind a screen so that all you can examine is those beams. They all emanate from, apparently, a single location - the grating. However, a close up of one beam will show a beam with parallel sides (same angle all the way along - ignore the diffraction due to the limited aperture). If you had to use that beam to estimate where the source is, you would say that it's at 'infinity'.
Thus is multi-slit diffraction and the beams have a small fixed angular width in the far field. This follows from the finite wavelength width of the laser......think about the color pattern produced by white light: it gets bigger with distance from grating. So with the grating a little outside focal point the lens makes this a little better. But you are thinking about a spectrometer with the "slit" effectively at infinity and you are indeed correct. But it really doesn't matter here.
What is hanging you up is essentially depth of field. If I shoot a laser beam through a lens it does indeed converge at the focus and then diverge....but by how much? The effective aperture of the lens for this circumstance is the size of the laser spot on the lens ! Maybe 1mm. So if f=10cm then the beam diverges 1/100 radians. So even for a fat lens the depth of field extends effectively to infinity.
I suggest the following experiment which requires a laser pointer and fat lens (I used my shop magnifier light) and a white card. Grab said items and play. I thought my analysis correct but found it a very useful 10 minute nonetheless.
All the beams will stay beams for our practical purposes.

sophiecentaur
Gold Member
2020 Award
Thus is multi-slit diffraction and the beams have a small fixed angular width in the far field. This follows from the finite wavelength width of the laser....
. . . which is what I already said when I drew a distinction between the laser and conventional rays. Conventional rays diverge in an obvious and finite way from where they are formed.
All the beams will stay beams for our practical purposes.
Yes I agree and their origins are at infinity and not at the plane of the grating. If, instead of a grating, you used a mirror to produce a single output beam, would you say that the Source Image position is in the plane of the mirror? Of course not. So why say that the image in this case is at the grating? The beams all converge on the grating but are you saying there is no difference between that and my theoretical filament lamp in a box of holes?
I guess my problem is only when the source position is needed or specified. And that applies here if we are discussing the 'Image Plane" in the diagram at the top of the thread. Where would a source have to be in order to be focussed at the Image Plane?

The depth of focus point you make has made me think and it's definitely relevant. But what is the OP talking about when he describes an Image being focussed at F? Perhaps what is being described is suffering from a poor laser source (a lot bigger than 1mm) and an obviously poor quality lens 'substitute'.

I have a slight glimmer about this. It is true to say that an observer moving from side to side would identify the direction of the source (when his eye was intercepting one of the beam) and that would lead him to deduce that the source was in fact at the grating (as long as the spread of the beams was not detectable). If the lack of spread was detectable then there would be a conflict. I have sort of said this already but, as you imply, it's all a matter of the actual numbers involved. There is a subtle difference between the coherent and non coherent cases but I think I can now accept that it could be OK to treat the grating as a source - for most purposes.

hutchphd
Homework Helper
Of course not. So why say that the image in this case is at the grating? The beams all converge on the grating but are you saying there is no difference between that and my theoretical filament lamp in a box of holes?
The filament lamp in a box of holes would be the same if the filament had zero size and the holes had no diffraction limits so that you generated good beams.

I guess my problem is only when the source position is needed or specified. And that applies here if we are discussing the 'Image Plane" in the diagram at the top of the thread. Where would a source have to be in order to be focussed at the Image Plane?
This is why I am discussing depth of field. To focus the individual beams (dots) is easy: the depth of field is very deep because the effective lens aperture is tiny for each beam. For the size of the dot pattern to remain independent of image distance, however, the source position must be near the "front" focus, because it requires the whole lens to produce the pattern (the geometry is obvious I think).

sophiecentaur
sophiecentaur
Gold Member
2020 Award
The filament lamp in a box of holes would be the same if the filament had zero size and the holes had no diffraction limits so that you generated good beams.
That's not actually true. The beams from holes in a box diverge from the point at the centre. (Angle proportional to d with filament at d=0, even for "good beams") The beams from a diffraction grating diverge only due to diffraction limiting by the grating aperture and the angular spread is much less so the virtual source position is way behind the grating. That is particularly obvious for a monochromatic source in both cases.

But I'd love to know what actual image is produced in the 'image plane' that's in the diagram. I can't even a hazard a guess.

hutchphd