Exposure.

[Jay A]

But your 1 degree spot meter will measure the light from a 4x larger area on the grey card if you use the spot meter at twice the distance.

And if I change it to a 5 degree spot meter, I STILL get the same result. So a different total area is measured with the same result!

10 feet, 1 degree reading, F8
10 feet, 5 degree reading, F8

So the area is irrelevant, no? (of course assuming the meter is still seeing the card and only the card.

 

[Jay A]

But distance doesn't matter in this case if you compensate by making the grey card bigger...

Anyway, back to the orginal thread about focal length...

Try to think about a pin hole, a light cone and moving a piece of photographic paper further away from the pin hole. It should be simple to see that as focal length (FL) increases (distance of the paper away from the pin hole defines the focal length) the area of the base of the light cone at that point increases in proportion to FL^2 (simple proof with similar triangles/area of a circle).
The light on the paper is the "area of paper/area of the base of the cone"
So the light on the paper is proportional to 1/(FL^2).

So if we want constant exposure as we "zoom" we have to compensate for this by making the physical aperture (pin hole size) let more light in.
The pinhole nees to be bigger in AREA by a factor of FL^2.
To do that we need to physical diameter needs to increase by a factor of the FL.

In other words as FL increases we need to make the physical aperture diameter (D) bigger by a factor of FL.

i.e. to maintain constant exposure the D needs to be proportional to the FL
in other words FL/D = constant.

FL/D is defined as the f/stop.

So to maintain constant exposure as FL length changes the f/stop must remain the same.

Andrew

Yes! Absolutely! But here you are moving the paper away from the cone (same as moving the film plane away from the lens). This is what I have been saying all along is what causes the change in exposure and thus the need for a bigger diaphragm. Yes a greater distance dictates the need for a longer focal length, but it's the relationship between the focal length and its distance from the film plane that is causing the loss of light, not the distance from the subject to the camera.

Someone please tell me...why is it so easy to observe the effect of the inverse square law when we move a light source away from a subject, but we cannot see any loss of light when we move ourselves away from the subject?

 

[Andre Affleck]

Here is the example Joe (Great Bustard) used, stating that the picture at 20ft (the one at f/4) would be dimmer:

Great Bustard wrote:

Shoot a subject 10 ft away at 50mm f/2 (aperture diameter = 50mm / 2 = 25mm). Then back up to 20 ft and shoot the same subject at 100mm f/4 (aperture diameter = 100mm / 4 = 25mm) and the same shutter speed.

Note that he is changing from f/2 to f/4 as he doubles the focal length to keep the aperture at a constant 25mm.

Do you agree that the picture at f/4 would be dimmer?

Of course the picture is going to be dimmer. You reduced the exposure by one stop. But ALL of the objects in that photo are going to be dimmer, not just the distant ones. What does that prove? Nothing other than if you reduce your exposure, your photo will be dimmer. This is more nonsense.

What happens when you shoot two objects at different distances in one photo? According to Joe, the distant one will be dimmer. This is what he as maintained, and it is incorrect!

 

[Great Bustard]

But your 1 degree spot meter will measure the light from a 4x larger area on the grey card if you use the spot meter at twice the distance.

And if I change it to a 5 degree spot meter, I STILL get the same result. So a different total area is measured with the same result!

10 feet, 1 degree reading, F8
10 feet, 5 degree reading, F8

So the area is irrelevant, no? (of course assuming the meter is still seeing the card and only the card.

A 1 degree AOV corresponds to a focal length of 2500mm. A 5 degree AOV corresponds to an AOV of 500mm. This means that a 1 degree AOV sees 1/25 the area 25 as a 5 degree AOV, so 1/25 as much light is seen with a 1 degree AOV compared to a 5 degree AOV for a given time interval and distance (assuming uniform illumination).

The aperture diameter for 2500mm f/8 is 2500mm / 8 = 312.5mm. The aperture diameter for 500mm f/8 is 500mm / 8 = 62.5mm. Thus, the aperture diameter for 2500mm f/8 is 5 times larger, so the area is 25 times larger.

Thus, 1/25 the area of the scene is cancelled out by an aperture area that is 25 times greater.

 

[Great Bustard]

Someone please tell me...why is it so easy to observe the effect of the inverse square law when we move a light source away from a subject, but we cannot see any loss of light when we move ourselves away from the subject?

Put a flashlight on the table and look right into it. Then back off and keep looking at it. The light appears dimmer.

It's not entirely that simple, however, since our eyes will adjust the aperture according to the brightness of the scene, but I hope the point is clear.

 

[Steen Bay]

But your 1 degree spot meter will measure the light from a 4x larger area on the grey card if you use the spot meter at twice the distance.

And if I change it to a 5 degree spot meter, I STILL get the same result. So a different total area is measured with the same result!

My guess is that the spot meter is 'aware' that you've changed to 5 degree (meaning app. 25x larger area) and takes that into consideration when calculating the correct exposure.

10 feet, 1 degree reading, F8
10 feet, 5 degree reading, F8

So the area is irrelevant, no? (of course assuming the meter is still seeing the card and only the card.

 

[Jay A]

But your 1 degree spot meter will measure the light from a 4x larger area on the grey card if you use the spot meter at twice the distance.

And if I change it to a 5 degree spot meter, I STILL get the same result. So a different total area is measured with the same result!

10 feet, 1 degree reading, F8
10 feet, 5 degree reading, F8

So the area is irrelevant, no? (of course assuming the meter is still seeing the card and only the card.

A 1 degree AOV corresponds to a focal length of 2500mm. A 5 degree AOV corresponds to an AOV of 500mm. This means that a 1 degree AOV sees 1/25 the area 25 as a 5 degree AOV, so 1/25 as much light is seen with a 1 degree AOV compared to a 5 degree AOV for a given time interval and distance (assuming uniform illumination).

The aperture diameter for 2500mm f/8 is 2500mm / 8 = 312.5mm. The aperture diameter for 500mm f/8 is 500mm / 8 = 62.5mm. Thus, the aperture diameter for 2500mm f/8 is 5 times larger, so the area is 25 times larger.

Thus, 1/25 the area of the scene is cancelled out by an aperture area that is 25 times greater.

Frankly I KNEW you were going to answer exactly as you did

Now tell me...why is it that when I look at a subject and move the light which is lighting up that subject away from it I can observe the light loss, but if I back away from a subject and look at it without changing the light which is lighting it up, I cannot?

 

[Jay A]

But your 1 degree spot meter will measure the light from a 4x larger area on the grey card if you use the spot meter at twice the distance.

And if I change it to a 5 degree spot meter, I STILL get the same result. So a different total area is measured with the same result!

My guess is that the spot meter is 'aware' that you've changed to 5 degree (meaning app. 25x larger area) and takes that into consideration when calculating the correct exposure.

10 feet, 1 degree reading, F8
10 feet, 5 degree reading, F8

So the area is irrelevant, no? (of course assuming the meter is still seeing the card and only the card.

Your guess?

 

[Great Bustard]

What happens when you shoot two objects at different distances in one photo? According to Joe, the distant one will be dimmer. This is what he as maintained, and it is incorrect!

According to joe, this is what happens when we shoot two objects at different distances:

http://forums.dpreview.com/forums/read.asp?forum=1032&message=36775789

The explanation comes in two parts. The first situation is for the same framing, but different distances. Let's consider the case of 5 ft 50mm f/2 and 10 ft 100mm f/2.

Both will have the same framing. The aperture diameter at 50mm f/2 is 50mm / 2 = 25mm. The aperture diameter for 100mm f/2 is 100mm / 2 = 50mm. Thus, the aperture at 100mm f/2 has four times the area as 50mm f/2.

The amount of light reaching the lens from the scene at 10 ft is 1/4 the amount of light reaching the lens at 5 ft (inverse square relationship for light). The validity of using the inverse square relationship is discussed at the end of this post as an addendum.

Since there is 1/4 as much light reaching the lens at 10 ft than at 5 ft, but the aperture has four times the area at 100mm f/2 than 50mm f/2, the two effects cancel each other out, and the same amount of light falls on the sensor. For a given sensor size, this results in the same density of light, which means the same exposure.

So, you are quite incorrect when you say, "According to Joe, the distant one will be dimmer", unless, of course, we use the same framing and aperture diameter, and then, of course, the more distant object will be dimmer.

 

[Jay A]

Put a flashlight on the table and look right into it. Then back off and keep looking at it. The light appears dimmer.

It's not entirely that simple, however, since our eyes will adjust the aperture according to the brightness of the scene, but I hope the point is clear.

I was the one who quoted wikepedia as saying that the inverse square law only applies to a point light source. I did see a reference in a google to 'inverse square law' and 'widepedia' someplace where it also stated that it does NOT apply to reflected light, but of course this has been rejected by the experts here.

 

[Andre Affleck]

Here is an example using actual images to prove the point. All photos shot in raw with identical parameters, imported to CS4 with no PP whatsoever.

First photo with two object made up of exactly the same material with nearly identcal reflectance, shot 5 feet away from the camera. Note the RGB values of each object:





Next, the same scene shot with exactly the same camera settings, but with one object twice the distance away (10 feet) Note the RGB values:





Next, the same scene shot with exactly the same camera settings, but with one object 25 feet away (Venus and Moon scenario but less exaggerated). Note the RGB values:





I think everyone can agree that the brightness of these objects are both visually and numerically equal in all three of these examples. How can this be? Because again, the light is reduced but so is the area, by exactly the same amounts, which leaves you with the same density of light (photons per unit area). The density of light is what defines brightness.

For the last time, Objects do not get dimmer with distance.

Why some people refuse to admit this, I don't now, but this example provides indisputable proof.

 

[Jay A]

Guys, I am going to respectfually bow out of this discussion. We have begun debating in circles and I don't think anyone is convincing anyone else of anything.

I honestly think that some of you are confusing a brightness discussion with a surface area discussion. I have cited websites which state things like "distance has no bearing on exposure," "the inverse square law applies only to light sources," I have cited a Rangefinder Magazine which states also that changing the distance from camera to subject has nothing to do with any change in exposure, etc, and some of you continue to cite nothing but this theory that you have which you think maintains that when you move further away from an object that that object reflects less light. Yes it takes up a smaller area on the sensor. Yes, when you compensate with a telephoto you gain back that area. But this has nothing to do with how bright it appears, nor how it affects the exposure. Just because there is more area of that object at the film plane does not mean that its brightness has changed. An object is lit by a third party, it's light source. Unless you change that light source the subject brightness will remain the same.

At any rate, I hope you guys don't take this as my way of saying, I am taking my football and going home. The above is just what I believe and continuing on and on with this discussion is not changing anything, plus it has taken away too much time from my job (yes I work on Saturdays)

I appreciate the spirited debate and maybe we will have another at some point.

Thanks guys.

Jay

 

[Steen Bay]

Someone please tell me...why is it so easy to observe the effect of the inverse square law when we move a light source away from a subject, but we cannot see any loss of light when we move ourselves away from the subject?

But you can see how the subject gets smaller. At twice the distance the apparent size (area) of the subject is only 1/4, and 1/4 the area with the same intensity of light per area unit, that means that 3/4 of the light is 'lost'.

 

[Steen Bay]

But your 1 degree spot meter will measure the light from a 4x larger area on the grey card if you use the spot meter at twice the distance.

And if I change it to a 5 degree spot meter, I STILL get the same result. So a different total area is measured with the same result!

My guess is that the spot meter is 'aware' that you've changed to 5 degree (meaning app. 25x larger area) and takes that into consideration when calculating the correct exposure.

10 feet, 1 degree reading, F8
10 feet, 5 degree reading, F8

So the area is irrelevant, no? (of course assuming the meter is still seeing the card and only the card.

Your guess?

Sorry, I should have said "Obviously the spot meter... "

 

[Great Bustard]

Here is an example using actual images to prove the point. All photos shot in raw with identical parameters, imported to CS4 with no PP whatsoever.

First photo with two object made up of exactly the same material with nearly identcal reflectance, shot 5 feet away from the camera. Note the RGB values of each object:





Next, the same scene shot with exactly the same camera settings, but with one object twice the distance away (10 feet) Note the RGB values:





Next, the same scene shot with exactly the same camera settings, but with one object 25 feet away (Venus and Moon scenario but less exaggerated). Note the RGB values:





I think everyone can agree that the brightness of these objects are both visually and numerically equal in all three of these examples. How can this be? Because again, the light is reduced but so is the area, by exactly the same amounts, which leaves you with the same density of light (photons per unit area). The density of light is what defines brightness.

For the last time, Objects do not get dimmer with distance.

Why some people refuse to admit this, I don't now, but this example provides indisputable proof.

Using the last pic in your examples (excellent examples, by the way!), the diameter of the near object in the photo is 4.5 times the diameter of the far object, so the far object is 4.5x further away, and the near object has 20 times the area.

The inverse square law says that 1/4.5^2 = 1/20 as much light is falling on the sensor from the far object as is falling on the sensor for the near object. However, the far object also takes up 1/20 the area on the sensor. So, 1/20 the light on 1/20 the area results in the same density of light (exposure).

 

[Jay A]

Excellent Andre!

 

[Great Bustard]

But your 1 degree spot meter will measure the light from a 4x larger area on the grey card if you use the spot meter at twice the distance.

And if I change it to a 5 degree spot meter, I STILL get the same result. So a different total area is measured with the same result!

10 feet, 1 degree reading, F8
10 feet, 5 degree reading, F8

So the area is irrelevant, no? (of course assuming the meter is still seeing the card and only the card.

A 1 degree AOV corresponds to a focal length of 2500mm. A 5 degree AOV corresponds to an AOV of 500mm. This means that a 1 degree AOV sees 1/25 the area 25 as a 5 degree AOV, so 1/25 as much light is seen with a 1 degree AOV compared to a 5 degree AOV for a given time interval and distance (assuming uniform illumination).

The aperture diameter for 2500mm f/8 is 2500mm / 8 = 312.5mm. The aperture diameter for 500mm f/8 is 500mm / 8 = 62.5mm. Thus, the aperture diameter for 2500mm f/8 is 5 times larger, so the area is 25 times larger.

Thus, 1/25 the area of the scene is cancelled out by an aperture area that is 25 times greater.

Frankly I KNEW you were going to answer exactly as you did

A feather in your cap.

Now tell me...why is it that when I look at a subject and move the light which is lighting up that subject away from it I can observe the light loss, but if I back away from a subject and look at it without changing the light which is lighting it up, I cannot?

If you move the light source away from the scene, less light is falling on the scene (inverse square law), so less light is being refelcted back to you.

If you back up from the scene, less light from the scene is reaching you (inverse square law), but the scene takes up a smaller area in your FOV, so the density of the light remains the same.

See the example and my response to it above:

http://forums.dpreview.com/forums/read.asp?forum=1032&message=36781106

 

[Jay A]

Using the last pic in your examples (excellent examples, by the way!), the diameter of the near object in the photo is 4.5 times the diameter of the far object, so the far object is 4.5x further away, and the near object has 20 times the area.

The inverse square law says that 1/4.5^2 = 1/20 as much light is falling on the sensor from the far object as is falling on the sensor for the near object. However, the far object also takes up 1/20 the area on the sensor. So, 1/20 the light on 1/20 the area results in the same density of light (exposure).

So if we take the far object and increase its size, it will get brighter?

(I'm just pulling your chain...I know what your answer will be)

 

[Great Bustard]

Using the last pic in your examples (excellent examples, by the way!), the diameter of the near object in the photo is 4.5 times the diameter of the far object, so the far object is 4.5x further away, and the near object has 20 times the area.

The inverse square law says that 1/4.5^2 = 1/20 as much light is falling on the sensor from the far object as is falling on the sensor for the near object. However, the far object also takes up 1/20 the area on the sensor. So, 1/20 the light on 1/20 the area results in the same density of light (exposure).

So if we take the far object and increase its size, it will get brighter?

(I'm just pulling your chain...I know what your answer will be)

Another feather in your cap! But, just to have it on the record:

If we made the far object 4.5x the diameter so that it had the same size on the sensor as the near object, the area would be 20x greater. That means that it would reflect (or emit) 20x as much light. However, it's 4.5x further away, so only 1/20 of that light is reaching the lens (inverse square law).

Thus, 20x the light at 1/20 the intensity results in the same amount of light, and the same amount of light on the same area results in the same exposure.

 

[Andre Affleck]

Post deleted.