Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

My question: why does light seem to occlude an object in front of it, especially when viewed from a distance? E.g. imagine a grid-like structure (for example an empty gasometer) viewed from a distance, with the sun behind it, or even the sun setting below the horizon. Light seems to overlap the object in front of it, as an aura or haze. I'm looking for a scientific explanation of this phenomenon.

Thanks.

[Edit:] Just for clarity's sake, here is a photo capturing exactly what I mean, where the black struts of the poles seem to become attenuated by the light behind them:

enter image description here

share|improve this question
    
I find it hard to decode what effect you actually mean. Do you mean e.g. these beams crossing the tree here? upload.wikimedia.org/wikipedia/commons/7/75/… If you do, then the light is actually coming from a different direction (Sun is not directly behind the tree), so that by going through holes between the trees, it may get in front of the tree. The beams are literally the places where the light goes - and is reflected from small droplets or dust. Note that these light rays are (nearly) parallel in 3D but optically come from 1 point (where the Sun is). –  Luboš Motl Jun 21 '13 at 14:29
    
What is occlude? I searched and I just obtained the closest definition to your scenario as "block". Light doesn't block anything. And how is a gasometer grid-like? It is just cylinder-like:upload.wikimedia.org/wikipedia/commons/2/2f/…, en.wikipedia.org/wiki/File:Gasometer_in_East_London.jpg, upload.wikimedia.org/wikipedia/commons/1/1e/…. –  Dimensio1n0 Jun 21 '13 at 14:49
1  
Thanks Luboš. I know I may not have explained it well, it's difficult to put into words, so your picture was helpful. In that image, I don't mean those shafts of light, but the points of light behind those dark trees on the right. Here's an image that depicts exactly what I mean: s1.ibtimes.com/sites/www.ibtimes.com/files/styles/… –  Greenchurch Jun 21 '13 at 14:50
    
Dimension10, there's nothing for you to see here. I'm sorry your English isn't up to the task. This is an example of an empty gasometer: builtstlouis.net/industrial/images/gasometer-fpse111.jpg. Thanks for looking anyway. –  Greenchurch Jun 21 '13 at 14:52
1  
I'd say it's a similar situation to when you shine a flashlight from behind a candle's flame. The flame seems to vanish (or it becomes harder to see in any case). The current answers already mention diffraction. Now I would argue that a second factor is saturation. Both in photographs and the human eye, saturation can occur. And the effect is especially strong/noticeable if there is a sharp contrast with a far less intense light source (such as the dark structure in the OP's exemplar). So diffraction + saturation with fairly high relative intensities would be my bet. –  Wouter Jun 21 '13 at 19:44

3 Answers 3

Image credit: By Mallowtek (Own work) [CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons

Image credit: By Mallowtek (Own work) [CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons

The phenomenon which the opening poster (incorrectly, in my view) described as "seems to occlude" (an occlusion is an obstruction to the passage of light) is known to photographers as "veiling glare".

The above link goes to a page on the website of a commercial solution provider for testing digital image quality. As the page notes,

Veiling glare is stray light in lenses and optical systems caused by reflections between surfaces of lens elements and the inside barrel of the lens.

and

It is a strong predictor of lens flare — image fogging (loss of shadow detail and color) as well as “ghost” images — that can degrade image quality in the presence of bright light sources in or near the field of view. It occurs in every optical system, including the human eye.

The second link goes to the Wikipedia page on lens flare, which says:

Lens flare is the light scattered in lens systems through generally unwanted image formation mechanisms, such as internal reflection and scattering from material inhomogeneities in the lens. These mechanisms differ from the intended image formation mechanism that depends on refraction of the image rays. Flare manifests itself in two ways: as visible artifacts, and as a haze across the image. The haze makes the image look "washed out" by reducing contrast and color saturation (adding light to dark image regions, and adding white to saturated regions, reducing their saturation).

(All emphases mine)

Measurement of scattered and reflected light

Image credit: By Bautsch (Own work) [CC0], via Wikimedia Commons

[Above diagram shows a] testing setup for determining undesired influence of "backlight". Grey screen at left is imaged optically via camera lens onto image plane of image sensor (or film). Red laser-light beam, source of which is 10° outside of field of view, ideally should not be imaged on image plane but absorbed inside the camera. [Translated from German Wikipedia.]

Measurement of backlight

Image credit: By Bautsch 14:20, 18. Jan. 2011 (CET).Bautsch at de.wikipedia [Public domain], from Wikimedia Commons

Photographic exposure taken by digital SLR camera, exhibiting undesired influence of backlight as per testing setup above. [Translated from German Wikipedia.]


Making this post Community Wiki, as I am neither a physicist nor a photographer, and I expect amplifications to be added so as to make this a more complete Answer.

share|improve this answer
    
Not sure this is consistent with the effect seen in the picture linked in the question. –  Kyle Jun 21 '13 at 17:51
1  
I like both Eugene and Kyle's answers, and they appear to be fully distinct mechanisms. Kyle: although the bright vertical lines crossing through the sun (in Eugene's picture) are certainly different than what Greenchurch asked about, it's unclear to me whether the circular glow covering the tree is the same or different. I think we're going to need an order of magnitude estimate of the diffraction effects to be sure. This requires knowing not just the wavelength of light and the size of the object, but also the relative degrees of brightness. –  Jess Riedel Jun 21 '13 at 18:12
    
@Jess Yes, the difference is the first Answer talks about diffraction at and around the object being photographed (trees, gasometer grid), while the second looks at reflections (and diffraction) taking place within the camera or eye. Maybe they are complementary? –  Eugene Seidel Jun 21 '13 at 18:19
1  
Thank you Eugene, let's agree to disagree over the occlusion! It's interesting that this is said to occur within any optical system, including the eye. I am certainly thinking about the eye, rather than photographs, and of course there are varieties of lens glare which we are more likely to see on photographs we find on the internet. As I said, I am not a physicist either, so I feel unqualified to accept any answer! Very useful to have these though, thank you. –  Greenchurch Jun 21 '13 at 18:24
1  
-1 those links are all describing various types on lens flare, which is not what OP asked about... –  BlueRaja - Danny Pflughoeft Jun 21 '13 at 20:07

Light is a wave; if you imagine a wave front passing through a buoy or a strut/leg of a pier, the wave will first be "blocked" by it, but then eventually "spread out" on the other side and fill in the hole the buoy/leg caused.

In optics we tend to refer to effects like these that result from the wave-like nature of light under the umbrella term "diffraction".

Here is an illustration to demonstrate this phenomenon; in your mind replace the block with a pier leg/buoy an the waves with water waves, or the block with your struts and the waves with light waves.

wave effects

share|improve this answer
    
Great diagrams Justin. I suggest editing your post to address the issues raised in the comments below Eugene and Kyle's answers. In particular, we are hoping an order-of-magnitude estimate can demonstrate that veiling glare is not responsible. –  Jess Riedel Jun 21 '13 at 18:52

I believe the answer is as simple as diffraction. The light diffracts a bit around the intervening object, so some light is deflected such that you perceive it as coming from a direction covered by the object. This happens all the time, but is more noticeable with a very bright light source.

EDIT: I'll attempt to estimate the magnitude of the diffraction effect. I assume the far field approximation throughout. This is a back-of-the envelope approach, so I'm liberally rounding and approximating.

Diffraction of light around a macroscopic object has some similarities to single slit diffraction with a slit larger than the wavelength being diffracted. The angular extent of the central intensity peak in single slit diffraction is approximately $2\lambda/a$ ($\lambda$ is the wavelength, $a$ is the size of the object), so roughly speaking the wave is deflected by half this angle at each edge of the slit (deflection = $\lambda/a$).

The angular size of the object is simply $a/R$ ($R$ is the distance to the object) provided that the object is far away. If the deflection angle is half the angular extent of the object, the intensity peak will totally cover the object (though not at uniform intensity!), so I'd argue that it would start to be quite noticeable if the deflection angle is 5-10% of the angular size of the object:

$.05\leq \frac{\lambda R}{a^2}$

Now I need to start guessing at the values of $\lambda$, $R$ and $a$. $\lambda$ is pretty easy, the Sun emits light with a peak wavelength of about $500\mathrm{nm}$. The width of the bars in the picture aren't too difficult either, I'd estimate $a\sim5\mathrm{cm}$. The trick part is the distance from which the picture was taken... it could be zoomed or cropped and there's nothing to give perspective... so I'll take a guess and say the photographer probably wasn't standing much closer than $100\mathrm{m}$ to the structure (getting further away will amplify the effect, so this seems like a good compromise). Putting in these values, I get:

$\frac{\lambda R}{a^2} = 0.02$

This means that about 4% of the width of the bars would be covered by diffraction effects. If the picture was taken from a bit further away (maybe 500m), then coverage is up to 20%.

Additionally, there are potentially secondary peaks in the diffraction pattern contributing; the secondary peaks are much dimmer than the primary, but the Sun is very bright and I'd guess the camera detector is saturated, so even a dimmer secondary peak could potentially still get a strong signal in the detector.

Still, unless this picture was taken from quite a distance (looks like maybe 80% coverage, so about 2km, maybe a little less if my other estimates are a little off), I don't think diffraction can entirely account for the observed effect.

share|improve this answer
    
Thank you Kyle, this sounds like it's the right answer, though I am no physicist. So I don't know whether it would be presumptuous for me to mark your answer as the correct one? Happy to do so if it doesn't violate etiquette. –  Greenchurch Jun 21 '13 at 16:46
    
Well if you believe the answer (I do, but then I have a bit of a bias wink) there's no harm marking it accepted. If someone else answers later with something you consider better/more informative you can unmark my answer and mark theirs instead. –  Kyle Jun 21 '13 at 16:53
1  
I believe there is some harm done accepting an answer that does not have much explanation, proof or support from other people who have some expertise. That's how you propagate wrong statements. Diffraction does seem possible here, but it could very well be other effects linked to the detector resolution of leakage from one pixel to another. To support diffraction, please add at least some basic estimates. –  fffred Jun 21 '13 at 17:00
    
Yes, well I'm not sure belief comes into it. I have no authority to accept it, though from perusal of the wikipedia article it makes sense to my layman's mind. Given that this is a resource others may consult, I will wait for now before accepting the answer, to see if anyone else wants to weigh in. Thanks again Kyle. –  Greenchurch Jun 21 '13 at 17:04
1  
Well, let me do a simple attempt to estimate the effect of diffraction. The angle separation between a non-diffracted ray and a diffracted ray is typically the square root of the wavelength divided by the distance to the object (see this). Here, it is about 0.01 degrees. The human eye resolution is actually close to that. So diffraction seems to be possible. –  fffred Jun 21 '13 at 17:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.