Author Topic: Schlieren images projected onto a wall (no camera)  (Read 1678 times)

Offline davekni

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Schlieren images projected onto a wall (no camera)
« on: December 28, 2020, 12:56:34 AM »
I made an optics setup that is a mix of conventional and projected Schlieren.  I think this was called "multiple Schlieren" somewhere, but can't find that reference.

This setup is close to conventional Schlieren, but with two key differences.  First, instead of a camera, a lens focuses the image onto a wall or screen.  The second difference is what makes it bright enough to allow projection.  Instead of a single point light source and single knife-edge, this uses a line-array light source and matching line-array shadow mask instead of a single knife edge.

The light source is a 6-watt white LED inside a 38mm diameter parabolic reflector.  A bright narrow-beam commercial flashlight will work just as well.  The camera-replacement is a 55mm diameter +1 closeup lens.  (It actually measures 1.2 diopter, but was sold as +1.  A true 1-diopter lens projects a somewhat larger image a bit farther away.)  Covering both the light source and lens is a vertical grid of lines at 1mm pitch, 0.5mm black and 0.5mm clear.  This shadow-mask was printed with a standard laser printer onto transparency film.

The most expensive piece is a telescope mirror, 8" (203mm) diameter in this case.  Cost was $110.  A spherical mirror is ideal for Schlieren.  This one is parabolic, which is close enough.  (Found a similarly-priced spherical mirror shortly after purchasing this one.)  The next most expensive item would be a bright LED flashlight, perhaps $30-$40 for a good one.  For this project I built my own with an LED, reflector, and heat-sink.  A +1 close-up lens can be found for under $10.  The rest is building materials, wood/plastic/screws/etc., and a few magnets.  The shadow-mask is a piece of transparency film printed with a laser printer.

The shadow-mask is mounted 1500mm from the mirror at the center of its curvature.  The magnets allow sliding the light and lens assembly to the center, as my wood frame is not exact.  The mirror has a 750mm focal length, with radius of curvature being twice focal length.  (It was listed as D203F800, but was really D203F750.)  The light source and lens mount behind the shadow mask.  Distance is not critical - gaps between the lens and/or light and shadow mask are fine.

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This printed shadow mask works fairly well, but isn't ideal.  Light passes twice through the transparency film, which isn't AR coated.  Also the lines are not as high optical density as I'd like.  Not sure how easy and/or expensive it would be to procure a better mask, either etched metal or chrome on AR-coated glass.  I tried an array of 0.5 x 3mm carbon-fiber strips.  That made artifacts due to it's 3mm depth and reflections of glancing-angle rays hitting the internal edges of the strips.  Extinction ratio was well better though.  I may try double-printing the transparency film, either repeat-printing the same side or duplex (printing the opposite side).  Issue will be alignment between the two printing passes.  Any suggestions for a line shadow mask?
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Offline Mads Barnkob

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Re: Schlieren images projected onto a wall (no camera)
« Reply #1 on: December 28, 2020, 12:31:02 PM »
Wow! Schlieren photography is something that I have wanted to do, but never gave myself the time to look into, just assumed too expensive :)

Such a cozy setting for this experiment and really shows that science can be done at home without a huge workshop and a lot of tools. That is a true quality of your demonstration here.

Do you have a smartphone? I use my smartphone as both primary and secondary camera, when my 5Dmk3 can not fit the scene or its impractical to get a huge DSLR somewhere :)

Regarding the grid mask, could you use something like stainless mesh for filtration? https://www.ebay.co.uk/itm/150-Micron-100-Mesh-316L-Stainless-Steel-Filter-Filtration-Oil-Screen/264870205171
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Offline MRMILSTAR

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Re: Schlieren images projected onto a wall (no camera)
« Reply #2 on: December 28, 2020, 04:39:32 PM »
I watched your Youtube video and it was fascinating. This inspires me to make my own apparatus.

I have one question though. In the video, you state that a spherical mirror was ideal but a parabolic mirror would be good enough. I would think that the opposite would be true. Wouldn't the spherical aberration of the spherical-ground mirror cause more image distortion than the parabolic mirror?
« Last Edit: December 28, 2020, 04:43:19 PM by MRMILSTAR »
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Offline davekni

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Re: Schlieren images projected onto a wall (no camera)
« Reply #3 on: December 28, 2020, 07:34:55 PM »
I don't have a smartphone or laptop or tablet.  Prefer a full-size keyboard and screen.  I could have connected a webcam to a USB extension cable and made a simultaneous low-quality video.  If I had better video editing skills, that low-quality video could have been an insert.

If wire placement is sufficiently uniform, a filter mesh could work.  Efficiency would be low, however, as the light (hole) and dark (wire) shapes aren't the same.  I have thought about making my own wire line screen with 0.5mm diameter wires held at their ends in the grooves of M6-1.0 threaded rods.

Parabolic mirrors are ideal for incoming parallel light rays such as from a distant star, causing them to converge at the focal point.  For conventional single-mirror Schlieren, the light source and camera lens are both close to the center of curvature of the mirror.  The light rays are fanning out from the center and ideally reflect back to the center.  That describes a sphere rather than a parabola.  A parabolic mirror used to reflect light back to its source has the same degree of error as does a spherical mirror used to focus parallel light rays.  However, this parabolic/spherical distortion is less than the distortion due to the non-point size and spacing of the light source and lens.  The mirror needs to be accurate relative to the 0.5mm shadow-mask lines, not close to diffraction limited.

There are two-mirror versions of conventional Schlieren where an off-axis parabolic shape is ideal.  The light source is converted to parallel rays, passes through the object plane, then is focused back from parallel to converging on the camera lens.  For this single-mirror Schlieren, an ellipse would be theoretically best, with one focus at the center of the light source and the other at the center of the lens.

Great to hear that this project might inspire other Schlieren builds!  That was my hope.  I think it could make a good classroom demonstration.  Direct viewing without a camera proves that there are no digital imaging tricks being played.
« Last Edit: December 28, 2020, 08:26:10 PM by davekni »
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Offline davekni

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Re: Schlieren images projected onto a wall (no camera)
« Reply #4 on: January 03, 2021, 04:31:07 AM »
Took 5 days instead of my planned 2, but I finally made some stroboscopic slow-motion Schlieren video of sound waves from a spark gap:
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This was made one frame at a time with a still camera.  The Schlieren illumination LED is pulsed for 1.33us as a strobe flash, delayed from the spark time.  Each frame has an additional 5us delay relative to the previous frame for the first loops, and 3.33us additional delay for the 10k:1 loop.

The most problematic bit was changing from my large 55mm diameter lens for projecting onto a wall to the camera's roughly 8mm aperture, especially with the camera aperture ~48mm behind the lens front, preventing it from being adjacent the shadow mask.  No matter where I placed the camera, from it's view the mirror was only partially illuminated.  Wasted a bunch of time with "quickie" fixes before stopping to think through the optics.  The camera's 8mm aperture 48mm behind the shadow mask reflects from the mirror as a 7.5mm aperture 48mm in front of the shadow mask (mirror side of the shadow mask).  The light source needs to be behind the shadow mask, and needs to uniformly illuminate the mirror through that virtual aperture on the other side of the shadow mask.  Ended up dropping the parabolic reflector and using two small convex lenses to focus an LED onto the virtual aperture.  Light efficiency was barely enough for the 1.33us strobe pulses and f2.6 camera with default ISO100 gain setting.
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Offline Uspring

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Re: Schlieren images projected onto a wall (no camera)
« Reply #5 on: January 03, 2021, 06:17:18 PM »
Beautiful videos.  ;D The later one reminds me of our discussion on air compression arising from TC arcs. I was surprised about the sensitivity of your setup. I had seen the usual images of air compression due to supersonic air flows. But there the compression is much larger than in acoustical waves. Sound pressure of 20 Pa is equivalent to already 120 dB and 20 Pa is only 0.02% of atmospheric pressure.

Do you have an idea of the sound pressure involved in your case? It is probably difficult to tell, since we are seeing only a projection of the spherical wave front.


Offline davekni

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Re: Schlieren images projected onto a wall (no camera)
« Reply #6 on: January 04, 2021, 04:06:03 AM »
Any estimate or guess on what fraction of spark energy becomes sound energy?  This spark gap is 2.6mm, ~8kV fed from 2nF, so 64mJ.  I suspect most of the electrical energy goes into the spark (only a few percent to dielectric and resistive losses).  Looking at images, I'd guess the pressure wave to be about 11mm long at 50mm diameter, so about 350cc volume.  I'm coming up with 1350Pa if all 64mJ went to sound.  Extrapolating from other experiments, I'm guessing about 100Pa would be visible.  That would allow viewing even if only 1% of spark energy became sound energy.

I think sound dB is based on RMS pressure.  Don't know what the averaging time is for RMS measurement.  That single short pulse may not register too high on even a peak-reading meter.  Listening from ~2m away didn't seem loud enough to warrant hearing protection.  Of course, pressure would drop by 1/40 compared to 50mm, and likely a bit more if the 11mm length spreads out any.
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Re: Schlieren images projected onto a wall (no camera)
« Reply #7 on: January 04, 2021, 07:17:41 PM »
An 11 mm wavefront is possibly mostly above the hearing range and you're right, CW sounds might be a different issue.

The sensitivity of your device still intrigues me. The refractive index of air is about 1.0003 and changing pressure by 100 Pa, i.e. 0.1%, would modify that value to 1.0003003. Not very much of a difference. I believe that in order to see this in your device, the pressure gradient must be almost orthogonal to the light rays.

Offline davekni

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Re: Schlieren images projected onto a wall (no camera)
« Reply #8 on: January 04, 2021, 07:58:25 PM »
Yes, I suspect the sound wave is visible because it is along the edge of an expanding sphere where light passes through quite a length of refractive gradient.  As the sphere expands, the gradient decreases, but the length increases.

I'm still contemplating ways to measure sensitivity.  Any thoughts?  I have argon available - need to find good data for its refractive index relative to air at the same temperature.  Humidity might mess that up a bit.

I adjusted the shadow mask to where it was blocking almost all the light, the most sensitive possible.  It left the images a bit under-exposed, which I fixed with image processing before posting the video.
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Re: Schlieren images projected onto a wall (no camera)
« Reply #9 on: January 05, 2021, 05:09:45 AM »
Ran a sensitivity test this evening.  Constructed a gently-heated stream of air rising from a round tube in the image plane (in front of the mirror).  Hung a fine (0.1mm wire) thermocouple in the air stream just above the tube end, then laterally outside the stream to measure temperature delta.  This cylindrical stream was barely visible with 1C delta.  It required 3-4C delta to make it as clear as the sound waves.  1C corresponds to roughly 330Pa and 3-4C to 1000-1330Pa (based on fraction of ~300K and fraction of 100kPa).  Looking through the edge of the expanding sound wave sphere gave something like a 4x length-to-width ratio, so ~1/4th the pressure.  Thus I'd estimate the actual sound pressure to be around 300Pa at 50mm, with minimum visible around 80Pa.
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Re: Schlieren images projected onto a wall (no camera)
« Reply #10 on: January 06, 2021, 06:52:52 PM »
Thank you for your measurement. I've been trying to work out, what your values mean in terms of refraction angles, as these are detected by your system:
The curvature of the light beam, i.e. the inverse of the curvature circle radius, is 1/n * dn/dx, where n is the index of refraction. x is here a coordinate orthogonal to the light beam.

A 1 C temperature difference corresponds to a delta n of about 1e-6. If I assume a delta x, i.e. the distance over which the temperature difference occurs, of 1 mm, I get a curvature radius of 1 km. Over, say 5 mm light travel length, the refraction angle would be 5e-6. That seems too small to be observable, since diffraction effects smearing out these kind of angles should be much greater. There is probably an easy answer to this, but I haven't found it yet.

As you mentioned, the contrast in the sound wave video does not decrease much when the wave reaches the edge. Pressure and pressure gradient drop as 1/r but the traversal distance, i.e., the length of the light beam within the gradient, increases proportional to r. So that should cancel. It doesn't quite cancel, though, another puzzle.

Offline davekni

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Re: Schlieren images projected onto a wall (no camera)
« Reply #11 on: January 07, 2021, 04:00:46 AM »
I haven't thought of a rigorous way to describe this (ie. no proof), but I think my array of light sources and slits makes the diffraction limit much less of a problem.  First consider the case of a single narrow line (say 2um wide) light source.  The 203mm diameter mirror 1500mm away can focus that light source back to the lens still only a few um wide due to diffraction.  That light could be passed by a narrow slit or blocked by a narrow dark line.  However, that narrow line at the lens will produce a fuzzy image at the image plane due to diffraction.

Now consider an array of 2um wide line light sources and an array of narrow slits.  The ability to pass or block light is the same, but the resulting image is now much sharper, but multiple copies spread out by the diffraction angles of the slit pitch.  With 0.5mm wide slits, the higher order diffraction images are suppressed.

BTW, I think there is a favorable factor of two due to light passing through the warm air column on its way to the mirror and again after bouncing off the mirror.  For an object plane close to the mirror (as here), the direct and reflected images of the object are close to overlapping.  The viewed image is a combination of the two.

1e-6 change in index of refraction makes a 4e-6 radian angle change through a cylinder.  Light at the center passes through 2r before reflecting from the mirror and another 2r after.  Light at the edge 1r away passes through none of the cylinder.  That 4e6 radians becomes 6um displacement at 1.5m.  Given the discussion above, 6um shift in the light pattern would be plenty visible if my shadow mask and focusing were perfect and adjusted to block most of the light.  Imperfections make it barely visible.

Thank you for making me think this through!  It is a fun challenge that I still haven't fully grasped.

Edit:  After another hour+ of thinking, I believe that conventional knife-edge Schlieren doesn't have a diffraction issue either.  As long as the camera lens/aperture is large, it still forms an accurate image based on the visible half not blocked by the knife.  Only if the knife edge were replaced with a narrow slit or pin-hole would there be a diffraction issue.

Edit2:  Changed a bit above, and thought of another way to look at diffraction.  With the shadow mask adjusted for an almost-black image, diffraction around anything in the object plane redirects light rays, making the image bright around the diffraction source.  I changed to a 0.5mm pitch (0.25mm lines and spaces) shadow mask.  Edges of opaque objects form quite bright outlines in the image now.  (Were before, but even more obvious now.)  The bright lines have clear steps at the expected diffraction angle of 0.5mm pitch lines, just under 2mm referenced back to the object (mirror) plane.  When viewing a 1C warmer cylindrical air column, the diffraction from that column should add to refraction in redirecting light rays and making the cylinder visible.

Here's a picture of a 70mm diameter opaque disk in front of the mirror using the new 0.5mm pitch shadow mask:
« Last Edit: January 07, 2021, 07:29:17 PM by davekni »
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Re: Schlieren images projected onto a wall (no camera)
« Reply #12 on: January 08, 2021, 06:29:08 PM »
I'm sorry about the delays in my postings. It is not, that I don't want to reply, it is because I'm not at all sure, that I understand your reasoning. Which certainly does not imply, that there is anything wrong with it.

I agree, that image resolution is not limited much by diffraction in the case of many slits or a knife edge, since both provide a wide pathway perpendicular to the light rays. But I think sensitivity is an issue of contrast, not of resolution. I like to think of the compressed air to be a source of light. This light source creates just the difference between field intensities with and without the air being compressed. When you setup the grating or knife edge in such a way, that it will block most of the undisturbed field, you can disregard that field and are left with the light from that difference light source.  Either diffraction or refraction, both of which bend the light rays near this source, can lead to light going past the knife edge.

In our case, refraction angles are only a few microradians, so it is not easy to see, how the light can get past the knife edge. In the case of diffraction, angles are much larger. Possibly the high sensitivity we see is due to diffraction.
« Last Edit: January 08, 2021, 07:02:19 PM by Uspring »

Offline davekni

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Re: Schlieren images projected onto a wall (no camera)
« Reply #13 on: January 08, 2021, 07:27:31 PM »
My estimate of 4 microradians makes a 6um displacement at the shadow mask 1.5m away.  My 1mm pitch (0.5mm lines and slits) shadow mask was adjusted to block ~90% of the light.  So the overlap between light lines and slits is only 50um.  A 6um shift is 12% of the remaining light intensity, which seems reasonable for what I was describing as barely visible.

I thought the puzzle was that diffraction from a 10mm diameter cylincder is much more than 6um (at a 1.5m distance), so would need to be displaced much more than 6um to make a significant intensity change.  That is what I was trying to puzzle out myself and came up with my previous post.  That post is a bit rambling and incoherent :) as I realized invalid aspects of my initial versions.
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Re: Schlieren images projected onto a wall (no camera)
« Reply #14 on: January 11, 2021, 06:05:15 PM »
I think, you are right by assuming diffraction angles to be much bigger than refraction angles. In the case, where you have a continuous change of refraction indices like e.g. a warmer region of air, you can calculate the refraction angle by the change of direction a wavefront takes. Consider e.g. a straight wavefront traversing adjacent regions of cold and warm air. The light in the warmer region will be a bit faster than in the cold one and will make the wavefront turn. The angle comes out to be L/n * dn/dx, n being the refraction index and L the length of the region. The coordinate x is to be taken parallel to the wave front. For e.g. a length L of 10 mm ( going through 5mm in each direction ), dn = 1e-6 and dx = 2 mm, the angle of refraction is 5 microradians.

Diffraction angles are of the order of lambda/dx for a slit of width dx. In our case, that would be 500 nm / 2 mm = 2.5e-4, which is 50 times more than the refraction angle. The ration of diffraction to refraction angles is independent of dx.

Whether the diffracted light will be visible, depends on its intensity. I think, that much of it makes it past the slits due to the large diffraction angles. As outlined in my previous post, I'm thinking of an extra light source to be the source of the diffracted light. The corresponding field amplitude is the difference between the amplitude in the warm and the cold air. The magnitude of the amplitude can be estimated from their relative phase shifts.

A 10 mm length of air contains about 20000 wavelengths. For a speed difference of 1e-6, the relative phase shift is 0.02 wavelengths or 0.12 radians. For small phase shifts, the difference between sine waves is a cosine wave with amplitude delta phi. Intensity, being the square of amplitude is thus only of the order of 1 %. Difficult to see, but perhaps possible, if most of the background light is blocked.

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Re: Schlieren images projected onto a wall (no camera)
« Reply #15 on: January 11, 2021, 11:21:10 PM »
Thank you for that brain-jog in the last paragraph!  I've been totally ignoring the obvious in all my analysis - that intensity is proportional to the square of field.

That explains what I'd subjectively observed.  Was questioning myself if it could be just non-linearity in human perception.  It seemed that the 3C-warmer air stream image was well more than 3x as visible as the 1C-warmer stream.  9x more might actually fit more closely to my observation.

One puzzle still:  With the shadow mask adjusted to block most (~90%) of the light, I'd expect diffraction to make bright areas, but not dark.  The part of the 500um wide light stripes that are diffracted will be spread out (fuzzy), so less blocked by the 500um wide shadow mask stripes.  Certainly that is the case when I add an opaque object - bright edges.  However, when I look at the sound wave images, left edges are dark and right edges are light.  To me that indicates refraction, at least a significant amount compared to diffraction.

The sound wavefront looks to be on the order 11mm wide, which I'd tried to roughly mimic with warm air cylinder diameter.  For round numbers, say 5mm radius and 0.5um light wavelength, for ~1e-4 radians diffraction.  At the shadow mask 1.5m from the wavefront, that's 150um spread.  50um from the edge of the spread 500um light stripe, the 150um spread should reduce intensity only ~30%.  A 6um refraction displacement would make a ~8% change in light intensity instead of ~12% (6um/50um) if diffraction were not considered.  Am I missing anything here?  So I'm back to thinking that refraction is the dominant effect being viewed.
« Last Edit: January 13, 2021, 07:31:48 PM by davekni »
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Re: Schlieren images projected onto a wall (no camera)
« Reply #16 on: January 15, 2021, 07:11:03 PM »
When doing a calculation to find out, what a Schlieren image looks like, I find, that it is somewhat difficult to attribute what we are seeing to either diffraction or refraction. Consider a wavefront a spherical mirror produces, when it focuses light to nearly a point at the sphere center. That wavefront is an ingoing spherical wave. Now assume something placed near the mirror, which modulates the wave, i.e. it attenuates and/or phase shifts it. So the field amplitude is multiplied by some spatially dependent value f(x,y), which is complex and so allows for both attenuation and phase shift. The amplitudes in the plane of the sphere center will then be the Fourier transform of f(x,y). I've found the derivation here: https://en.wikipedia.org/wiki/Fourier_optics#Fourier_transforming_property_of_lenses somewhat opaque, but it is not difficult to homebrew that by thinking of Huygens principle. The idea also applies to spherically shaped mirrors.

On a screen further behind, the Fourier values are transformed back and you will have the image described by f(x,y) back. The upshot of all this is, that placing a knife edge near the center of the mirror sphere will block a certain region in spatial frequency space, e.g. those with negative frequencies if the edge is placed exactly touching the focal point.

In order to see, what happens to the image I find it simpler to look at this frequency filtering process in the spatial domain. That can be described by a convolution kernel. It has the form (one dimensional for simplicity):

K(x) = sin(x)/x   +    j*(1-cos(x))/x

The first part just blurs the image a bit. The imaginary second part is antisymmetric with respect to x. If used as a convolution kernel it approximates a derivative. So the Schlieren setup adds a spatial derivative of f to the image amplitude.

To get back to the puzzle you mention: In the case of purely diffraction, the function f is strictly real, which makes the imaginary part of the kernel strictly imaginary. When adding that to something real, it can only increase amplitude. But for compressed air, f will have a phase shifting, i.e. imaginary part. That can interfere then either positively or negatively with the sin(x)/x part of the kernel.

Wrt to refraction being the dominant part: If you look at the cause of the Schlieren in the case of compressed air, that is caused by phase and not by attenuation, since compressed and non compressed air are similarly transparent. And phases are the source of refraction. But to discuss this in terms of ray bending to get by the knife edge is somewhat questionable as we are in the frequency domain near that edge, but the location of light rays certainly belong into the spatial domain.


Offline klugesmith

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Re: Schlieren images projected onto a wall (no camera)
« Reply #17 on: January 16, 2021, 06:23:10 PM »
I love Dave's projection feature, so viewer's eye (or camera) doesn't need to be in a special place.

My own schlieren imaging project, now dormant for 14 years, did require eye or camera in a special place.
But got pretty sharp pictures of some pretty small things.

Just downloaded a pair from my place on Flickr, found by searching for "shock diamonds".
We're looking at fast and slow jets of "duster spray" gas, from nozzle diameter 0.7 mm (0.028").

https://www.flickr.com/photos/r_k_f/4645245687

This morning I moved some junk in the garage, and dust covers, to reveal the optical bench where it has remained since 2006.  Concave spherical mirror, from ebay, is about 6 inches in diameter and mounted a lot like Dave's.   Initially used to look at rising air plumes from soldering irons, warm hands, etc. placed close to the mirror.   Then used to look at small subjects close to the lamp or to image of lamp, I don't remember which.

Near the other end are remaining parts that used to hold lamp, knife-edge, camera guide, and optional subject.


Couple of other points.
1. As a schlieren subject, non-luminous shock diamonds are common in compressed air jets, as from ordinary shop blow-guns.  Can be seen in shadow of jet in ordinary sunlight, or more sharply when light source is reflection of sun in a convex mirror.  Enlarged image can be projected on a wall by using point source of light, such as laser pointer without its lens.
On my Flickr page is a video of duster spray shock diamonds. Light source is a little flashlight bulb. Shadow of the jet (slightly enlarged) falls on sensor of a webcam from which lens has been removed.

2. Looking forward to joining conversation about the mathematics of schlieren and shadowgraph imaging, which use different mechanisms to translate refractive index gradients into bright and dark areas.

3. Schileren is an ordinary noun, not a proper noun. So in English it should not be capitalized, except in titles and at the beginning of sentences.

[edit] In response to where this thread was going before my post, I think most schlieren images can be explained without any consideration of light wavelengths and diffraction.  It's just about ray deflection due to refractive index gradients in the medium.   Schlieren methodology generates brightness differences according to direction of the ray deflection WRT the knife edge or other mask.  One variant can give a color contrast according to one "knife edge" axis, and orthogonal color contrast according to an orthogonal "knife edge" axis.
Let's tip our hats to a couple of great contributors & teachers here, Andrew Davidhazy and Gary Settles.
https://scholarworks.rit.edu/article/365/
« Last Edit: January 16, 2021, 07:59:54 PM by klugesmith »

Offline davekni

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Re: Schlieren images projected onto a wall (no camera)
« Reply #18 on: January 16, 2021, 09:27:25 PM »
Yes, color looks like fun too.  Should be fairly simple with my setup.  I just need to print some colored shadow-masks.  The printed colors aren't as smooth as good color filter film, so some light will be lost (redirected) due to refraction and diffraction from the toner layer non-uniformity.

Thank you for clarification on capitalization.  I found articles written both ways, so took a random guess.

Diffraction isn't the desired effect in schlieren optics.  It is present and places limits on sensitivity and resolution.  I've been struggling with simplified conceptual models to estimate diffraction effects without making a computer model of intensity and phase across the shadow mask.  Might make the computer model some day for my shadow mask version.

I made a 0.5mm pitch (250um lines and spaces) shadow mask.  With that, diffraction from the mask becomes quite noticeable, as in my previously posted image of an opaque disk.  Most of the above discussion is about diffraction from the density object being imaged (warm air stream or whatever).  Diffraction around the edges of an opaque object are visible with any shadow mask or knife edge.  The fuzziness of the diffracted light around the disk edges is due to more diffraction from the shadow mask.
David Knierim

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Re: Schlieren images projected onto a wall (no camera)
« Reply #19 on: January 17, 2021, 05:12:45 PM »
klugesmith, be gentle on me. Schlieren is a word of my native language and a noun and as such always capitalized. My mistake in carrying that over into English.

Wrt to the math: I've found it of interest to see, that diffraction and refraction formally can be incorporated into one complex number, the real part causing diffraction and the imaginary refraction. So technically they seem related. But perhaps, there is no real meaning behind this.

I've wondered, that in wikipedia articles about schlieren photography Fourier optics wasn't mentioned. I've derived the convolution kernel in my last post following Fourier optics principles and I believe (still), that to be correct. The kernel shows features of smearing and also of differentiation, which is in agreement to Davids observation of sound waves. Note that brightness is inverted when comparing the right and left side of the images. That is, because we are seeing derivatives of the refraction index and not the index itself.

Qualitatively, the kernel keeps its form in the limit of zero wavelength. Snell's law of refraction also does not depend on wavelength. Probably there is a way to calculate schlieren images without resorting to waves, but I'm not aware of them.
« Last Edit: January 17, 2021, 05:27:19 PM by Uspring »

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Re: Schlieren images projected onto a wall (no camera)
« Reply #19 on: January 17, 2021, 05:12:45 PM »

 


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