Why can I see a light from farther away than it shines?

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vestureofblood
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Why can I see a light from farther away than it shines?

I understand that when you shine a light toward an object, that only a small portion of that light is reflected off most surfaces.
I understand that what light is reflected off those surfaces is scattered and only a portion of it bounces back to the users eyes.
I understand that only part of the light from the source makes it to the object because of inverse square.

WHAT I DO NOT UNDERSTAND is why if I am standing the receiving end, outside the range of the flashlight far far away I can still see a very bright light when I look back to ward it.

If I was on a mountain top in the desert and 20 miles away in the darkness a car was driving across the valley floor I would be able to see it perfectly. WHY?

I cant get my head around why inverse square law would not seem to apply here. At 20 miles away if I was holding a light meter I would be getting ZERO lux from those headlights.

It seems like since I was outside the range of the light none would be hitting my retinas. But that is not the experience I am having.

In Him (Jesus Christ) was life; and the life was the light of men. And the light shineth in darkness; and the darkness comprehended it not.
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Edited by: vestureofblood on 07/19/2022 - 18:22
zoulas
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I think it has to do with usable light but I will let the experts answer this one. Similar explanation why you can see a star from earth but if you were standing on that star you could not see the earth.

ggf31416
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Your eyes focus the light from the flashlight into a small point, and your eyes are dark adapted. You can easily see many individual stars if the sky if there is little ambient light (think of every star as a distant flashlight) but even all the stars combined are not enough to walk without tripping (Wikipedia lists moonless clear night sky with airglow as 0.002 lux, so the starts themselves are even fainter).

turkeydance
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roundtrip?

light from headlight make one trip.
reflected light makes two and it
significantly dimmer when it
starts back.

zoulas
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This is truly a question for Carl Sagan.

My opinion is that light always makes two trips. If it only made one, everything would be blackness.

If you shined a light on the moon, it would have to reach the moon and reflect back to you. If it only reached the moon but did not reflect back to you, you would not see the moon at all.

Scallywag
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turkeydance wrote:
roundtrip?

light from headlight make one trip.
reflected light makes two and it
significantly dimmer when it
starts back.


This one. It’s reflected inefficiently and making a round trip, or a one-way trip with no reflection
Oli
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https://aty.sdsu.edu/mirages/FM/lights.html Nocturnal Lights. I found this article over a year ago. I’ll have to go and read it again now, but I think this is the one that talks about surveyors using fire on hilltops at night to get compass readings.

You can't compare the big flashlight in the sky to the little flashlight in your hand.

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SammysHP
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Your eyes are much more sensitive than the light meter. They receive the same lux as the meter, but are still able to detect the light. On the other hand the same lux scattered and mostly absorbed from an object returns almost nothing to your eye. You can shine a thrower at a wall and look at the spot. Now try it the other way around (and post a photo of it on r/flashlight because it’s trending over there at the moment).

Unheard
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The illuminated object eats energy and scatters the light.

The human eye can see single photons.

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flashflood
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The main issue is that the reflected light also follows an inverse square law, so unless you are illuminating a mirror or retroreflective surface, the returned light is more like inverse fourth power. I first learned about this while working with a company that makes depth sounders. A basic illustration can be found here: https://www.edn.com/inverse-fourth-power/

zoulas
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This was a huge gimmick in an infomercial some time ago. Instead of stating throw distance, they would say “this flashlight is so powerful, it can be seen from 10 miles away.”

This woulds lead many people to believe that the light has a 10 mile throw. Meanwhile in reality that light only has a 10 foot throw.

CRC2
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I love that the person in the diagram is actually you lmao

kennybobby
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It depends upon whether you consider light to be a wave or a particle—it is neither but has properties of both.

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If you exposed some film to a distant light source, it would not show a point of light like our eyes see, just an inappreciable amount of basically uniform exposure caused by the dim light. The same thing happens with a light meter when the light is too dim to pick up. This is because light travels in waves and much like waves on the ocean they seem to have no point of origin. But our eyes don’t work like a piece of film or a light meter. If you put the film in a camera, the lens focuses the parallel waves of light into a point, concentrating the dim light so it stands against a dark background in a photo. Our eyes have lenses that work much the same as a camera lens. Since the light is concentrated into a point by the lens in our eyes, we can see it at great distances (billions of light years for stars).

When we shine a flashlight on something in the distance, we see only the light reflected back to us. Not only do we have to deal with the light lost on the return trip, but the reflected light is scattered and is no longer in parallel waves and will no longer focus into a distinct point by the lenses in our eyes or in the lens of a camera.

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In a dark place, you can see a light on a sub-lumen moonlight setting that is pointed at you from quite a distance. Even if the same light is used in a totally dark room the eye cannot see any light being cast of a white wall 10 feet away. Sorta the same thing at work with brighter lights and greater distances.

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The story on battlefields used to be an enemy could see a cigarette being lighted by a Zippo from miles away.

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lampliter
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At a certain distance that light would disappear. If you didn’t move the light source or your body; how long would it take to “see the light”?

https://www.space.com/speed-of-light-properties-explained.html

"The marvel of all history is the patience with which men and women submit to burdens unnecessarily laid upon them by their governments.” ~ U.S. Senator William H. Borah (1865-1940) affectionately known as the "Lion of Idaho"

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flashflood wrote:
The main issue is that the reflected light also follows an inverse square law, so unless you are illuminating a mirror or retroreflective surface, the returned light is more like inverse fourth power. I first learned about this while working with a company that makes depth sounders. A basic illustration can be found here: https://www.edn.com/inverse-fourth-power/

The illustration shows two point light sources while reflected light is none.

We discussed this already:

https://budgetlightforum.com/comment/1772035#comment-1772035

Edit: I made this observation when I had a flashlight mounted on binoculars. The magnified image of an illuminated object was way dimmer than when looking with the naked eye. This was counterintuitive and so I made that model.

Spitzbube.

flashflood
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Unheard wrote:
flashflood wrote:
The main issue is that the reflected light also follows an inverse square law, so unless you are illuminating a mirror or retroreflective surface, the returned light is more like inverse fourth power. I first learned about this while working with a company that makes depth sounders. A basic illustration can be found here: https://www.edn.com/inverse-fourth-power/

The illustration shows two point light sources while reflected light is none.

We discussed this already:

https://budgetlightforum.com/comment/1772035#comment-1772035

Edit: I made this observation when I had a flashlight mounted on binoculars. The magnified image of an illuminated object was way dimmer than when looking with the naked eye. This was counterintuitive and so I made that model.

The inverse square law applies to an omnidirectional (or at least conical, which is a fixed fraction of a sphere) light source. Perfectly collimated light, like a laser beam, doesn’t diminish at all. This principle applies to both the source and the reflector. A flashlight, unless it’s a laser, is a conical light source. The reflecting surface is typically omnidirectional (roughly hemispherical) as well. To see this, imagine shining a flashlight at a white wall, and walking around the room looking at the wall from different angles. You will be able to see the hotspot from anywhere, because the wall mainly scatters the light rather than reflecting it. This, incidentally, is why retroreflective road signs look so much brighter than everything else at night — because they are reflecting, not scattering, your headlights.

Unheard
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flashflood wrote:

The inverse square law applies to an omnidirectional (or at least conical, which is a fixed fraction of a sphere) light source.

It applies to a point light source, which reflected light is not.

That’s why a phtographer can measure light from an illuminated object at any distance. Try for yourself and measure with your phone an illuminated wall. Halve the distance. What are the figures you get?

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flashflood
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Unheard wrote:
flashflood wrote:

The inverse square law applies to an omnidirectional (or at least conical, which is a fixed fraction of a sphere) light source.

It applies to a point light source, which reflected light is not.

That’s why a phtographer can measure light from an illuminated object at any distance. Try for yourself and measure with your phone an illuminated wall. Halve the distance. What are the figures you get?

It applies to an omnidirectional point source. If the point is directional, like single photon, there is no loss of intensity at any distance — that’s just conservation of energy. The correct way to think about the problem is to look at the statistical behavior of single photons. Your emitter sends photons onto some two-dimensional area (hence inverse square distribution), and then each photon is scattered randomly by the surface (unless it is a true reflector), which means it can go in any direction. Each photon’s point of impact thus behaves roughly like an omnidirectional point source, and therefore the number of scattered photons returning to the two-dimensional area of your retina also follows an inverse square law.

Unheard
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Just do the proposed experiment.

Edit: I did Big Smile . Interesting.

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flashflood
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Unheard wrote:
Just do the proposed experiment.

If I double my distance from a uniformly illuminated wall, I’m getting 1/4 the intensity from each square inch of wall, but my lens is taking in 4 times the area, so there’s no net effect. That experiment doesn’t tell us anything. Or are you proposing something else?

ggf31416
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flashflood wrote:

The inverse square law applies to an omnidirectional (or at least conical, which is a fixed fraction of a sphere) light source. Perfectly collimated light, like a laser beam, doesn’t diminish at all. This principle applies to both the source and the reflector.

For large enough distances lasers still follow the square distance law as we still have difraction, so you can’t expect to shine a laser to the Moon and receive it back from a mirror without attenuation. For close enough distances the dispersion is much smaller than the original beam width so the square distance law may be ignored.

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One of my viewers suggested I use an eye chart and sit my camera off to the side, then shine a flashlight at it from various distances to measure usable light. My thoughts would be that the camera is then only seeing the light one way and would therefore be brighter and inaccurate, yes? I’d need to take the photo from where I’m shining the flashlight from.

Piercing The Darkness YouTube channel – https://www.youtube.com/c/PiercingTheDarkness

Unheard
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flashflood wrote:
Unheard wrote:
Just do the proposed experiment.

If I double my distance from a uniformly illuminated wall, I’m getting 1/4 the intensity from each square inch of wall, but my lens is taking in 4 times the area, so there’s no net effect. That experiment doesn’t tell us anything. Or are you proposing something else?


We’re talking about brightness – luminosity.

See:

Camera in manual mode, histogram plots, distance halved.

I see the brightness doesn’t change significantly, no matter the distance.

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ggf31416
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The JPG already has some processing applied, they only have 256 possible brightness levels so the scale is compressed in a non proportional way (for example 210 may be 4000 and 220 8000), you would need to shoot in RAW if your camera supports it and disable as much processing as possible.

https://www.cambridgeincolour.com/tutorials/gamma-correction.htm

Unheard
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ggf31416 wrote:
The JPG already has some processing applied, they only have 256 possible brightness levels so the scale is compressed in a non proportional way […]

No, it’s not. Be rest assured I know what I’m doing.

Please provide your data if you doubt the results.

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flashflood
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Cool — that’s exactly what you’d expect from theory, but it’s always good to have experimental confirmation.

zoulas
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Talk about getting off topic.

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