same f-number on different focal length - similar exposure?

jrtrent said:

bclaff said:

jrtrent said:

jrtrent said:

... Focal length is defined by Nikon as:

Click to expand...

"It is not a measurement of the actual length of a lens, but a calculation of an optical distance from the point where light rays converge to form a sharp image of an object to the digital sensor or 35 mm film at the focal plane in the camera."

So the light rays of a 100 mm lens converge twice as far away from the sensor than they do with a 50 mm lens. This makes the light 1/4 as bright as the 50 mm lens due to the extra distance the light travels to reach the sensor, so the area of the aperture has to be 4 times bigger to give the same light intensity.

The area of the aperture of a 50 mm lens at f/8 is about about 30.67 square mm; the area of the aperture of a 100 mm lens at f/8 is about 122.66 square mm--four times bigger to compensate for the two times greater distance the light travels to the sensor.

This is an explanation I've seen published fairly often, but I know that in similar threads, other members have had different explanations they like better; unfortunately, I've never been able to understand those other explanations.

Click to expand...

Despite the source this is wrong and a bit bizarre.

For example, the sharp image of the object is at the focal plane; so that distance is zero!

You may want to see this .

In any case, the "inverse law" would apply to the exit pupil, which is not directly tied to focal length.

However, this (the "inverse law") is why effective f# is f# * (1 + m/p)
where m is magnification and p is pupil magnification.

Click to expand...

Thanks. I don't think I've seen that formula before. Could you explain further what is meant by magnification and pupil magnification, and the difference between effective f# and regular f#?

Click to expand...

Magnification is (size on the sensor) / (actual size)
At infinity magnification is zero.

Pupil magnification is (size of exit pupil) / (size of entrance pupil)
Many of the formulas you are familiar with are simplifications that assume the pupil magnification is unity.

When you are focused at other than infinity the amount of light reaching the sensor is determined by the Numeric Aperture (NA) or alternatively the effective f#
Normally magnification is pretty small so this is often only of consequence in close-up (macro) work.

BTW, manufacturers treat this differently.
For example, Nikon reports effective f# whereas many others don't.
This is good but can confuse some close-up photographers because when they put their f/2.8 macro lens on a Nikon body at f/2.8 and focus closer like to 1x the body (correctly) reports something like f/4.8

jrtrent said:

Libraries in my area are still closed due to coronavirus, so I can't go get the book it came from, but below is a paragraph from an older thread of the type of explanation I've read concerning how f-stops work for different focal lengths:

One interesting illustration in an old book showed a camera with a 2-inch lens and a camera with a 6-inch lens; because of their different formats, both covered the same field of view. Though they're taking the same picture at the same subject distance, to be able to achieve the same brightness the 6-inch lens needed an aperture opening with an area 9 times greater than the 2-inch lens due to the 3 times greater distance of the lens from the film plane, which the inverse square law indicates results in an intensity 1/9 as great. This difference is what the f-stop system takes care of so neatly for us, with f/8, for example, giving the same intensity of illumination at the film plane for the camera with the 2-inch lens (area of the aperture at f/8 calculates to .0490894 square inches) as it does for the camera with the 6-inch lens (area of .441786 square inches, which is 9 times bigger).

An explanation like the one above is the only one I've seen in the books I've been able to find that deal with the issue, and it has the benefit of being simple enough for me to grasp. I hope I can learn how your alternate explanation works. If you can explain the terms in the formula, I think I have a chance.

Click to expand...

I won't directly address that and I'm not 100% sure what you're after but consider this:

Light grasp (it goes by a lot of other names) is the area of the entrance pupil times the solid angle of the lens.

The area of the entrance pupil relates directly to focal length and f#.

The solid angle relates directly to angle of view.

So, for example, if focal length goes up by a factor of 3 and angle of view goes down by a factor of 3 you have the same light grasp.
(I simplified this slightly but don't worry this is the concept.)

Hope this helps.

bclaff said:

jrtrent said:

bclaff said:

jrtrent said:

bclaff said:

... Focal length is defined by Nikon as:

Click to expand...

"It is not a measurement of the actual length of a lens, but a calculation of an optical distance from the point where light rays converge to form a sharp image of an object to the digital sensor or 35 mm film at the focal plane in the camera."

So the light rays of a 100 mm lens converge twice as far away from the sensor than they do with a 50 mm lens. This makes the light 1/4 as bright as the 50 mm lens due to the extra distance the light travels to reach the sensor, so the area of the aperture has to be 4 times bigger to give the same light intensity.

The area of the aperture of a 50 mm lens at f/8 is about about 30.67 square mm; the area of the aperture of a 100 mm lens at f/8 is about 122.66 square mm--four times bigger to compensate for the two times greater distance the light travels to the sensor.

This is an explanation I've seen published fairly often, but I know that in similar threads, other members have had different explanations they like better; unfortunately, I've never been able to understand those other explanations.

Click to expand...

Despite the source this is wrong and a bit bizarre.

For example, the sharp image of the object is at the focal plane; so that distance is zero!

You may want to see this .

In any case, the "inverse law" would apply to the exit pupil, which is not directly tied to focal length.

However, this (the "inverse law") is why effective f# is f# * (1 + m/p)
where m is magnification and p is pupil magnification.

Click to expand...

Thanks. I don't think I've seen that formula before. Could you explain further what is meant by magnification and pupil magnification, and the difference between effective f# and regular f#?

Click to expand...

Magnification is (size on the sensor) / (actual size)
At infinity magnification is zero.

Pupil magnification is (size of exit pupil) / (size of entrance pupil)
Many of the formulas you are familiar with are simplifications that assume the pupil magnification is unity.

When you are focused at other than infinity the amount of light reaching the sensor is determined by the Numeric Aperture (NA) or alternatively the effective f#
Normally magnification is pretty small so this is often only of consequence in close-up (macro) work.

BTW, manufacturers treat this differently.
For example, Nikon reports effective f# whereas many others don't.
This is good but can confuse some close-up photographers because when they put their f/2.8 macro lens on a Nikon body at f/2.8 and focus closer like to 1x the body (correctly) reports something like f/4.8

bclaff said:

Libraries in my area are still closed due to coronavirus, so I can't go get the book it came from, but below is a paragraph from an older thread of the type of explanation I've read concerning how f-stops work for different focal lengths:

One interesting illustration in an old book showed a camera with a 2-inch lens and a camera with a 6-inch lens; because of their different formats, both covered the same field of view. Though they're taking the same picture at the same subject distance, to be able to achieve the same brightness the 6-inch lens needed an aperture opening with an area 9 times greater than the 2-inch lens due to the 3 times greater distance of the lens from the film plane, which the inverse square law indicates results in an intensity 1/9 as great. This difference is what the f-stop system takes care of so neatly for us, with f/8, for example, giving the same intensity of illumination at the film plane for the camera with the 2-inch lens (area of the aperture at f/8 calculates to .0490894 square inches) as it does for the camera with the 6-inch lens (area of .441786 square inches, which is 9 times bigger).

An explanation like the one above is the only one I've seen in the books I've been able to find that deal with the issue, and it has the benefit of being simple enough for me to grasp. I hope I can learn how your alternate explanation works. If you can explain the terms in the formula, I think I have a chance.

Click to expand...

I won't directly address that and I'm not 100% sure what you're after

Click to expand...

I haven't had time to absorb what you've written yet, but what I'm after is the same thing the OP asked about. Why does f/8 give the same brightness for both a 100mm lens and a 50mm lens when they have quite different aperture diameters? The explanation I offered, and which I've seen in several books and articles, explains this by applying the inverse square law to the focal length; that is, the distance light travels from the lens to the sensor or film (which you described in that other thread you gave a link to as being from H to F'). I've been told in the past that this explanation was in error, but I still don't see why or how. Maybe I will after I read this post of yours a few more times.

bclaff said:

but consider this:

Light grasp (it goes by a lot of other names) is the area of the entrance pupil times the solid angle of the lens.

The area of the entrance pupil relates directly to focal length and f#.

The solid angle relates directly to angle of view.

So, for example, if focal length goes up by a factor of 3 and angle of view goes down by a factor of 3 you have the same light grasp.
(I simplified this slightly but don't worry this is the concept.)

Hope this helps.

Click to expand...

jrtrent said:

jrtrent said:

...

Click to expand...

I haven't had time to absorb what you've written yet, but what I'm after is the same thing the OP asked about. Why does f/8 give the same brightness for both a 100mm lens and a 50mm lens when they have quite different aperture diameters? The explanation I offered, and which I've seen in several books and articles, explains this by applying the inverse square law to the focal length; that is, the distance light travels from the lens to the sensor or film (which you described in that other thread you gave a link to as being from H to F'). I've been told in the past that this explanation was in error, but I still don't see why or how. Maybe I will after I read this post of yours a few more times.

Click to expand...

It has nothing to do with the inverse square law. That's a coincidence.
It has to do with the angle of view.

bclaff said:

jrtrent said:

bclaff said:

...

Click to expand...

I haven't had time to absorb what you've written yet, but what I'm after is the same thing the OP asked about. Why does f/8 give the same brightness for both a 100mm lens and a 50mm lens when they have quite different aperture diameters? The explanation I offered, and which I've seen in several books and articles, explains this by applying the inverse square law to the focal length; that is, the distance light travels from the lens to the sensor or film (which you described in that other thread you gave a link to as being from H to F'). I've been told in the past that this explanation was in error, but I still don't see why or how. Maybe I will after I read this post of yours a few more times.

Click to expand...

It has nothing to do with the inverse square law. That's a coincidence.
It has to do with the angle of view.

Click to expand...

So the angle of view is what I will get at with this formula you gave?

f# is f# * (1 + m/p)

At first blush angle of view wouldn't seem to matter much; after all, in the example of the 2" lens and the 6" lens on different film sizes, the angle of view was the same and the brightness was equal, and brightness is still equal when using a 100 mm lens at f/8 and a 50 mm lens at f/8 on the same camera despite having different angles of view. Well, I'll keep working at it, but probably not until tomorrow--right now, my head hurts.

jrtrent said:

bclaff said:

jrtrent said:

jrtrent said:

...

Click to expand...

I haven't had time to absorb what you've written yet, but what I'm after is the same thing the OP asked about. Why does f/8 give the same brightness for both a 100mm lens and a 50mm lens when they have quite different aperture diameters? The explanation I offered, and which I've seen in several books and articles, explains this by applying the inverse square law to the focal length; that is, the distance light travels from the lens to the sensor or film (which you described in that other thread you gave a link to as being from H to F'). I've been told in the past that this explanation was in error, but I still don't see why or how. Maybe I will after I read this post of yours a few more times.

Click to expand...

It has nothing to do with the inverse square law. That's a coincidence.
It has to do with the angle of view.

Click to expand...

So the angle of view is what I will get at with this formula you gave?

f# is f# * (1 + m/p)

At first blush angle of view wouldn't seem to matter much; after all, in the example of the 2" lens and the 6" lens on different film sizes, the angle of view was the same and the brightness was equal, and brightness is still equal when using a 100 mm lens at f/8 and a 50 mm lens at f/8 on the same camera despite having different angles of view. Well, I'll keep working at it, but probably not until tomorrow--right now, my head hurts.

Click to expand...

No, that's "apples and oranges".
Angle of view doesn't relate to effective f#

Angle of view has to do with how much light is gathered by the entrance pupil.
All light gathered by the entrance pupil is delivered to the exit pupil.
The exit pupil then projects that toward the image plane.
(Which is where effective f# comes in.)

--
Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )

Consider a motion picture or slide projector with screen, You can change the projected image size by changing the screen to projector distance. Move the projector further from the screen results in a bigger but dimmer image. The image is dimmer because the light from the projector's light bulb must now cover more screen area. Conversely, if you move the projector closer to the screen the image shrinks however it get brighter.

This is the stuff that the the f-number system solves. It's math intertwines magnification which is a product of focal length with the limitation of the lens to pass light due to the surface area of the working lens opening (aperture).

Keep in mind, the camera lens is a projection system that follows a law in physics, the law of the inverse square. Double the projection distance doubles the image image size. This 2X change in magnification results in a 4X change in image area. Thus image brightness drops by a factor of 4.

Thank goodness the f-number system comes to our rescue.

Tourlou said:

Leonard Migliore said:

JohnDope said:

Hi,

It might sound like a trivial question for most of you, but I'm gonna ask anyway.

Let's say I have 2 lenses, 50mm and 100mm (FF, for argument's sake)

if I set my aperture the same on both lenses on the same camera, will the exposure be identical?

the f number is the ratio of the focal length to the diameter of the entrance pupil.

Hence, if I set e.g. f8 on both these lenses, the size of the entrance pupil will be different.

8=50/E.P --> E.P= 50/8.

and

8=100/E.P --> E.P=100/8

if the enterance pupil is not the same size, the amount of light (or the flux of photons) hitting the sensor would be different. Does that mean the exposure would be different as well?

Click to expand...

The exposure will be the same. That's what f/numbers are for.

The entrance pupil of the 100mm lens is twice the diameter of the entrance pupil of the 50mm lens. This means it has four times the area and collects four times the light.

Now, imagine you are imaging a circle with both lenses. The image of the circle using the 100mm lens will be twice the diameter that it would be with the 50. So it also has four times the area.

The increased magnification exactly compensates for the increased light collection, giving the same exposure. Such is the glory of f/numbers.

Click to expand...

You wrote it down the wrong way around!!! The 100mm lens has twice the aprture, but is looking at a circle half as big as the 50mm, hence the equivalence. It's a good simple way to explain it Leonard.

Click to expand...

For distant objects, doesn't a 100mm lens produce an image twice the size of one generated by a 50mm lens? Isn't that why we use telephotos?

Well, I misread. I'm looking at what's in front of the lens, you are looking at what's behind the lens. The explaination is the same one way or the other.

Regards.

Ron

The more the area, the more light it captures

bclaff said:

jrtrent said:

bclaff said:

jrtrent said:

bclaff said:

...

Click to expand...

I haven't had time to absorb what you've written yet, but what I'm after is the same thing the OP asked about. Why does f/8 give the same brightness for both a 100mm lens and a 50mm lens when they have quite different aperture diameters? The explanation I offered, and which I've seen in several books and articles, explains this by applying the inverse square law to the focal length; that is, the distance light travels from the lens to the sensor or film (which you described in that other thread you gave a link to as being from H to F'). I've been told in the past that this explanation was in error, but I still don't see why or how. Maybe I will after I read this post of yours a few more times.

Click to expand...

It has nothing to do with the inverse square law. That's a coincidence.
It has to do with the angle of view.

Click to expand...

So the angle of view is what I will get at with this formula you gave?

f# is f# * (1 + m/p)

At first blush angle of view wouldn't seem to matter much; after all, in the example of the 2" lens and the 6" lens on different film sizes, the angle of view was the same and the brightness was equal, and brightness is still equal when using a 100 mm lens at f/8 and a 50 mm lens at f/8 on the same camera despite having different angles of view. Well, I'll keep working at it, but probably not until tomorrow--right now, my head hurts.

Click to expand...

No, that's "apples and oranges".
Angle of view doesn't relate to effective f#

Click to expand...

The question wasn't about effective f-stop, but I think that has to do with the idea that f-stops assume infinity focus, and that illumination is decreased for close-up photography because the lens has to be extended further from the film plane to maintain focus.

"The lens, projecting the image, acts as the source of the image-forming light. And with increasing distance to the image plane, the area over which its 'output' is spread increases too (and, by the way, the size of the image itself does as well: magnification increases). Consequently there is less light per unit of area. The projected image is less bright.

The relation between image distance and illumination is a quite simple one too: illumination decreases with the square of the distance. Double the lens to film distance, and the illumination in the projected image decreases by a factor 4 (2²)." http://www.hasselbladhistorical.eu/HT/HTComp.aspx

bclaff said:

Angle of view has to do with how much light is gathered by the entrance pupil.
All light gathered by the entrance pupil is delivered to the exit pupil.
The exit pupil then projects that toward the image plane.
(Which is where effective f# comes in.)

Click to expand...

Sorry, but I'm left not knowing what parts of what you've written relate to why a 100mm lens at f/8 gives the same brightness as a 50mm lens at f/8.

I did re-read what you wrote above to Leonard:

"The amount of light collected is the product of the area of the entrance pupil and the solid angle of the light collected. The solid angle is in turn a function of the angle of view.

So the area and angle are what cancel out.
eg. 100mm has a larger area and an offsetting smaller angle of view."

Doing a search for solid angle, I found:

Light collection

The light collection efficiency is the solid angle that an optic makes with an object. The f-number describes this angle:
f-number: f/# = l/d
where l is distance and d is diameter of lens.

The solid angle that a lens collects is approximately:
OMEGA = pi d² / 4 l²
The fraction of light that an optic collects is this solid angle divided by the total 4 pi steradians:
fraction collected = OMEGA /4pi = pi d² / 16 pi l² = 1 / 16 (f/#)². http://elchem.kaist.ac.kr/vt/chem-ed/optics/lenses.htm

If I'm understanding the formula given above for the solid angle, and it's entirely possible I'm not, then when set to the same f-stop, all lenses regardless of focal length will have the same solid angle.

Your solid angle should include the sine of half the angle of view.
If it doesn't it's an approximation.

The half-angle at 100mm is 12.21 degrees and at 50mm it's 23.40 degrees
sin(12.21) is 0.21 and sin(23.40) is 0.40

Note this isn't exactly a ratio of 2 but so close that you'd never see an exposure difference.

bclaff said:

Your solid angle should include the sine of half the angle of view.
If it doesn't it's an approximation.

Click to expand...

Thanks. Perhaps that's why the author of that article wrote:

The solid angle that a lens collects is approximately:
OMEGA = pi d² / 4 l²

bclaff said:

The half-angle at 100mm is 12.21 degrees and at 50mm it's 23.40 degrees
sin(12.21) is 0.21 and sin(23.40) is 0.40

Note this isn't exactly a ratio of 2 but so close that you'd never see an exposure difference.

Click to expand...

Should I look for a different formula for solid angle, or is the one above from that article close enough to avoid gross errors? I have to admit that I do like that it's simple!

jrtrent said:

bclaff said:

Your solid angle should include the sine of half the angle of view.
If it doesn't it's an approximation.

Click to expand...

Thanks. Perhaps that's why the author of that article wrote:

The solid angle that a lens collects is approximately:
OMEGA = pi d² / 4 l²

jrtrent said:

The half-angle at 100mm is 12.21 degrees and at 50mm it's 23.40 degrees
sin(12.21) is 0.21 and sin(23.40) is 0.40

Note this isn't exactly a ratio of 2 but so close that you'd never see an exposure difference.

Click to expand...

Should I look for a different formula for solid angle, or is the one above from that article close enough to avoid gross errors? I have to admit that I do like that it's simple!

Click to expand...

I'm not sure you really need to do a calculation.
Isn't it the principle you want to understand?

Solid angle is basically sin(half-angle)^2

So entrance pupil changes by 4x and solid angle changes by 1/4x.

bclaff said:

jrtrent said:

bclaff said:

Your solid angle should include the sine of half the angle of view.
If it doesn't it's an approximation.

Click to expand...

Thanks. Perhaps that's why the author of that article wrote:

The solid angle that a lens collects is approximately:
OMEGA = pi d² / 4 l²

bclaff said:

The half-angle at 100mm is 12.21 degrees and at 50mm it's 23.40 degrees
sin(12.21) is 0.21 and sin(23.40) is 0.40

Note this isn't exactly a ratio of 2 but so close that you'd never see an exposure difference.

Click to expand...

Should I look for a different formula for solid angle, or is the one above from that article close enough to avoid gross errors? I have to admit that I do like that it's simple!

Click to expand...

I'm not sure you really need to do a calculation.
Isn't it the principle you want to understand?

Click to expand...

Yes, but until I work through the steps, I don't think I'll understand. The other explanation, the one included in the article from Hasselblad that I quoted, makes sense to me. I can easily visualize and understand why having the lens farther from the film plane reduces the illumination.

The lens, projecting the image, acts as the source of the image-forming light. And with increasing distance to the image plane, the area over which its 'output' is spread increases too (and, by the way, the size of the image itself does as well: magnification increases). Consequently there is less light per unit of area. The projected image is less bright.

The relation between image distance and illumination is a quite simple one too: illumination decreases with the square of the distance. Double the lens to film distance, and the illumination in the projected image decreases by a factor 4 (2²).

The explanation you gave to Leonard, "the amount of light collected is the product of the area of the entrance pupil and the solid angle of the light collected. The solid angle is in turn a function of the angle of view," is one I can't visualize or understand yet. I'll have to work with it some more.

bclaff said:

Solid angle is basically sin(half-angle)^2

So entrance pupil changes by 4x and solid angle changes by 1/4x.

Click to expand...

Thanks.

I wouldn't agonize over it since for your purposes the inverse square law works.

But, technically, if you use the solid angle, doubling the distance will not exactly work our as 1/4. As the angle gets larger the discrepancy gets greater.

The trigonometry the problem is that

sin(a)^2 != 4 * sin (a / 2)^2

(Many people drear this kind of math!)

bclaff said:

I wouldn't agonize over it since for your purposes the inverse square law works.

But, technically, if you use the solid angle, doubling the distance will not exactly work our as 1/4. As the angle gets larger the discrepancy gets greater.

The trigonometry the problem is that

sin(a)^2 != 4 * sin (a / 2)^2

(Many people dread this kind of math!)

Click to expand...

And I'm one of them! Thanks for all the help; much appreciated.

JohnDope said:

Hi,

It might sound like a trivial question for most of you, but I'm gonna ask anyway.

Let's say I have 2 lenses, 50mm and 100mm (FF, for argument's sake)

if I set my aperture the same on both lenses on the same camera, will the exposure be identical?

Click to expand...

Standing the same 8 foot distance from the focus point , 90mm and 135mm lenses, yes. If I meter off a grey card after framing and focusing I will get the same exposure setting within 1/5th of a stop, negligible in most cases. f5.6 ½ 1/500th @80

Distance to focus target matters. My 135 requires more light at minimum focus distance. So a ¾ shot on the 90 in landscape orientation from 8 feet will require less light than just the face filling the frame at minimum focus distance shooting with the 135mm. All that internal repositioning of lenses to facilitate close focus sucks up light.

JohnDope said:

the f number is the ratio of the focal length to the diameter of the entrance pupil.

Hence, if I set e.g. f8 on both these lenses, the size of the entrance pupil will be different.

8=50/E.P --> E.P= 50/8.

and

8=100/E.P --> E.P=100/8

if the enterance pupil is not the same size, the amount of light (or the flux of photons) hitting the sensor would be different. Does that mean the exposure would be different as well?

Thanks.

Click to expand...

FingerPainter said:

Bob said:

Exposure consists of f-stop + shutter + ISO.

It's the combination of those three things that will produce middle gray when shooting a white wall or middle gray when shooting a black wall or middle gray when shooting a gray card.

Focal length is not part of the equation.

Click to expand...

C'mon, Bob! You know better than that.

Exposure is the amount of light falling on the sensor per unit area while the shutter is open.

The three parameter of exposure are:

1. t-stop (approximated by f-stop)
2. shutter speed
3. scene luminance

Together, these are the only three parameters affecting how much light falls on the sensor per unit area. ISO is no more a part of exposure than is focal length.

Also, middle grey is not an exposure - it is a lightness.

Click to expand...

Technically yes, but in practice and usually when "exposure" is brought up in photography one is referencing the "exposure triangle". While exposure is how much (aperture) and how long (shutter speed) the sensor is exposed to light, ISO (sensitivity), plays are large roll in the output appearance of the final image combined with exposure. ISO can't be ignored when considering exposure, just because technically it's not part of the exposure process. It is part of the imaging process.

--
https://500px.com/jhmac

James H MacAllister said:

FingerPainter said:

Bob said:

Exposure consists of f-stop + shutter + ISO.

It's the combination of those three things that will produce middle gray when shooting a white wall or middle gray when shooting a black wall or middle gray when shooting a gray card.

Focal length is not part of the equation.

Click to expand...

C'mon, Bob! You know better than that.

Exposure is the amount of light falling on the sensor per unit area while the shutter is open.

The three parameter of exposure are:

1. t-stop (approximated by f-stop)
2. shutter speed
3. scene luminance

Together, these are the only three parameters affecting how much light falls on the sensor per unit area. ISO is no more a part of exposure than is focal length.

Also, middle grey is not an exposure - it is a lightness.

Click to expand...

Technically yes, but in practice and usually when "exposure" is brought up in photography one is referencing the "exposure triangle".

Click to expand...

Only if you're a devotee of the "exposure triangle", which many of us believe to be more a source of confusion than illumination.

James H MacAllister said:

While exposure is how much (aperture) and how long (shutter speed) the sensor is exposed to light,

Click to expand...

plus how much light is coming from the scene. The combination of f-number and exposure time is called 'exposure value' (EV), not 'exposure'.

James H MacAllister said:

ISO (sensitivity), plays are large roll in the output appearance of the final image combined with exposure. ISO can't be ignored when considering exposure, just because technically it's not part of the exposure process.

Click to expand...

It's not only 'technically' not part of the exposure 'process', it's not part of exposure. It cannot be, because ISO is defined in terms of exposure. If it were part of exposure, its definition would be circular.

James H MacAllister said:

It is part of the imaging process.

Click to expand...

That depends on which imaging process you're talking about. 'ISO' is not an essential part of making an image from a digital camera. It's a convenience, designed to allow photographers to operate digital cameras in much the same way as they operated film cameras. Also, th identification of ISO as 'sensitivity' is a recipe for confusion, you need to be very careful about what you mean by 'sensitivity' and what the other half of the conversation might infer from the word.

--
Is it always wrong
for one to have the hots for
Comrade Kim Yo Jong?

bobn2 said:

James H MacAllister said:

FingerPainter said:

Bob said:

Exposure consists of f-stop + shutter + ISO.

It's the combination of those three things that will produce middle gray when shooting a white wall or middle gray when shooting a black wall or middle gray when shooting a gray card.

Focal length is not part of the equation.

Click to expand...

C'mon, Bob! You know better than that.

Exposure is the amount of light falling on the sensor per unit area while the shutter is open.

The three parameter of exposure are:

1. t-stop (approximated by f-stop)
2. shutter speed
3. scene luminance

Together, these are the only three parameters affecting how much light falls on the sensor per unit area. ISO is no more a part of exposure than is focal length.

Also, middle grey is not an exposure - it is a lightness.

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Technically yes, but in practice and usually when "exposure" is brought up in photography one is referencing the "exposure triangle".

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Only if you're a devotee of the "exposure triangle", which many of us believe to be more a source of confusion than illumination.

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There is no place for beliefs in science. The exposure triangle is a fact, as the ISO themselves make clear in their relevant standards - not to mention the manufacturers of every photographic exposure meter, whether internal or external, ever made. You set the sensitivity (ISO) point the meter at the scene or light-source and the meter gives you an exposure value in the form of a range of combinations of shutter speed and aperture settings that match that EV..

bobn2 said:

bobn2 said:

While exposure is how much (aperture) and how long (shutter speed) the sensor is exposed to light,

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plus how much light is coming from the scene. The combination of f-number and exposure time is called 'exposure value' (EV), not 'exposure'.

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The exposure is literally when the sensor is exposed to the image projected onto it by the lens. The term "the exposure" is also used more loosely by photographers to refer to the result of that exposure.

bobn2 said:

bobn2 said:

ISO (sensitivity), plays are large roll in the output appearance of the final image combined with exposure. ISO can't be ignored when considering exposure, just because technically it's not part of the exposure process.

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It's not only 'technically' not part of the exposure 'process', it's not part of exposure. It cannot be, because ISO is defined in terms of exposure. If it were part of exposure, its definition would be circular.

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That is not true according to the relevant ISO standards.

bobn2 said:

bobn2 said:

It is part of the imaging process.

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That depends on which imaging process you're talking about. 'ISO' is not an essential part of making an image from a digital camera. It's a convenience, designed to allow photographers to operate digital cameras in much the same way as they operated film cameras. Also, th identification of ISO as 'sensitivity' is a recipe for confusion, you need to be very careful about what you mean by 'sensitivity' and what the other half of the conversation might infer from the word.

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Why not simply let the ISO themselves explain what the so-called ISO setting on your digital camera does?
I quote directly from their catalogue:
ISO 12232:2019
Photography — Digital still cameras — Determination of exposure index, ISO speed ratings, standard output sensitivity, and recommended exposure index
This document specifies the method for assigning and reporting ISO speed ratings, ISO speed latitude ratings, standard output sensitivity values, and recommended exposure index values, for digital still cameras. It is applicable to both monochrome and colour digital still cameras.

And there you have it straight from the horses mouth.