Change of exposure with magnification

[TonyW]

This is related to some discussion in other threads.

Yesterday I did an experiment on the exposure that is to obtained with different magnifications. I set my camera on manual with SS 1/10, f/8, ISO 500. The camera is a Canon 5diii with a 100mm macro lens. I set up a target and photographed it at distances of about 3 metres down to about 400mm (target to sensor plane). The maximum magnification was around 2/3.

I found that the image was underexposed at 400mm by about 1.5 stops compared with 3m. The camera metre told me that as well as the final image.

I was not surprised by the amount of change but I could not have predicted the direction of change. If I really set my mind to it I could probably work out why it went that direction but it is not immediately obvious. Am I being very stupid and missing something very simple?

[William W]

[My bold and sub-editing]

In order for a camera lens to focus progressively closer, the lens apparatus has to move further from the camera's sensor (called "extension"). For low magnifications, the extension is tiny, so the lens is always at the expected distance of roughly one focal length away from the sensor. However, once one approaches 0.25-0.5X or greater magnifications, the lens becomes so far from the sensor that it actually behaves as if it had a longer focal length. At 1:1 magnification, the lens moves all the way out to twice the focal length from the camera's sensor.

The most important consequence is that the lens's effective f-stop increases. (*1)

This has all the usual characteristics, including an increase in the depth of field, a longer exposure time and a greater susceptibility to diffraction. In fact, the only reason "effective" is even used is because many cameras still show the uncompensated f-stop setting (as it would appear at low magnification). In all other respects though, the f-stop really has changed.

A rule of thumb is that at 1:1 the effective f-stop becomes about 2 stops greater than the value set using your camera.

An aperture of f/2.8 therefore becomes more like f/5.6, and f/8 more like f/16, etc. However, this rarely requires additional action by the photographer, since the camera’s metering system automatically compensates for the drop in light when it calculates the exposure settings.


Footnote:

(*1) The reason that the f-stop changes is because this actually depends on the lens's focal length. An f-stop is defined as the ratio of the focal length to aperture diameter. A 100 mm lens with an aperture diameter of 25 mm will have an f-stop value of f/4, for example. In the case of a macro lens, the f-stop increases because the effective focal length increases — not because of any change in the aperture itself (which remains at the same diameter regardless of magnification).

REF: Cambridge In Colour Tutorials; Macro Camera Lenses; https://www.cambridgeincolour.com/tu...cro-lenses.htm

WW

[TonyW]

Thanks, Bill. I was being stupid and it is obvious now.

I have one quibble with the terminology though (and I know it is not yours but generally used). In high school physics I remember the formula 1/a + 1/b = 1/f, for an idealised thin lens. The focal length f is constant in this formular but the distance b of the film/sensor plane from the lens is what is relevant in my question. It is misleading to call this the focal length.

[george013]

Thanks, Bill. I was being stupid and it is obvious now.

I have one quibble with the terminology though (and I know it is not yours but generally used). In high school physics I remember the formula 1/a + 1/b = 1/f, for an idealised thin lens. The focal length f is constant in this formular but the distance b of the film/sensor plane from the lens is what is relevant in my question. It is misleading to call this the focal length.

Names are sometimes used different. It's often a struggle to figure out what is meant.

The lightmeter meters the amount of light and calculates a camerasetting so that the average light-intensity is set to the point that represents the middle of the histogram. The light that hits the lightmeter comes through the lens and is reflected light. Choising another focal length means changing your view and thus changing the amount of light, It will be different if your subject is in a dark surrounding or in a light surrounding. Don't forget it's all reflected light. My thoughts.

George

[William W]

You are not "stupid" - but I do understand your meaning.

*

. . . I have one quibble with the terminology though . . . the formula 1/a + 1/b = 1/f, for an idealised thin lens. The focal length f is constant in this [formula] but the distance b of the film/sensor plane from the lens is what is relevant in my question. It is misleading to call this the focal length.

I agree that it is misleading "to call this [distance 'b', the distance from lens to film plane] the “focal length".

But I do not read the CiC Tutorial as misleading in this respect.

Note the words which are used, (my bold and underlined for emphasis):

"However, once one approaches 0.25-0.5X or greater magnifications, the lens becomes so far from the sensor that it actually behaves as if it had a longer focal length. At 1:1 magnification, the lens moves all the way out to twice the focal length from the camera's sensor."

In the above para: the words "behaves as if" qualify the compound noun “focal length'.

In the above para: the words "moves all the way out" describe a point at a distance which is (linearly) at two times the focal length - that is to say the compound noun “focal length” is being used as a definitive reference length.

"A 100 mm lens with an aperture diameter of 25 mm will have an f-stop value of f/4, for example. In the case of a macro lens, the f-stop increases because the effective focal length increases — not because of any change in the aperture itself (which remains at the same diameter regardless of magnification)."

In the above para: "effective" is a qualifier for the compound noun “focal length”.

***

With respect, George, I don't understand how your comment is relevant to the Original Post about a Macro Lens’s magnification differences and the related Exposure Compensation requirements for same: the Original Post has nothing to do with light meters.

WW

[george013]

With respect, George, I don't understand how your comment is relevant to the Original Post about a Macro Lens’s magnification differences and the related Exposure Compensation requirements for same: the Original Post has nothing to do with light meters.

WW

From TS.

I found that the image was underexposed at 400mm by about 1.5 stops compared with 3m. The camera metre told me that as well as the final image.

I was not surprised by the amount of change but I could not have predicted the direction of change. If I really set my mind to it I could probably work out why it went that direction but it is not immediately obvious. Am I being very stupid and missing something very simple?

And I use a lightmeter to get the "right" exposure. And when your magnification is getting more, you will lose some border , so your average light-intensity is changing. His question is how a different magnification results in a different exposure and I assume under the same light conditions.

And what about the terminolgy I agree with Tony. I mentioned it before.The image-distance is renamed in something as "effective focal length". When the subject is not at infinity, the image distance is by definition different from the focal length.

George

George

[TonyW]

Thanks again, Bill. I think you are right about the CiC tutorial and if I had read it properly in the first place everything would have been clear.

George, I am not sure I understand all that you say but I was careful in my experiment to choose a target that was uniform on average so that as the view of it became larger, its average brightness did not change. This does emphasise the advantage of through the lens metering as opposed to the old fashioned external light meters.

[george013]

Thanks again, Bill. I think you are right about the CiC tutorial and if I had read it properly in the first place everything would have been clear.

George, I am not sure I understand all that you say but I was careful in my experiment to choose a target that was uniform on average so that as the view of it became larger, its average brightness did not change. This does emphasise the advantage of through the lens metering as opposed to the old fashioned external light meters.

Since you didn't show an example we all don't know the circumstances.
Put a dark object in front of a white wall. Position your camera so that that object covers half of the view. This results in a certain exposure value. When you zoom in, a part of the white wall will disappear, changing the average light-intensity. So another exposure value.

I still have some problems with the tutorial.

George.

[DanK]

Position your camera so that that object covers half of the view.

I think it is clear that Bill's comment refers to a uniformly illuminated surface. That is the assumption one has to make to answer the OP's question, which is asking about the effects of high magnification on exposure. If you have a surface that is not uniformly illuminated, you can't isolate the effect of magnfication.

[xpatUSA]

It was interesting to read that the exposure changed at all. Perhaps there was a factor not accounted for?

As a quick and dirty test, I created a white image on my monitor and set it to full screen. I put my cam+70mm macro lens almost up to the screen and set the exposure such that it read 0.0 EV. I moved back, checking the metering as I went. The metering only changed when black parts of the monitor began to appear in the viewfinder.

So, why does this thread expect a change under the conditions of shooting a large white evenly-lit surface and moving back and forth?

(hopefully, nobody will quote the inverse square law)

[DanK]

It was interesting to read that the exposure changed at all. Perhaps there was a factor not accounted for?

As a quick and dirty test, I created a white image on my monitor and set it to full screen. I put my cam+70mm macro lens almost up to the screen and set the exposure such that it read 0.0 EV. I moved back, checking the metering as I went. The metering only changed when black parts of the monitor began to appear in the viewfinder.

So, why does this thread expect a change under the conditions of shooting a large white evenly-lit surface and moving back and forth?

(hopefully, nobody will quote the inverse square law)

That's not what I got. I used this for a uniform target (useful for finding dust bunnies): http://www.pbase.com/copperhill/image/95174363/original

I then focused at perhaps 3+ ft (as far back as I could get without adding background) and at MWD. The difference was about what the tutorial predicted: a bit under 2 stops.

The only explanation I have is the one in the tutorial: a change in effective f/stop resulting from the internal forward movement of the elements of the macro lens at close distances.

[george013]

That's not what I got. I used this for a uniform target (useful for finding dust bunnies): http://www.pbase.com/copperhill/image/95174363/original

I then focused at perhaps 3+ ft (as far back as I could get without adding background) and at MWD. The difference was about what the tutorial predicted: a bit under 2 stops.

The only explanation I have is the one in the tutorial: a change in effective f/stop resulting from the internal forward movement of the elements of the macro lens at close distances.

The so called effective f-stop doesn't explain anything. It's a tool to use. The explanation is on the other side of the lens, the subject side. If the surface of the view is getting smaller, then the amount of reflecting and thus captured light is also getting smaller. Thus to gain the same exposure, your ev must be larger.
There is also another method to make the view smaller, that's zoom in. But don't forget then that when you zoom in, the diameter of the diafragma is growing too. The f-number stays constant.
My thoughts.
George

[xpatUSA]

That's not what I got. I used this for a uniform target (useful for finding dust bunnies): http://www.pbase.com/copperhill/image/95174363/original

I then focused at perhaps 3+ ft (as far back as I could get without adding background) and at MWD. The difference was about what the tutorial predicted: a bit under 2 stops.

The only explanation I have is the one in the tutorial: a change in effective f/stop resulting from the internal forward movement of the elements of the macro lens at close distances.

Aha! You focused the lens, but I did not change the focus at all while moving about, therefore no lens element or barrel movement took place and therefore the aperture size remained constant. All is well in the world, eh?

So, with a lens like yours, I could have just sat at constant distance, twiddled the focus ring and watched the metering change as I did so, IMHO.

[xpatUSA]

The explanation is on the other side of the lens, the subject side. If the surface of the view is getting smaller, then the amount of reflecting and thus captured light is also getting smaller. Thus to gain the same exposure, your EV must be larger.

You may have mis-read where it says: "as far back as I could get without adding background" which means that the subject is a plane surface of constant luminance. In Dan's experiment, the illuminance at the sensor does not vary with distance (assuming that he does not change the focus this time).

[george013]

You may have mis-read where it says: "as far back as I could get without adding background" which means that the subject is a plane surface of constant luminance. In Dan's experiment, the illuminance at the sensor does not vary with distance (assuming that he does not change the focus this time).

Call it something like the inverse inverse square law. The sensor needs a total amount of light deliverd by the surface of the view. If you half it, then you need twice the time.

It's not completly as I put it but close. I think.

George

[DanK]

Aha! You focused the lens, but I did not change the focus at all while moving about, therefore no lens element or barrel movement took place and therefore the aperture size remained constant. All is well in the world, eh?

Bingo.

The sensor needs a total amount of light deliverd by the surface of the view.

no, for constant exposure, it needs a constant amount of light per unit area. Ted's experiment demonstrates that.

Imagine this little thought experiment. You are looking n A4 sheet of paper, illuminated evenly. Now, you interpose a piece of cardboard with a small rectangle cut out, being careful not to get in the way of the light, and you look at the small area you can see through the rectangle. Would the rectangle look darker with the cardboard screen because you are looking only at a fraction of the sheet? No, because the illumination per unit area is unchanged. In essence, that is exactly what Ted did. I did something different: I refocused the lens, causing a change in effective aperture and hence a change in illumination per unit area.

[xpatUSA]

Call it something like the inverse inverse square law. The sensor needs a total amount of light deliverd by the surface of the view. If you half it, then you need twice the time.
George

Consider a large surface emitting or reflecting light. Consider your camera; with a fixed exposure setting, it receives an illuminance from the surface proportionally to the angle or the field of view. Place the camera at a distance for 1:1 magnification. The sensor, say 36x24mm in size, receives light from 36x24mm at the surface.

Now step backwards to get 1:2 magnification. The sensor now receives light from 72x48mm at the surface. By this means, the illuminance remains constant so long as the whole surface fills the frame. Since the illuminance remains constant, the exposure EV remains constant and does not change as you suggested above.

Hope this helps,

[george013]

Consider a large surface emitting or reflecting light. Consider your camera; with a fixed exposure setting, it receives an illuminance from the surface proportionally to the angle or the field of view. Place the camera at a distance for 1:1 magnification. The sensor, say 36x24mm in size, receives light from 36x24mm at the surface.

Now step backwards to get 1:2 magnification. The sensor now receives light from 72x48mm at the surface. By this means, the illuminance remains constant so long as the whole surface fills the frame. Since the illuminance remains constant, the exposure EV remains constant and does not change as you suggested above.

Hope this helps,

The sensor is buiold to collect a certain amount of energie, fotonen or whatever they are called. With double surface with the same illuminance you archieve that in half the time.

George

[xpatUSA]

The sensor is built to collect a certain amount of energy, photons or whatever they are called. With double [the] surface area with the same illuminance you achieve that in half the time.

George

Please, please, please give us an example with numbers and credible formulae that prove your assertion instead of just these general statements.

Meanwhile, please consider this statement from the Light Measurement Handbook by Ryder:

Radiance is independent of distance for an extended area source, because the sampled area increases with distance, canceling inverse square losses.

An 'extended area source' is what we are talking about in this thread. If you don't understand that, this makes it more clear:



Please observe that the illuminance is exactly the same at the two distances illustrated. The units of 'asb' may look odd to you but they are valid illuminance units nevertheless.

This post, complete with it's credible reference proves your statement to be incorrect, sorry.

[tao2]

Please, please, please give us an example with numbers and credible formulae that prove your assertion instead of just these general statements.

AAhhh Expat , would that ye had made that request at post #2....

Nae offence tae Tony...