John Fisher wrote:
The smaller sensor doesn't NEED fewer photos, it GETS fewer photons. So, the only way to get to a given brightness in the final image from a cropped sensor (relative to full frame) is to either boost the signal (and the noise), or to increase the number of photons hitting the sensor by opening up the aperture (increasing the f stop).

John, am I correct in understanding that the fundamental root of the disagreement, is that you don't accept that the exposure is based on the amount of light hitting a given point on the sensor, but that it really is the grand total over the entire sensor.  Again we're not talking about quality or image noise differences in sensor sizes, but the actual exposure value (we used f-stop as an example) changes depending on the sensor size?  I.E. That's why you feel that the cropped sensor has to "boost the signal" to get the same exposure?

IF I understand you correctly, and please correct me if I'm wrong, then I can't help wondering about the following theoretical situation...

If you have 10 identical cameras set up on a grid taking the same photo at the same time, then you can say that the total amount of light being captured by all 10 camera sensors is 10 times what a single camera is.  Or if you have a 100 cameras on the grid it would be 100 times the total amount of light captured by the sensors.  Yet the correct exposure would be identical whether you have 1 camera or 1000 cameras.

But if the GRAND TOTAL of light hitting the ENTIRE sensor (or number of "sensors") doesn't matter for exposure (but IS a factor in image quality, noise and is part of the depth of field equation depending on the area size, etc), but the amount of light hitting a given point on the sensor does, that would explain why the camera exposure doesn't change regardless of sensor size (or the number of theoretical camera sensors taking the same photo).

Again, we're not talking about the quality differences or noise differences in sensor sizes, but whether the smaller sensor requires more light (or a "boosted signal") to get the correct exposure.  Remember that the earlier source article used the ISO difference (for the mythical "equivalency") as a rating of equivalent noise / image quality, when comparing sensor sizes, but not as a requirement for correct exposure.

And my apologies John, if I have misunderstood the "heart" of the disagreement between the two of you.  If I haven't understood your position correctly, then my example (of course!) wouldn't apply...

John Fisher wrote:
The smaller sensor doesn't NEED fewer photos, it GETS fewer photons. So, the only way to get to a given brightness in the final image from a cropped sensor (relative to full frame) is to either boost the signal (and the noise), or to increase the number of photons hitting the sensor by opening up the aperture (increasing the f stop). And that is why that 70-200 f2.8 on your full frame sensor will APPEAR to be a 100-300 (field of view) f4.5 (depth of field and brightness/with same noise) on your cropped sensor camera.

Anyway, this makes sense to me.
...

By "Boost" I think you mean "amplify".   I'm not sure this argument applies to digital prints.   Bigger prints require more ink, not more "boost".


Be that as it may, amplifying or "boosting" the image in this way does not affect the perceived noise.   The issue is not the level of the noise, but the signal to noise ratio.   


When it comes to sensor size the two main issues that affect noise in the final print are:
1) Bigger pixels tend to have less noise than smaller pixels
2) Higher pixel counts yield Increased pixels per inch in the final print.  More pixels per inch in the print results in a reduction of noise per unit area due to pixel averaging.

If sensor size is constant, the above tend to balance each other out.  Assuming similar technology and the same frame size, you should see similar noise levels in a print from a 15 megapixel sensor as a 30 megapixel sensor.

Larger sensors produce less noisy prints because they allow you to increase the pixel count (which reduces noise by averaging), or to use larger pixels (which are inherently less noisy), or some combination of the two.

Michael Fryd wrote:
Your issue seems to be that smaller sensors get fewer total photons, therefore the smaller sensor acts like a larger sensor but at a smaller aperture.

What's so special about aperture?  Instead of talking about equivalent apertures why not talk about equivalent time?

Silly me! Indeed,when discussing lenses (which is what this discussion is about) what is so special about aperture? Hmmm, let me see, oh yes, I remember now, lenses have an aperture that can be controlled, but you can't control time with a lens! I am waiting to see your new 70-200 1/200th lens, I'll bet it looks different than a 70-200 f2.8!

You backed into this alley. That gear marked R? It stands for reverse.

(For those not familiar with Frick and Frack, the South Beach Paul Buff flacks, we have been having this "discussion" for some time now. Mike actually brought up "what is special about aperture" last week. Next we are going to hear about 8X10 sheet film and 35mm, using a projector to put an image on the wall, etc. etc. The problem is that Mike will postulate some givens, then argue to a reasonable conclusion. I on the other hand, don't accept the givens, so not surprisingly I will arrive at a different conclusion, Suggesting we should be talking about time when you are discussing the light density provided by a lens on a sensor is an example of this.

Here is another example of what Mike and I have been discussing, and it is what makes arguing with him so painful. Every word he says below is absolutely correct, and completely in line with what I've been saying (I mean we can get rid of the term ISO, but I can live with that here). But does that mean we have an understanding...... oh be still my silly heart, no. Mike will then say, "You misunderstand what I'm saying..............." and off we go again like a dog and a cat caught in the same bush.)

Mikey McMichaels wrote:
...
I can't think of any scenario where photons needed per unit area isn't more relevant than total photons.
...

Michael Fryd wrote:
Here's a scenario where total photons captured is more relevant:

As a general rule, if all other factors are equal, the more total photons you capture, the less noise you will see in your final image.  It is a useful point of view if you are trying to understand some of the signal to noise issues that affect the final print.

Looking at it this way, it is clear why higher ISO setting yield more noise (you are capturing fewer total photons).  It is also clear why larger sensors tend to have less noise (larger sensors capture more total photons).

Using this perspective we can determine how much we need to reduce ISO if we want to crop an image without increasing noise in the final print.

Imagine you are shooting with a full frame camera at ISO 3200.   You make a 16 by 20 print, and the noise level is just barely acceptable.   You take a small portion of the image  (about 1/4 the areas of the whole image) and blow it up to 16 by 20, but you find that the noise level is now a bit too high.

If you consider total photons captured, you can determine how many more photons you would need to capture in order for the total in the cropped portion to match the total in the full original frame.  If you want to get the same number of photons onto 1/4 the area, you need 4 times as many photons per unit area.  You dial your camera down to ISO 800 and reshoot.  You should find that the enlargement from the small area shot at ISO 800 has about the same noise as the original full image shot at ISO 3200.


So you see, there are different ways of looking at an issue.  Generally the best way depends on which aspect of the issue we are concerned about.  If we are concerned about exposure, we generally are concerned about photons per unit area.  If we are concerned about noise in the final print, we might care more about total photons captured.

Everyone have a great Memorial Day weekend, and do take a minute (just one) to remember why we have this holiday.

John
--
John Fisher
700 Euclid Avenue, Suite 110
Miami Beach, Florida 33139
(305) 534-9322
http://www.johnfisher.com

Michael Fryd wrote:

Here's a scenario where total photons captured is more relevant:

As a general rule, if all other factors are equal, the more total photons you capture, the less noise you will see in your final image.  It is a useful point of view if you are trying to understand some of the signal to noise issues that affect the final print.

Looking at it this way, it is clear why higher ISO setting yield more noise (you are capturing fewer total photons).  It is also clear why larger sensors tend to have less noise (larger sensors capture more total photons).

Using this perspective we can determine how much we need to reduce ISO if we want to crop an image without increasing noise in the final print.

Imagine you are shooting with a full frame camera at ISO 3200.   You make a 16 by 20 print, and the noise level is just barely acceptable.   You take a small portion of the image  (about 1/4 the areas of the whole image) and blow it up to 16 by 20, but you find that the noise level is now a bit too high.

If you consider total photons captured, you can determine how many more photons you would need to capture in order for the total in the cropped portion to match the total in the full original frame.  If you want to get the same number of photons onto 1/4 the area, you need 4 times as many photons per unit area.  You dial your camera down to ISO 800 and reshoot.  You should find that the enlargement from the small area shot at ISO 800 has about the same noise as the original full image shot at ISO 3200.


So you see, there are different ways of looking at an issue.  Generally the best way depends on which aspect of the issue we are concerned about.  If we are concerned about exposure, we generally are concerned about photons per unit area.  If we are concerned about noise in the final print, we might care more about total photons captured.

Here's what I'm missing - can't you summarize everything you just wrote into "If you want to get the same number of photons onto 1/4 the area, you need 4 times as many photons per unit area."?

That's still looking only at photons per unit area.

Michael Fryd wrote:

Although it may seem that way, it is all interrelated.

Consider two sensors of identical size and similar technologies.  One has 20 million photo sites, and one has 50 million.  Let's call the 20 million camera the "large pixel" camera, and the 50 million on the "small pixel" camera.

If we examine the images on our screen, mapping one sensor pixel to one screen pixel, we will see more noise in the pixels from the small pixel camera.  However, that is not the same as saying the small pixel camera produces images with more noise.

When we are looking at the final print, we really aren't concerned with noise at the pixel level, we are concerned with noise per sqare mm on the print.   Prints from both cameras will have about the same noise levels.  The difference is that the print from the small pixel camera will show a little more resolving power (detail).

Imagine a crop factor camera with 20 megapixels a full frame camera with 50 megapixels.  The individual photo sites for each camera are about the same size.  If photo site size was the important factor, then prints from both cameras would have equal noise levels.  In reality, the prints from the 50 megapixel camera will show less noise because we get a noise reduction by mapping individual sensor pixels to a smaller area on the print.


The reason that crop factor cameras produce noisier images really does boil down to the fact that smaller sensors capture fewer total photons.  Higher pixel counts (i.e. smaller pixels) give us more resolving power (ultimately limited by the noise in the sensor).  More total photons give us less noise.

Everything you wrote makes sense, but you've made one assumption, which is that the way to compare noise in the prints is to print them at the same size.

If you print the cropped image smaller so that the ratio of sensor to sensor is the same as print to print, the noise will match. That's the appropriate way to compare them.

John Fisher wrote:

The smaller sensor doesn't NEED fewer photos, it GETS fewer photons. So, the only way to get to a given brightness in the final image from a cropped sensor (relative to full frame) is to either boost the signal (and the noise), or to increase the number of photons hitting the sensor by opening up the aperture (increasing the f stop). And that is why that 70-200 f2.8 on your full frame sensor will APPEAR to be a 100-300 (field of view) f4.5 (depth of field and brightness/with same noise) on your cropped sensor camera.

Anyway, this makes sense to me.

John
--
John Fisher
700 Euclid Avenue, Suite 110
Miami Beach, Florida 33139
(305) 534-9322
http://www.johnfisher.com

But that's not accurate.

10,000 photons hitting a FF sensor will give you the same exposure as 5,000 hitting a 1.5 crop.



If you opened the aperture wider for a cropped sensor and had the same exposure, you'd narrow the DoF and match the FF DoF which isn't what happens.

Michael Fryd wrote:

Your issue seems to be that smaller sensors get fewer total photons, therefore the smaller sensor acts like a larger sensor but at a smaller aperture.

What's so special about aperture?  Instead of talking about equivalent apertures why not talk about equivalent time?

If you're shooting 100mm, f/4 at 1/100 on a full frame, why not say those settings act like 150mm f/4 at 1/256 on the crop factor?  The crop factor results in the 100mm lens having a field of view like a 150mm on a full frame, and the crop factor also makes the 1/100 shutter speed gather the same number of photons as 1/256 would on a full frame.

Both points of view equally express the concept of differences in total light.     Of course both "equivalencies" break down if we try to use them for computing the ability to stop motion, or depth of field.

And that's the fundamental problem with equivalencies, you either need to accept that they only apply to one small aspect of the situation, or you need to go in "whole-hog" and essentially define a complete and consistent measurement system.   Generally it's easier to keep to actual measurements, and common units.


Getting past the "equivalent aperture" issue, there is an underlying concept - bigger sensors tend to capture more total photons, and this typically results in an image that has less noise than an image from a smaller sensor.   There is nothing new about this, and the concept is certainly not unique to digital.  Anyone who has shot multiple formats of film knows that larger film sizes result in prints with less grain. 


When it comes to film, no one talks about equivalent apertures, equivalent shutter speeds, or equivalent ISO speeds.  The interesting thing here is that Canon actually made an EOS crop factor film SLR (EOS IX 7).  It took the full line of EOS EF lenses.  It came with an EF 22-55mm f/4-5.6 USM lens.  The lens was not described as a 27-69mm "equivalent".  The manual did include a short section "About the Effective Field Angle of EF lenses".  The manual explains that the frame is smaller than a typical 35mm frame, and therefore the effective field angle of the lens is about what you would expect from a lens that was 1.25 times the focal length on a full frame camera.

It wasn't until digital SLR cameras came out that companies started this "effective focal length" nonsense.

The transition from film to digital has not changed the meanings of  aperture, shutter speed, ISO and focal length,

Isn't the result going to be that a large negative makes a print with smaller grain or "less visible" grain? That's different than fewer instances of grain throughout the print and it's a function of requiring less enlarging of the large negative if you're making identical sized prints.


Another way to look at it would be to compare contact sheets of from different sized negatives shot on the same film. The contact sheets would have identical grain, but different sized images.

LightDreams wrote:
John, am I correct in understanding that the fundamental root of the disagreement, is that you don't accept that the exposure is based on the amount of light hitting a given point on the sensor, but that it really is the grand total over the entire sensor.

This is my interpretation of what's being misunderstood.

John Fisher wrote:
But the real question is do you also multiply the f stop by 1.6 as well (so f2,8 becomes f4.5)?

Going back to this - the answer is that it depends on what you want to describe. If you want to describe the DoF, yes, but no the exposure.

If the crop affected the exposure, you'd have to set the sensor size or film stock size when using a handheld light meter.

Mikey McMichaels wrote:
...
Isn't the result going to be that a large negative makes a print with smaller grain or "less visible" grain? That's different than fewer instances of grain throughout the print and it's a function of requiring less enlarging of the large negative if you're making identical sized prints.


Another way to look at it would be to compare contact sheets of from different sized negatives shot on the same film. The contact sheets would have identical grain, but different sized images.

Mikey McMichaels wrote:
Everything you wrote makes sense, but you've made one assumption, which is that the way to compare noise in the prints is to print them at the same size.

If you print the cropped image smaller so that the ratio of sensor to sensor is the same as print to print, the noise will match. That's the appropriate way to compare them.

It all boils down to what's important to you.  There are lots of ways to compare various sensor sizes, but I personally prefer to compare the results, not the intermediate steps.

In my mind, the important question is "assuming I can the lens I feel is appropriate for the camera/shot, If I shoot some images with a full frame, and some with a crop body, what difference will I see in the 16 by 20 prints I sell to my customers?"   For instance if I am shooting in my studio, and I have a crop body with a 40mm f/2.8 lens, and a full frame with a 60mm f/2.8, how will the prints from one camera compare to another.  Perhaps I am strange, but If I am shooting with both cameras, I don't want to present smaller proof prints for the crop body images.

As an academic exercise, let's consider what happens if we use the same "magnification" (ratio of print size to sensor size) for both a full frame and a crop factor camera.   Our print from the full frame will be 8 x 12, and our print from the crop factor will be about 5 x 7.5.  To make the comparison easier let's make that 8 x 12 feet and 5 x 7.5  feet.

If we get out our loupe and take a close look, each print will have about the same amount of noise per square inch.  That's because a square inch on either print corresponds to equal size areas on the sensor. 

But let's stop for a moment and consider whether or not this is merely an academic comparison, or a practical one.

In the real world, a larger print will have a larger viewing distance.  People will typically be further away from the 8 x 12 foot print than from a 5 x 7.5 print.  Noise (and details) in the 5 x 7.5 print will be more noticeable as the viewer is closer.  Thus in order for a typically views to perceive equal quality, the 5 by 7.5 print will need to be better quality per square inch than the larger print.  But we know that it is the same quality per square inch, and therefore the print from the larger sensor will generally look better.

By the way, if you doubt that larger prints can have poor print quality per square inch and still look good, get a close look at a large highway billboard.  They are typically printed a 10ppi or less.   On a 4" by 6" print, this would look like a blurry mess.  When the viewing distance is large, it looks sharp.

So, let's step back and ask why we care about image quality and noise from a camera.  I tend to the practical.  I care about how it affects my final product (usually a print).   In this context I need to look at the whole system that maps photons on the sensor to my final print. 

But that's just my viewpoint.  Perhaps you carry around both a full frame and a crop body?  Perhaps you use the crop body when you want 5 x 7 prints, and the full frame when you want 8 x 10 prints?  If this is your workflow, then your comparison criteria may differ from mine.

John Fisher wrote:
The smaller sensor doesn't NEED fewer photos, it GETS fewer photons. So, the only way to get to a given brightness in the final image from a cropped sensor (relative to full frame) is to either boost the signal (and the noise), or to increase the number of photons hitting the sensor by opening up the aperture (increasing the f stop). And that is why that 70-200 f2.8 on your full frame sensor will APPEAR to be a 100-300 (field of view) f4.5 (depth of field and brightness/with same noise) on your cropped sensor camera.

Anyway, this makes sense to me.

John

Mikey McMichaels wrote:
But that's not accurate.

10,000 photons hitting a FF sensor will give you the same exposure as 5,000 hitting a 1.5 crop.



If you opened the aperture wider for a cropped sensor and had the same exposure, you'd narrow the DoF and match the FF DoF which isn't what happens.

And this understanding and explanation which makes so much sense to you is precisely what gets me and Michael running around the May pole. It is a film based understanding of exposure. Photons per unit area (which is what you are postulating) works when the final output is film (either negative or positive). A given ISO film when properly processed always gives the same brightness (or exposure) regardless of the size of the film (ISO 100 is the same for sheet film or 35mm). A sensor has no ISO, so smaller sensors have to have more signal amplification (relative to large sensors) to produce the image we see on the screen. Clearly you would agree that a smaller sensor being hit by fewer photons can't possibly have the same native electrical output as a larger sensor hit by twice as many photons.

I would go into more detail, but there it is, I don't accept the givens. To you, 10,000 photons hitting a give area  will give you the same exposure as 5,000 photons hitting half the surface area. This is true for film, it is not true for digital sensors.

John
--
John Fisher
700 Euclid Avenue, Suite 110
Miami Beach, Florida 330139
(305) 534-9322
http://www.johnfisher.com

John Fisher wrote:
I would go into more detail, but there it is, I don't accept the givens. To you, 10,000 photons hitting a give area  will give you the same exposure as 5,000 photons hitting half the surface area. This is true for film, it is not true for digital sensors.

I think that confirms the heart of the issue.  John believes exposure is based on the total of the photons hitting the larger sensor area (so the size of the area affects exposure and not just quality / depth of field, etc) rather than Michael's view that exposure is defined by the amount of light hitting any given point on a sensor (which means that the total area doesn't come into the exposure part of the equation).

At least if I read their statements correctly.


[EDIT] Correction. Ii might depend on how John is making use of the term "exposure" (as per Michael's post below). So I should stay out of it, watch and learn! [/EDIT]

John Fisher wrote:
And this understanding and explanation which makes so much sense to you is precisely what gets me and Michael running around the May pole. It is a film based understanding of exposure. Photons per unit area (which is what you are postulating) works when the final output is film (either negative or positive). A given ISO film when properly processed always gives the same brightness (or exposure) regardless of the size of the film (ISO 100 is the same for sheet film or 35mm). A sensor has no ISO, so smaller sensors have to have more signal amplification (relative to large sensors) to produce the image we see on the screen. Clearly you would agree that a smaller sensor being hit by fewer photons can't possibly have the same native electrical output as a larger sensor hit by twice as many photons.

I would go into more detail, but there it is, I don't accept the givens. To you, 10,000 photons hitting a give area  will give you the same exposure as 5,000 photons hitting half the surface area. This is true for film, it is not true for digital sensors.

John,

I think we agree on a few things and disagree on others.

The reason 10,000 photons hitting a give area  will give you the same exposure as 5,000 photons hitting half the surface area is that "exposure" is defined in terms of photons per unit area.  Therefore, by definition, they are the same exposure.   

In the context of photography, "exposure" is traditionally discussed in terms of photons per unit area.   If you want to discuss photons in the entire frame, I don't think you should call this "exposure".  Applying these two similar (but different) meanings to the word "exposure" leads to confusion.


I agree that total photons captured plays an important theoretical role in how noisy/grainy the image will be.

I disagree that this issue is somehow unique to digital.  This issue has been around since film days, and explains why an image captured on a 4 by 5 piece of Plus-X captures more detail then an image captured on a standard 35mm frame of that same Plus-X film.


I suggest a different term, perhaps "Total Photons Captured"  (TPC).  This will make it clear whether we are discussing exposure (photons per unit area) which is independent of frame size, or total photons captured (which is the exposure times the area of the film/sensor).

Now let's talk about "aperture".   Traditionally, this is expressed as the ratio of the focal length of the lens to the diameter of the opening in the lens.   We use this system because exposure (photons per unit area) is dependent on this ratio.  Two lenses set to f/4 will yield the same exposure, even if they have different focal lengths.

Aperture affects a number of aspects of the image:  Auto focus ability, depth of field, diffraction issues, and (as you have mentioned) brightness.  Generally, if you want talk about "equivalent" aperture, you will come up with different numbers depending on which issue you are trying to make "equivalent"

Now lets do some math, and see how we need to adjust aperture if we want to maintain TPC when switching from full frame to crop factor.

The number of photons that passes though the lens is proportional to the area of the opening. Thus if you halve the diameter, you reduce the photons by a factor of 4 (area is proportional to the square of the diameter).

The dimensions of a traditional 35mm SLR frame are 24mm by 36mm, which gives us an area of 864 sq. mm.
The dimension of a Canon crop factor sensor are about 18.6mm by 27.9, which give us an area of about 519 sq. mm.
The area of the full frame sensor is about 1.6 times the area of the crop sensor.

In order to get the same TPC on the crop body, we need about 1.6 times as many photons per unit area.  This means the area of the aperture must be 1.6 times larger on the crop body.  Remember, aperture ratios are based on diameter, but area is based on the square of the diameter.  In order to increase the aperture area by 1.6,  the aperture diameter must increase by a factor of 1.26 (the square root of 1.6).   This is a difference of about 0.9 stops.



Of course, there are practical considerations that don't allow you to match TPC on a crop factor to TPC on a full frame.


Let's assume you were getting a beautifully exposed shot on a full frame camera at f/3.2 with the ISO set to 100 and a shutter speed of 1/1000.

You decided to take a second shot with your crop factor camera.  If you shoot at the same f/3.2, 1/1000 with the ISO set to 100, you will get an equally perfectly exposed image.  The crop factor did capture fewer total photons, and the resulting image may be a bit noisy.   You decide to that you want the crop factor camera to give you the same image quality as the full frame, so you leave all the settings the same, except that you open the aperture from f/3.2 to f/2.5.  Both cameras are now capturing the same total number of photons (the crop factor has a higher density over a smaller area).

Theoretically, both should have equivalent noise levels.  In practice, the image shot at f/2.5 will be almost one stop over exposed.  Image quality may suffer as highlights may be blown out.

It turns out the pixels are not perfect.   There is a practical minimum and a maximum number of photons that can be measured.  If too few photons hit the pixel, they will be lost in the noise.   If too many hit the pixel, it will be maxed out. 


It turns out that it we want to get the best quality image out of a particular sensor we want to get the photons per unit area (exposure) into the working range of the sensor.  Too many photons are just as bad as too few.

Thus if I am getting a perfect exposure on one camera at ISO 100, f/3.2 at 1/1000, I should also get the same perfect exposure on any other camera at those settings, no matter what the sensor size.  If I increase the photons per unit density on a smaller sensor, I will max out the pixels and lose my highlight detail.

By the way, this is the benefit of working in photons per unit area rather than TPC.  With photons per unit area, the settings are independent of frame size.

Michael Fryd wrote:
I disagree that this issue is somehow unique to digital.  This issue has been around since film days, and explains why an image captured on a 4 by 5 piece of Plus-X captures more detail then an image captured on a standard 35mm frame of that same Plus-X film.

This is factually incorrect. If a 4x5 camera and a 35mm camera are set up side by side to shoot a resolution target and a  1x1 cm square cut from the center of each of the negatives, the 4x5 negative will not show more detail when examined with a loupe than the 35mm negative.  Often it will show less detail because while the film resolving power is the same for both formats, 35mm lenses often have higher resolving power due to better design.

The reason an 8x10 print made from a 4x5 negative is sharper is because it's only a 2x enlargement while the same 8x10 print from a 35mm negative is approximately an 8x enlargement.

Grayscale Photo wrote:

This is factually incorrect. If a 4x5 camera and a 35mm camera are set up side by side to shoot a resolution target and a  1x1 cm square cut from the center of each of the negatives, the 4x5 negative will not show more detail when examined with a loupe than the 35mm negative.  Often it will show less detail because while the film resolving power is the same for both formats, 35mm lenses often have higher resolving power due to better design.

The reason an 8x10 print made from a 4x5 negative is sharper is because it's only a 2x enlargement while the same 8x10 print from a 35mm negative is approximately an 8x enlargement.

Well we both agree that an 8 x 10 print from a 4 x 5 negative is typically sharper than an 8 x 10 print from a 35mm negative.  I think you would agree that the print from the 4 x 5 will show finer film grain.

It's perfectly reasonable to state that the lower sharpness from the smaller negative is due to the higher magnification needed for an 8 x 10 print.   This same phenomena explains why larger digital formats tend to produce sharper (and less noisy) prints.  Larger sensors require smaller print size to sensor size ratios (i.e. "magnification").

Hoever this is not the only way to look at the situation.  If you apply information theory to the system, you will come to the same conclusion based on the fact that the larger sensor/film captures more total photons.

Claude Shannon is generally considered to be the father of information theory.  These theories allow one to look at systems (such as analog film or digital sensors) and compute how film/sensor size affects noise levels.

Michael Fryd wrote:
Well we both agree that an 8 x 10 print from a 4 x 5 negative is typically sharper than an 8 x 10 print from a 35mm negative.  I think you would agree that the print from the 4 x 5 will show finer film grain.

The final 8x10 print will show finer grain from a 4x5 negative because the magnification is much lower (2x versus about 8x).  The grains are the same size on both negatives if the emulsion is the same (i.e. plus-x).

Information theory will tell you the amount of information recorded on a 1x1 sq. cm of either size film will be approximately the same, again assuming the same emulsion.  The same theory will tell you a 4x5 negative is capable of recording a lot more information in its entirety than a 35mm negative.  Square inches count.  However, 4x5 film isn't inherently capable of recording finer detail on the film itself, which is usually what we mean when we say more detail."

Grayscale Photo wrote:

The final 8x10 print will show finer grain from a 4x5 negative because the magnification is much lower (2x versus about 8x).  The grains are the same size on both negatives if the emulsion is the same (i.e. plus-x).

Information theory will tell you the amount of information recorded on a 1x1 sq. cm of either size film will be approximately the same, again assuming the same emulsion.  The same theory will tell you a 4x5 negative is capable of recording a lot more information in its entirety than a 35mm negative.  Square inches count.  However, 4x5 film isn't inherently capable of recording finer detail on the film itself, which is usually what we mean when we say more detail."

I guess it depends on what you are concerned about.  From an academic perspective it's interesting to discuss the detail per unit area in the sensor/film.  As a photographer, I am concerned with the detail per unit area in the 16 by 20 print I deliver to the client.  It is a matter of personal preference which you find more interesting.

Michael Fryd wrote:
As a photographer, I am concerned with the detail per unit area in the 16 by 20 print I deliver to the client.

In that case, large format films rule.  A 16x20 print from a 4x5 negative is only a 4x magnification.

Grayscale Photo wrote:

In that case, large format films rule.  A 16x20 print from a 4x5 negative is only a 4x magnification.

In terms of minimizing grain, absolutely.  The larger the negative the less visible grain in the final print.

From a practical standpoint one needs to find the right balance between price, convenience, quality and other factors.

John Fisher wrote:
Clearly you would agree that a smaller sensor being hit by fewer photons can't possibly have the same native electrical output as a larger sensor hit by twice as many photons.

Yes, but there's no relevance. The amplification isn't done based on a total, is based on photons per unit area.

The percentage of amplification in FF and crop sensors is identical at identical ISOs.

John Fisher wrote:
I would go into more detail, but there it is, I don't accept the givens. To you, 10,000 photons hitting a give area  will give you the same exposure as 5,000 photons hitting half the surface area. This is true for film, it is not true for digital sensors.

I can't figure out why you think it's not true for digital sensors.

Michael Fryd wrote:
Presumably, there are more photons coming off the final print, then were captured by the sensor.  Let's call this increase in photons the "boost".  As the smaller sensor captured fewer photons, it need a higher "boost" in order to get to the final print.   Obviously this boost increases the noise as well as the image, and therefore one might think that the higher the "boost", the noisier the image,

I was about to respond to this as if you were comparing the smaller sensor to the larger sensor, but I think you're comparing it to the print, right?

The thing that's throwing this whole discussion off is the idea of a boost. It's not the way to think about it.


Let's look at it in terms of sensor size and a context where the total number of photons doesn't change proportionately to the sensor size.


Imagine a shot down the center of a street at night. Assuming there are no cars, the center of our FF shot will be fairly dark - let's say there's one street light visible in the distance fairly close to the center of the frame.

Let's also imagine there are a few buildings and street lights very close to the camera's position and they're visible around the edges of the frame.

Let's say we're able to select an exposure that places the darkest parts in zone two and the brightest spots in zone 8 - including the distant light near the center of the frame.


Now if replace the camera with a cropped sensor, the sensor is only going to see the center of the frame, which is mostly dark except for the street light that's close to the center.

With the exact same exposure settings for each set up, the cropped sensor photo will be identical to the same are within the FF photo. The total number of photos will have dropped more than proportionately to the sensor size, but the exposure will still be correct. With less surface area and the smaller field of view, you don't need to have the total number drop proportionately to match the same surface area and field of view from the FF sensor.

If you did the same thing but shot a grey card that filled the frame of both cameras, the total number of pixels would drop proportionately to the sensor size.


Maybe we're overlooking what's being cropped by the sensor - it's the number of photons that's being cropped.


If you open a lens to a specific aperture the number of pixels that pass through in a given time period will be the same regardless of the size of the sensor in the camera or even if the lens is on a camera.

A FF sensor will crop pixels because the projected image is round and larger than the FF sensor. A cropped sensor will crop more pixels because it's got a smaller surface area. But the number of photons needed for that specific surface area - not the size, but the specific area within the specific image - is identical on FF and a crop.

So there's no boost needed between a crop and FF.


As far as a boost when comparing a sensor and the photons reflected off of a print, it makes sense that the much larger surface area would have the potential to reflect more photons. The question is the light within the scene that was shot and the viewing area. It would be possible to shoot a photo at mid day and have more photons pass through the lens and be captured by the sensor than reflected by a giant print because the print doesn't have a defined illumination and most likely isn't being viewed in mid day sun. I would speculate that normal interior lighting would reflect more total photons, but not low light.

If you specify that the lighting conditions are identical for the sensor capture and print viewing, then, of course total photons will be a function of surface area.

But there's still no need for the idea of a "boost".

LightDreams wrote:
I'm really reluctant to say this as I'm sure it will be misinterpreted...

Doing research on both John and Michael's comments, I slowly started to realize that there is indeed (at least usually) some additional signal boosting going on with crop sensor cameras.

BUT, let me be very clear, the exposure settings that you set with your camera still remain the same and I strongly agree with Michael's "photons per unit area" concept. I.E. that the light REACHING the sensor at a given point, is the same, regardless of the size of the sensor.

So where does signal boosting come in?  Well it turns out that when you get down to the INDIVIDUAL megapixel level, there is a mix of a space with a sensor reading the light and circuitry supporting it.  And, if the individual megapixel "light capturing" sensor area is smaller (such as when you're fitting in more megapixels on a smaller crop sensor), then you have to add some signal boost to compensate or use some other form of increased digital sensitivity. All so that the ISO sensitivity registers to the same level across the various cameras.

The amount of that "boost" (which is already factored into the exposure) varies completely from sensor to sensor depending on the particular megapixel design, sensitivity and physical size.

Example 1.  You make a crop sensor that is just a chopped down version of a full frame sensor.  Each megapixel (and supporting circuitry size for each megapixel) has not changed at all. There are just fewer of them (it's basically a "chopped down" full frame sensor).  In this extreme example there would be absolutely no additional signal boost built in or required.

Example 2.  You take a 42 megapixel full frame sensor and scale absolutely everything down (in perfect scale) to fit into a crop frame sensor.  The size of the sensor ON EACH INDIVIDUAL MEGAPIXEL SENSOR (we're NOT talking about the entire sensor size) is smaller catching less light.  So the electrical signal from each megapixel is boosted so that it gets up to the same exposure value.  Note this is even though the light hitting that spot on the sensor didn't change with the change in formats.

Now in real world practice, they design the sensors to get the optimum performance out of the space and the total amount of megapixels, etc, etc, that they need.for the design.  So how much signal boosting is done (to get it up to the same standard exposure sensitivity) or not done would vary from sensor to sensor  / case by case.

But it's all behind the scenes.  The light reaching a given point on the sensor hasn't changed (Michael's Photons per Unit Area) and the proper exposure settings for your camera hasn't changed.  But there's likely some degree of additional signal boost going on to some degree (varies dramatically) depending on the specific design of the sensor and the smaller total available space to fit each individual megapixel sensor into.

I would also note that it's quite likely that even on a full frame format camera, if they up the megapixel count they likely are also boosting the signal, depending on the design of the (now smaller) individual megapixel sensors being fit into the same space.

I can't help wondering whether this is what John is referring to, as far as the exposure "being different" and needing "a signal boost". And the difference between film and digital sensors.  It's possible that some part of the disagreement may come from is (as Michael previously alluded to) making sure that everyone is talking in the same terms...?

I think the issue with gain in the pixels is dependent on pixel size, not sensor size.  Thus the gain would be the same for a 19.5 megapixel crop sensor as a 50 megapixel full frame.

I think the "boost" issue has to do with the ratio of the print size to the sensor size, and is independent of pixel size.

A smaller sensor captures fewer total photons than a larger sensor.  When you make a 16 by 20 print, it takes a certain number of photons coming from a print in order to see it.   The ratio of the number of photons it takes to see the image to the number of photons captures is the "boost".   It theory is that larger "boosts" (as from a smaller sensor) increases noise as well as the signal, and therefore prints from a crop sensor camera are noisier.

The flaw in this theory is that it assumes absolute noise level is significant.  It turns out that it is not the absolute signal level that matters, but the ratio of the signal level to noise level.   Enlargement of an image in the digital domain does not change the signal to noise ratio, and therefore does not change the "noisiness" of the image.   

When you set the ISO level of you camera to 100, you adjusting the parameters of the entire system such that if the image was properly exposed with ISO 100 film, you can use the same shutter speed and aperture to get proper levels in your JPEG file.   Whether the adjustments apply at the hardware, firmware or software level is not particularly relevant to the usefulness of ISO on a digital camera system.   As ISO is based on sensitivity per unit area, it is independent of sensor size.

LightDreams wrote:
I'm really reluctant to say this as I'm sure it will be misinterpreted...

Doing research on both John and Michael's comments, I slowly started to realize that there is indeed (at least usually) some additional signal boosting going on with crop sensor cameras.

BUT, let me be very clear, the exposure settings that you set with your camera still remain the same and I strongly agree with Michael's "photons per unit area" concept. I.E. that the light REACHING the sensor at a given point, is the same, regardless of the size of the sensor.

So where does signal boosting come in?  Well it turns out that when you get down to the INDIVIDUAL megapixel level, there is a mix of a space with a sensor reading the light and circuitry supporting it.  And, if the individual megapixel "light capturing" sensor area is smaller (such as when you're fitting in more megapixels on a smaller crop sensor), then you have to add some signal boost to compensate or use some other form of increased digital sensitivity. All so that the ISO sensitivity registers to the same level across the various cameras.

The amount of that "boost" (which is already factored into the exposure) varies completely from sensor to sensor depending on the particular megapixel design, sensitivity and physical size.

Example 1.  You make a crop sensor that is just a chopped down version of a full frame sensor.  Each megapixel (and supporting circuitry size for each megapixel) has not changed at all. There are just fewer of them (it's basically a "chopped down" full frame sensor).  In this extreme example there would be absolutely no additional signal boost built in or required.

Example 2.  You take a 42 megapixel full frame sensor and scale absolutely everything down (in perfect scale) to fit into a crop frame sensor.  The size of the sensor ON EACH INDIVIDUAL MEGAPIXEL SENSOR (we're NOT talking about the entire sensor size) is smaller catching less light.  So the electrical signal from each megapixel is boosted so that it gets up to the same exposure value.  Note this is even though the light hitting that spot on the sensor didn't change with the change in formats.

Now in real world practice, they design the sensors to get the optimum performance out of the space and the total amount of megapixels, etc, etc, that they need.for the design.  So how much signal boosting is done (to get it up to the same standard exposure sensitivity) or not done would vary from sensor to sensor  / case by case.

But it's all behind the scenes.  The light reaching a given point on the sensor hasn't changed (Michael's Photons per Unit Area) and the proper exposure settings for your camera hasn't changed.  But there's likely some degree of additional signal boost going on to some degree (varies dramatically) depending on the specific design of the sensor and the smaller total available space to fit each individual megapixel sensor into.

I would also note that it's quite likely that even on a full frame format camera, if they up the megapixel count they likely are also boosting the signal, depending on the design of the (now smaller) individual megapixel sensors being fit into the same space.

I can't help wondering whether this is what John is referring to, as far as the exposure "being different" and needing "a signal boost". And the difference between film and digital sensors.  It's possible that some part of the disagreement may come from is (as Michael previously alluded to) making sure that everyone is talking in the same terms...?

I think a lot of what you said, probably all, is correct, but the unit area we're discussing is really a single pixel if we're talking about ISO.

Once you look at it that way, the rest isn't very relevant.

Michael Fryd wrote:

I think the issue with gain in the pixels is dependent on pixel size, not sensor size.  Thus the gain would be the same for a 19.5 megapixel crop sensor as a 50 megapixel full frame.

I think the "boost" issue has to do with the ratio of the print size to the sensor size, and is independent of pixel size.

A smaller sensor captures fewer total photons than a larger sensor.  When you make a 16 by 20 print, it takes a certain number of photons coming from a print in order to see it.   The ratio of the number of photons it takes to see the image to the number of photons captures is the "boost".   It theory is that larger "boosts" (as from a smaller sensor) increases noise as well as the signal, and therefore prints from a crop sensor camera are noisier.

The flaw in this theory is that it assumes absolute noise level is significant.  It turns out that it is not the absolute signal level that matters, but the ratio of the signal level to noise level.   Enlargement of an image in the digital domain does not change the signal to noise ratio, and therefore does not change the "noisiness" of the image.   

When you set the ISO level of you camera to 100, you adjusting the parameters of the entire system such that if the image was properly exposed with ISO 100 film, you can use the same shutter speed and aperture to get proper levels in your JPEG file.   Whether the adjustments apply at the hardware, firmware or software level is not particularly relevant to the usefulness of ISO on a digital camera system.   As ISO is based on sensitivity per unit area, it is independent of sensor size.

By boost you're really just talking about enlargement.

You can take the same photo an print it in different sizes which will reflect more total photons, but you haven't needed to boost anything.

Michael McGowan wrote:
...

ISO measures the sensitivity to light. But the number of photons falling on the 35mm film is obviously less than the number falling on the 120. Thus, your "logic" involving photons is specious at best. Perhaps it would help to think of ISO as the sensitivity of a single site or a single grain on film.

Actually, ISO measures sensitivity per unit area.   Two different ISO 100 films can have very different grain sizes.   Two different ISO 100 sensors can have very different pixel sizes (the pixels on my old 3 megapixel D30 were much larger than the pixels on my 7D, but they both have a base ISO of 100).

Michael McGowan wrote:
...
Aperture is what it is. It's a measure of how open a lens is. And DOF at a given aperture for a given focal length doesn't change.
...

Actually, "aperture" is typically expressed as the ratio of the focal length to the diameter of the aperture opening.

The reason is that the light per unit area falling on the sensor/film is controlled by the ratio of focal length to aperture diameter.   Increase both the aperture diameter and focal length by the same factor, and the light on the sensor remains the same.


Suppose you are getting a correct exposure with a 25mm aperture diameter on a 50mm lens (f/2).

That same same 25mm aperture diameter will get you two stops underexposed with a 100mm lens (f/4). 

You would need an aperture diameter of 50mm on a 100mm lens (f/2) to get the same exposure as a 25mm aperture on a 50mm lens.