Inverse Square Law - I have always wondered...

[rpcrowe]

The inverse-square law for light intensity states: "The intensity of illumination is proportional to the inverse square of the distance from the light source." The light from the source to the subject will fall off as per the Inverse Square Law...

However, in this case, why do I not need increased exposure when I increase my lens to subject distance?

If the light arriving on a subject is decreased in intensity at the inverse square of the distance from the light source, why does the light reflected from the subject to the camera not fall off as it travels from the subject to the camera?

I am sure that there is a simple reason for this but, at the moment I cannot wrap my mind around it...

[Venser]

Here you go.

[PhotomanJohn]

Richard - Yes there is a simple answer. The amount of light from the subject does fall off by the square of the distance but so does the area of the subject. So you end up with less light coming from a proportionally smaller subject so the intensity (exposure or light per unit area) is the same.

Does that help?

John

[Mike Buckley]

I'm not smart enough to think of Richard's question. I'm also not smart enough to understand John's response but I am smart enough to take his word for it.

[revi]

simple enough, though:
if you double the distance to an object, its (linear) size appears to halve.
That means its area (hor. size x vert. size) is reduced to a quarter (0.5x0.5).
Total amount of light received from that object is also reduced to a quarter.
So, the amount of light received per unit surface area (say a pixel) stays the same.

(as an aside, the inverse square law is in theory only valid for point sources, i.e. zero size...
In practice, it holds up pretty well for a naked speedlite at normal working distances)

[pnodrog]

The law only relates to a point source from which the light is radiating equally in all directions. The law does not apply to light reflected from a surface.

[davidedric]

Photoman John has explained it with admirable economy

[benm]

For natural light: the change in lens-to-subject distance is negligible compared to the sun-to-subject distance.
For flash: exposure does need to increase when lens-to-subject distance changes (aperture = GN/distance).

[PhotomanJohn]

For natural light: the change in lens-to-subject distance is negligible compared to the sun-to-subject distance.
For flash: exposure does need to increase when lens-to-subject distance changes (aperture = GN/distance).

The first sentence is true but doesn't have anything to do with the OP's question. The second sentence is not true. As an example, if one has the flash on a light stand and then moves only the camera farther away from the subject, the exposure doesn't change. This is why GN refers to FLASH to subject distance not lens to subject distance.

The point of this whole discussion is that lens to subject distance does not affect the exposure.

John

[DanK]

Another way to put photoman John's answer is that if you move back a distance that reduces the light from an object of any given size by any factor, say X, then the area captured by the lens will increase by the same factor, and for the same reason.

[tao2]

Well, the wagons are circling but arrows are not quite hitting the mark...

why do I not need increased exposure when I increase my lens to subject distance?

Aperture and focal length -

Even though the diaphragm opening is larger - at a given f-number, on longer lenses, the light transmitting ability is the same. It's an accurate generalisation (given lens structures and efficiencies.)

The greater amount of light transmitted by the larger opening is offset by the distance travelled.

Proof

e.g. With a 200mm lens focused on a distant subject, the beam travels 200mm fae the rear of the lens tae the film/sensor in the focal plane. Similarly, a 50mm lens has tae project its beam over 50mm. This is when the ISL kicks in.

If the 200mm lens transmits the same amount of light as the 50mm, then the intensities are in a 16:1 ratio. The 200mm must transmit 16 times as much light as the 50mm. because of the 4x longer light path. So.... its aperture must have a 16x greater area.

Setting f4 on a 200mm lens means a 50mm opening and a 12.5mm opening on a 50mm. The diameter of the effective beam on the lens surface is 4x greater on a 200mm. than on a 50mm. - at the same f-number. Using ᴨ r² for both openings gives areas of 1963.50 and 122.72 respectively. Giving the 16:1 ratio required.

"At any given f-number., the light-transmitting ability of the lenses is mathematically the same"

This applies tae all f-numbers and all lenses.

Bearing in mind that, in photography, maths doesn't always tie-in neatly with the practicalities; nor does the inverse square law...

[xpatUSA]

An interesting subject and the difference between a point source and area source is defined in this graph in terms of deviation from the aforesaid Inverse Square Law:



It is explained fully in this excellent book by Ryder which I have put up on my site for y'alls reference:

http://kronometric.org/phot/xfer/lighting%20handbook.pdf

Won't be there for long . . . it's not supposed to be there at all ;-)

I was going to provide a link to where I got it from but they've changed the rules to where you have to enter all your personal info and sign up for junk email and unsolicited phones, etc, you know the drill . . . .

[pnodrog]

Thanks Ted.

I was refraining from introducing Lambert's cosine law and how it applies when in close proximity to light reflected from ceilings and walls etc but your graph covers it and most other none point forms of lighting nicely. From the point of view of lighting a subject it can be important but it can more or less be ignored in reference to subject from camera distance.

Lets not start talking about collimated and laser lighting or even some reflections....

[Panama Hat & Camera]

The inverse-square law for light intensity states: "The intensity of illumination is proportional to the inverse square of the distance from the light source." The light from the source to the subject will fall off as per the Inverse Square Law...

However, in this case, why do I not need increased exposure when I increase my lens to subject distance?

If the light arriving on a subject is decreased in intensity at the inverse square of the distance from the light source, why does the light reflected from the subject to the camera not fall off as it travels from the subject to the camera?

I am sure that there is a simple reason for this but, at the moment I cannot wrap my mind around it...

The answer is very simple.
The Inverse Square Law is valid for punctual sources of light, but can be a fairly good approximation in some cases.
The reason why you don't need to increase exposure when you move away from the subject that is beeing photographed is because the camera receives less light of the subject and more light of the background.
If the background is dark (an empty place in the night or a black wall) and you are taking photos of a small bright subject, you will need to increase exposure when the distance between the camera and the subject is increased.
Here is an example how to make an experiment to comprove the Inverse Square Law: try taking photos (at various distances) of a sheet of white paper fixed on a black panel. You will see that you will have to increase exposure when the distance between the camera and the subject is increased.
I hope this explanation be useful for you.
Cheers,
Antonio.

[PhotomanJohn]

The reason why you don't need to increase exposure when you move away from the subject that is beeing photographed is because the camera receives less light of the subject and more light of the background.
If the background is dark (an empty place in the night or a black wall) and you are taking photos of a small bright subject, you will need to increase exposure when the distance between the camera and the subject is increased.
Here is an example how to make an experiment to comprove the Inverse Square Law: try taking photos (at various distances) of a sheet of white paper fixed on a black panel. You will see that you will have to increase exposure when the distance between the camera and the subject is increased.

Antonio - What you have stated does not correctly answer the OP's question. What you are really describing is how the exposure metering system in a normal camera works (other than spot metering). In your example, if the camera is in an auto mode (auto, program, aperture priority, shutter priority), as you back up the metering system is averaging in more of the black background so it is INCREASING the exposure and you would need to dial in some MINUS compensation to keep the exposure of the white paper the same. If the camera was in manual exposure mode you would see no difference in the exposure of the white paper on the black background as you backed up.

Posts 3, 5 and 10 have correct responses.

John

[Panama Hat & Camera]

Antonio - What you have stated does not correctly answer the OP's question. What you are really describing is how the exposure metering system in a normal camera works (other than spot metering). In your example, if the camera is in an auto mode (auto, program, aperture priority, shutter priority), as you back up the metering system is averaging in more of the black background so it is INCREASING the exposure and you would need to dial in some MINUS compensation to keep the exposure of the white paper the same. If the camera was in manual exposure mode you would see no difference in the exposure of the white paper on the black background as you backed up.

Posts 3, 5 and 10 have correct responses.

John

John,
I can't agree with you.
The camera doesn't measure the light emitted by the source (the answers in posts 3, 5 and 10 consider that it is true). The camera measures the light that reaches its sensor (and the light spreads in many directions).
Imagine a translucent ball (inside illuminated) located in an absolutely dark place and think that you are taking photos of the ball with the camera at 10 feet, 20 feet, 40 feet, 80 feet distant of the ball.
Even with spot metering, when the distance from the camera to the subject were increased it would be necessary increasing exposure for compensating the less amount of light that reaches the sensor.
The only exception occurs when a spotlight with a parabolic mirror (and the lamp in the focus of the parabola) is emitting a parallel beam of light straight to camera lens. In this case, the amount of light and the exposure would remain the same.
Cheers,
Antonio.

[xpatUSA]

Imagine a translucent ball (inside illuminated) located in an absolutely dark place and think that you are taking photos of the ball with the camera at 10 feet, 20 feet, 40 feet, 80 feet distant of the ball.

Yes, Tony, but we must also know the size of the ball and (to a lesser extent) it's surface texture. If, for example, the ball was 160 feet in diameter, the amount of light (illuminance) reaching the sensor would not change for any of those distances. Please see my post #12 in this thread for more information.

Even with spot metering, when the distance from the camera to the subject were increased it would be necessary increasing exposure for compensating the less amount of light that reaches the sensor.

I am not sure that is true in every case, sorry.

Now it is time for someone to mention the camera/lens angle of view . . .

[Panama Hat & Camera]

Yes, Tony, but we must also know the size of the ball and (to a lesser extent) it's surface texture. If, for example, the ball was 160 feet in diameter, the amount of light (illuminance) reaching the sensor would not change for any of those distances. Please see my post #12 in this thread for more information.

Ted,
You're right. I was supposing that the subject (the ball) only would occupy the frame height in the photo that the camera is nearest from the ball.

I am not sure that is true in every case, sorry.

Now it is time for someone to mention the camera/lens angle of view . . .

There is a noticeable exception to what I wrote on my posts before:

It is the case that the framing is always the same (perhaps I didn't understand Richard's question).
If you frame the subject and takes a photo, then moves the camera away from the subject, but you increase the zoom factor of the lens to get the same framing, the exposure will be the same (if the subject maintains its illuminance). This happens because, although the f-number be the same, the diameter of the lens diaphragm will increase proportionally with the focal length of the lens. In this case, posts 3, 5 and 10 have correct responses.
Cheers,
Antonio.

[PhotomanJohn]

Antonio - Maybe this article will help.

http://www.scantips.com/lights/flashbasics1b.html

John

[ajohnw]

I'm sort of inclined to say really to some of the comments but this is not an area that I have ever been really interested in other that when illuminating something - solid angles etc.

Taking one suggestion a white sheet on an impossible dead black surface that reflects no light at all but this time using a spot meter that never overlaps the size of the white sheet.. Take a reading - move further away and take another - I suspect providing that the lens isn't changed the readings will be identical as the size of the sheet on the sensor will reduce as the camera is moved further away. I may be wrong on this but intuitively speaking I don't think so.

Another situation - taking a shot of stars - point sources at very great distances. If the ISO is fixed the faintest star that will be recorded is set purely by the effective diameter of the lens and if visual via an eyepiece also by the diameter of the pupil of the eye. This is a pure solid angle aspect. From practical experience I don't agree with the pupil of the eye aspect as I find the size of the exit pupil coming out of the eyepiece is more important. Must be some sort of contrast effect related to magnification.

Normal sources of illumination might be the sun or the moon. Their distance is so great in terrestrial terms that object distances on the planet are more or less irrelevant. Not so flash guns and brolly lighs etc.

John
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