Anyone keen to photograph some candles, for scientific experiment?

[amolitor]

E = t π B / (4 N^2)

Note the absence of any term relating to the distance to the object.

It's not that it contradicts, the question is whether it has anything to do with inverse square law, which is all about the distance.

Helen's point is that image illumination (E) has nothing to do with distance. You claim that image illumination (E) does vary with distance.

One of you is right, and the other one is wrong. Guess which one?

 

[fjrabon]

christop, I acknowledge what you said in that other thread, ASCII stuff and all that. I addressed it with the question to fjrabon in my previous post.

E = t π B / (4 N^2)

Where
E is the image illumination,
t is the lens transmittance (dimensionless),
π is Pi (dimensionless),
B is the object illumination, and
N is the f-number (dimensionless).

Note the absence of any term relating to the distance to the object.

The above equation does not contradict the inverse square law at all.

It's not that it contradicts, the question is whether it has anything to do with inverse square law, which is all about the distance.

It's not that it has to do with the inverse square law, it's that a properly focused image in some sense does away with the inverse square law in a 'brightness per inch' sense and instead turns the inverse square law into what we call 'perspective'. When you've properly focused on an image, it no longer gets dimmer as it gets farther, it gets smaller when it gets farther. Unless you have a lens doing this, it gets dimmer, but more spread out.

 

[amolitor]

Hey Tris, could we skip past this boring "establishing some facts" part and get right to the part where you present your no doubt hilarious theory?

 

[tris_d]

You seem to be confusing light fields and points of light as resolved by a lens or an eye.

Lenses take light from a source that is hitting your eye, all across retina, and focuses it into a coherent image. When you look at a candle, the light from that candle isn't hitting a small part of your eye, its hitting your whole eye, your whole body, the whole room. Your eye focuses it. Whatever size the flame looks like is based upon your distance to it, as the further you move away, the smaller part of your visual frame it takes up. However, because your eye (or camera) focused the diffuse light back into a coherent image, you no longer get the dimming effect. Your eye sees the candle as brightly as it would see it a few feet away.

So, if you focus the light you get the smaller effect, but not the dimmer effect. If you were not to focus the light (essentially what holding a sheet of paper over the light does) you would get the dimming effect, but a corresponding increase in the 'spread' of the light (which is the whole reason the inverse square law works to begin with).

This is why we kept telling you that you could represent stars as dimming, or getting smaller, but not both.

So the amount of light that hits the lens drops off with the distance, but the brightness stays the same as it proportionally gets focused on a smaller area on the image. Correct?

Now, measuring the amount of light falling on an area from a light is a totally different matter. If you're in a dark room and you have a light meter, as you move a flash light closer to the light meter, it will register more light. As you move it away, it will register less. However, if you took pictures of the same flashlight, the actual flashlight would seem equally as bright no matter the distance. However, in the further picture it would be taking up less space in the visual frame. What that means is that if you were to calculate the value the sensor read for the flashlight and multiply it by the area it took up in the frame, the flashlight that was further away would follow the rules of the inverse square law, because the less light is being illustrated by the light taking up less space in the frame.

again, indicating that you can illustrate the inverse square law as lights being dimmer, or lights being smaller, but not both. Which was your original problem.

All right. I think we finally sorted this out. The thing is no one mentioned any lens or focus until Unpopular replayed to me in that "stupid physics" thread. I suppose it was just plain obvious to you, but it never occurred to me, it is not something I could find in my physics text books in relation to inverse square law, so you should cut me some slack about previous misunderstandings.

The reason why stars appear to get dimmer the further they get is because they are too far away for our eyes to focus on them. We can only focus at an arbitrarily far point into space, and after that, things just look equally small. Because we can't really focus on the objects in space, as they get further away, they simply look dimmer. If, however, you had a telescope accurate enough to focus on them, they would look just as bright as anything else, even if they were very tiny.

Ah, the stars. Now we can finally get back to Olbers' paradox. So, when the light source get so far away that its focused projection covers no more than one pixel on the image, then the inverse square law starts to apply and the brightness drops off with the square of any further distance from that point on, ok?

 

[tris_d]

Hey Tris, could we skip past this boring "establishing some facts" part and get right to the part where you present your no doubt hilarious theory?

Ok, here is question for you: does magnification make distant stars appear brighter?

 

[amolitor]

Hey Tris, could we skip past this boring "establishing some facts" part and get right to the part where you present your no doubt hilarious theory?

Ok, here is question for you: does magnification make distant stars appear brighter?

Define "appear brighter".

 

[unpopular]

provided that the optical system's f-ratio is similar across magnification, the amount of light reaching the sensor will be similar. so no, magnification does not increase the appearance of brightness.

 

[tris_d]

Define "appear brighter".

Ok, for simplicity lets just talk about gray-scale images, so the brightness is defined from 0 to 100 (in Photoshop), where 0 is black, 100 is white, and in between are shades of gray. And when talking about magnification of some star and its consequent brightness, I think what I am asking is just about the pixel in the very center of the "blob". -- By the way, do human eyes have some magnification property built-in, and can it vary for example as we focus to closer and further away objects?

 

[christop]

So the amount of light that hits the lens drops off with the distance, but the brightness stays the same as it proportionally gets focused on a smaller area on the image. Correct?

Yep!

All right. I think we finally sorted this out. The thing is no one mentioned any lens or focus until Unpopular replayed to me in that "stupid physics" thread. I suppose it was just plain obvious to you, but it never occurred to me, it is not something I could find in my physics text books in relation to inverse square law, so you should cut me some slack about previous misunderstandings.

I think that's fair enough. This is a photography forum, and you started a discussion about the physics of light, so I think most of us assumed that you already had some understanding of optics.

Ah, the stars. Now we can finally get back to Olbers' paradox. So, when the light source get so far away that its focused projection covers no more than one pixel on the image, then the inverse square law starts to apply and the brightness drops off with the square of any further distance from that point on, ok?

Yeah, if a star covers less than a whole pixel, its size cannot be measured, but the brightness of the pixel becomes basically an average of the brightness of the star itself and the surrounding blackness of space. The pixel's brightness can then be used to estimate the angular size of the star relative to the size of a pixel and therefore its distance can be estimated. (I'm not sure if imaging sensors are used this way in reality, but I think the concept is correct.)

 

[amolitor]

Ok, I think that sounds good to me.

Greater magnification is a somewhat problematic phrase as well, but I think we can work with it. There are two factors in play here, focal length, and aperture. Aperture is normally measured as the ratio of the lens opening to the focal length. So, if when we increase the focal length, we normally decrease the aperture. The answer to your question is, it depends.

Greater magnification is associated with longer focal lengths. When imaging the sky, this simply means that we're placing less of the area of the sky onto the sensor.

If you double the focal length (increasing magnification) while keeping aperture the same (for example, for from a 300mm f/2.8 lens to a 600mm f/2.8 lens), several things occur:

- less of the sky is imaged on to the sensor, 1/4 as much, to be exact.
- the size of the lens opening has doubled, which means the area of that opening has quadrupled, which means the light-gathering power has quadrupled.

The effect is that the apparent brightness of the stars will indeed quadruple. This is why building a bigger telescope is actually a helpful thing to do. Stars not previously visible are now visible, if you can build a telescope with a bigger hole for light to pass through, with more light gathering capability.

If you double the focal length but do not change the light gathering power of the system (for instance, inserting elements into a telescope system to increase the magnification) you will effectively cut the aperture in half. The will also have some effects:

- less of the sky is imaged on the sensor, 1/4 as much. Same as last time.
- the size of the lens opening is the SAME, the light gathering power is the SAME. The infinitesimal points which are the images of stars will remain at the same brightness.


This is why, in order to see deeper into the night sky, you cannot simply add elements on to the viewing end of the telescope. You must build a bigger telescope, with more light gathering power -- with a larger physical aperture through which light can pass.

 

[christop]

If you double the focal length (increasing magnification) while keeping aperture the same (for example, for from a 300mm f/2.8 lens to a 600mm f/2.8 lens), several things occur:

- less of the sky is imaged on to the sensor, 1/4 as much, to be exact.
- the size of the lens opening has doubled, which means the area of that opening has quadrupled, which means the light-gathering power has quadrupled.

The effect is that the apparent brightness of the stars will indeed quadruple. This is why building a bigger telescope is actually a helpful thing to do. Stars not previously visible are now visible, if you can build a telescope with a bigger hole for light to pass through, with more light gathering capability.

Hold on... the light-gathering power with a physically larger aperture (but with the same f-number) is exactly counteracted by the increased "spreading" (or magnification) of the light on the imaging plane (the sensor). That's why the image doesn't change in brightness as you zoom in and out while keeping a fixed aperture.

 

[amolitor]

Stars don't spread, they're infinitesimal points, for our purposes.

If you're taking a picture of a light bulb or an onion or a cow, then the image spreads out and smears the extra light across more sensor, and you are perfectly correct. Stars being imaged on a digital sensor won't make it past boundaries of the pixel, so the recorded number will be bigger for that single cell of the sensor's array. This is why I asked about "apparent brightness"

With film.. I suspect the same thing will occur. My sense is that, in general, the star's image still is pretty much a point so it's just going to bang harder on the same single silver halide crystal, and render it more and more broken down, and hence darker in the negative. There might be something else going on there, though. Maybe the Airy Disk (sp?) becomes "resolvable" in an interesting way such that it makes sense to think of it as "bigger" rather than "brighter" in the land of astro film. I'm not an astro guy, I can just handle a little maths from time to time.

 

[christop]

Stars don't spread, they're infinitesimal points, for our purposes.

If you're taking a picture of a light bulb or an onion or a cow, then the image spreads out and smears the extra light across more sensor, and you are perfectly correct. Stars being imaged on a digital sensor won't make it past boundaries of the pixel, so the recorded number will be bigger for that single cell of the sensor's array. This is why I asked about "apparent brightness"

Oh, I see. In that case you'd be correct.

 

[tris_d]

So the amount of light that hits the lens drops off with the distance, but the brightness stays the same as it proportionally gets focused on a smaller area on the image. Correct?

Yep!

All right. I think we finally sorted this out. The thing is no one mentioned any lens or focus until Unpopular replayed to me in that "stupid physics" thread. I suppose it was just plain obvious to you, but it never occurred to me, it is not something I could find in my physics text books in relation to inverse square law, so you should cut me some slack about previous misunderstandings.

I think that's fair enough. This is a photography forum, and you started a discussion about the physics of light, so I think most of us assumed that you already had some understanding of optics.

Ah, the stars. Now we can finally get back to Olbers' paradox. So, when the light source get so far away that its focused projection covers no more than one pixel on the image, then the inverse square law starts to apply and the brightness drops off with the square of any further distance from that point on, ok?

Yeah, if a star covers less than a whole pixel, its size cannot be measured, but the brightness of the pixel becomes basically an average of the brightness of the star itself and the surrounding blackness of space. The pixel's brightness can then be used to estimate the angular size of the star relative to the size of a pixel and therefore its distance can be estimated. (I'm not sure if imaging sensors are used this way in reality, but I think the concept is correct.)

Well, your patience paid off. I appreciate it, all of it, thank you very much. Thank you everyone else. And it's nice when people agree, makes me feel warm inside. It's just that... it's BORING!! So prepare to hate me once again as I'm now ready to prove the whole world wrong, or so I shall try. Brace yourself, here I come!

 

[runnah]

Can't you accomplish the same effect with cars on a highway at night?

http://image1.masterfile.com/em_w/00/53/02/700-00530279w.jpg

Or am I missing the point?