Re: My hyper-confused self and the f-stop
Quote:
Originally Posted by
BennoLF
...
Let's say we have a wall 4x5 feet with an even luminous density of 30 cdl/sq. ft., suspended in an infinite blackness. ... If the camera was fixed at a point where the wall filled the frame of the 300mm lens, the 90mm lens from the same distance would show tons of black around it, and would need a longer exposure to attain an average zone V, right? Because the bright wall would take up a smaller percentage of the frame? ...
Benno
Benno,
I've highlighted where your logic breaks down. Yes, you are right, the "average" amount of light will be different in the scenario you describe.
Your mistake is that, when photographing, we don't care about averaging the amount of light hitting the film surface. We care about getting the proper amount of light for each point of the film surface.
Think of it this way: with the frame filled by your wall of light, every square mm on the film surface gets just the right amount of light to expose the wall correctly.
With the 90mm lens, not every square mm gets this much exposure, but every sq. mm of the image of the wall of light does. Yes, if you add up the total light hitting the film in the two exposures, the total amount will be different. But, the exposure for the wall is the same because the intensity of light for the area of the image of the wall is the same. Clear?
For your sky exposure, the light balancing mechanism is different. It's the aperture: The 300mm lens sees less light total than the 90mm due to the of the angle of view if and only if the aperture the light is passing through is the same physical size. At f/8 for both lenses, the aperture is a different size. The f/8 aperture on the 300mm lens is larger (more "light efficient") and lets in more light than the smaller aperture on the 90mm lens set at f/8 (which is less efficient and lets in less light). Voilà, the amount of light ends up being the same and the exposure is the same.
Your conclusions in the paragraphs about the two f/2 lenses are simply wrong. The illumination is the same for both lenses, because at f/2, the apertures for the two lenses are different sizes, which balances the difference in the amount of light gathered by the different angles of view, evening them out. "Speed is intensity X area" is wrong. Intensity x area = total light for light projected through lenses if and only if the aperture the light is passing through is the same physical size. You're effectively leaving the aperture size and the whole idea of f-numbers out of your thinking here.
Once you get your head around this, you'll kick yourself :)
Best,
Doremus
Re: My hyper-confused self and the f-stop
Quote:
Originally Posted by
C. D. Keth
No. Exposure and field of view are different. We are not dealing with automatic or through the lens metering. I can expose a photograph precisely the same from the same vantage point whether I do it on a 15mm lens or a 500mm lens. The perception of brightness may be different, however. A black dog in a snowy field will look different if he fills 80% of the frame than when he fills 2% of the frame, even when exposed the same.
Isn't that based on a 'correct' metering system though, i.e. a system looking to put faces on zone VI, deepest shadow detail on zone III, etc? Something looking for avg. zone V would say a black dog filling 80% of the frame is underexposed while a 2% dog would be underexposed. Obviously for real photographs we wouldn't give a fig about whether or not the dog fills 80% of the frame exposure-wise. We want the dog to be occupying a certain zone. But the computer wants a mathematical balance, and can't tell if it looks good.
Quote:
Originally Posted by
C. D. Keth
Lets continue to imagine this clear sky scenario with a 25mm lens and a 50mm lens. The field of view of the 25mm lens will be twice the angular field of view both horizontally and vertically.
Yes, the 25mm lens does gather light from a larger area of sky, 4x larger by area in fact. That does mean that it gathers 4x more light as a gross amount of photons. However, it then passes that light through an aperture that is 1/4 the area of the longer lens' aperture.
Lets do the math with theoretical f/2 figures, to make relatively small numbers:
area of the entrance pupil, a circle, is found with the equation area=pi*(r^2) where r is the radius of the circle
pi*(12.5^2)=491 square mm area for the 25mm lens
pi*(25^2)=1963 square mm area for the 50mm lens
Let’s check the relationship between our f2 entrance pupils. Remember that the areas of sky photographed were different by a factor of 4x, the 25mm lens “saw” 4x more area of sky.
491*4=1963
The 25mm lens that saw 4x more sky let only 1/4 of the gross light through its aperture than did the longer lens.
Net result? The same amount of light gets through each lens and is recorded onto the film. Being a completely featureless, clear, blue sky the images look exactly the same.
Wait shoot I feel like what you've said would mean that I'm right. Same amount of light from smaller angle of view would mean more light/degree, right? Gahh.
For a starry night the 50mm lens would be the better choice, no? More 'light efficiency'?
Quote:
Originally Posted by
Doremus Scudder
Benno,
I've highlighted where your logic breaks down. Yes, you are right, the "average" amount of light will be different in the scenario you describe.
Your mistake is that, when photographing, we don't care about averaging the amount of light hitting the film surface. We care about getting the proper amount of light for each point of the film surface.
Think of it this way: with the frame filled by your wall of light, every square mm on the film surface gets just the right amount of light to expose the wall correctly.
With the 90mm lens, not every square mm gets this much exposure, but every sq. mm of the image of the wall of light does. Yes, if you add up the total light hitting the film in the two exposures, the total amount will be different. But, the exposure for the wall is the same because the intensity of light for the area of the image of the wall is the same. Clear?
^but if we're talking about an average of zone V isn't total amount of light exactly what we care about?
Also, I think I messed up my example/I'm confused enough to disagree with my own example.
The total for the two exposures will be the same, but the average is different, no?
600 foot-candles filling the frame or 600 foot-candles from a single point is still 600 foot-candles, and the infinite blackness doesn't add any light. Adding a bunch of 'zero-values' doesn't change the total but it does change the average. Holy man I'm even more lost.
Quote:
Originally Posted by
Doremus Scudder
For your sky exposure, the light balancing mechanism is different. It's the aperture: The 300mm lens sees less light total than the 90mm due to the of the angle of view if and only if the aperture the light is passing through is the same physical size. At f/8 for both lenses, the aperture is a different size. The f/8 aperture on the 300mm lens is larger (more "light efficient") and lets in more light than the smaller aperture on the 90mm lens set at f/8 (which is less efficient and lets in less light). Voilà, the amount of light ends up being the same and the exposure is the same.
Ok so the total amount of light is the same. That makes sense to me, at least for now haha.
But a longer lens is still being more light-efficient, letting a higher proportion of the available light through the back? So for something like astro that's what we would want, right? Maximum efficiency because for most of us the stars aren't that bright?
Quote:
Originally Posted by
Doremus Scudder
Your conclusions in the paragraphs about the two f/2 lenses are simply wrong. The illumination is the same for both lenses, because at f/2, the apertures for the two lenses are different sizes, which balances the difference in the amount of light gathered by the different angles of view, evening them out. "Speed is intensity X area" is wrong. Intensity x area = total light for light projected through lenses if and only if the aperture the light is passing through is the same physical size. You're effectively leaving the aperture size and the whole idea of f-numbers out of your thinking here.
That paragraph was from the petapixel thingy. I was just saying speed=intensity x area because I thought their use of the word speed was ambiguous, but I might've made it worse.
Near the end they said:
"So in terms of exposure value, the 24mm lens will produce equivalent brightness images for any given ISO and shutter speed........The long lens collects more light at a time from a smaller area of the scene while the short lens collects less light at a time from a larger area of the scene."
Isn't the above correct?
Less light at a time from a wide angle should be the same as more light at a time from a narrow angle, assuming our proportions work out.
Quote:
Originally Posted by
Doremus Scudder
Once you get your head around this, you'll kick yourself :)
I sure hope I do, especially after all the trouble y'all've been to to punch it into my thick skull. I'm getting the same feeling I get when someone's walking me through a math proof or some moral argument and the light's beginning to dawn.
My hyper-confused self and the f-stop
Quote:
Originally Posted by
BennoLF
Isn't that based on a 'correct' metering system though, i.e. a system looking to put faces on zone VI, deepest shadow detail on zone III, etc? Something looking for avg. zone V would say a black dog filling 80% of the frame is underexposed while a 2% dog would be underexposed. Obviously for real photographs we wouldn't give a fig about whether or not the dog fills 80% of the frame exposure-wise. We want the dog to be occupying a certain zone. But the computer wants a mathematical balance, and can't tell if it looks good.
Wait shoot I feel like what you've said would mean that I'm right. Same amount of light from smaller angle of view would mean more light/degree, right? Gahh.
For a starry night the 50mm lens would be the better choice, no? More 'light efficiency'?
Read my post again slowly and carefully. Don’t skim it. I feel as though you have read only 2/3 of the words.
The dog thing, it’s not based on ANY metering system. As I said when I wrote it that is a perceptual effect dealing with how your eyes perceive contrast. I probably should have left it out to avoid this confusion.
With the lens comparison, read that again, too. Do the math. I have. Let me tell you there’s no way around physics no matter how you try. If the lenses are at the same f-stop, they will allow the same intensity of light through because the iris is constructed to be inversely proportional to the field of view. That offsets the “wider lenses gather more light” thing you’re hung up on.
There is a book called “Basic Photographic Materials and Processes.” It’s by Nanette Salvaggio. She teaches in the photo tech dept. at RIT. The best thing you could do is find any edition of that book and read it.
Re: My hyper-confused self and the f-stop
If you have 2 cameras next to each other, a 3¼X4¼ with a 90mm lens and an 11X14 with a 300mm lens would you meter different? They both are seeing the same image. f8 on the little camera little lens is the same as f8 on the big camera big lens. Trust me (I'm old and I know of what I speak). You'd meter one time for both cameras assuming both are shooting the same asa film.
1 Attachment(s)
Re: My hyper-confused self and the f-stop
Quote:
Originally Posted by
C. D. Keth
With the lens comparison, read that again, too. Do the math. I have. Let me tell you there’s no way around physics no matter how you try. If the lenses are at the same f-stop, they will allow the same intensity of light through because the iris is constructed to be inversely proportional to the field of view. That offsets the “wider lenses gather more light” thing you’re hung up on.
The fact that the math works out is why I'm asking all these questions. If the math didn't work out then I'd just throw my hands up, meter something, and hope my negatives come out.
Quote:
Originally Posted by
Jim Galli
If you have 2 cameras next to each other, a 3¼X4¼ with a 90mm lens and an 11X14 with a 300mm lens would you meter different? They both are seeing the same image. f8 on the little camera little lens is the same as f8 on the big camera big lens. Trust me (I'm old and I know of what I speak). You'd meter one time for both cameras assuming both are shooting the same asa film.
No, because they're seeing the same thing..........this is why I'm confused. I'm not asking about practical terms here, I'm just puzzling around.
Attachment 201724
(pardon my awful penmanship)
So in this chart I'm trying to illustrate my conundrum.
From C.D. Keth's math, it seems that they both pass the same amount of light. Cool.
However, if two lenses are taking in a different amount but passing the same amount through the back, then musn't one be passing a greater proportion of the incoming light?
Gaahhhhh
Re: My hyper-confused self and the f-stop
Quote:
Originally Posted by
BennoLF
WARNING!!
If you want to close the tab and go about your day conniption-free, please do.
Thanks!
Benno
The best advice so far. Unsubscribed.
Re: My hyper-confused self and the f-stop
Benno,
You're right about the total amount of light fooling an averaging meter. And, it does make a difference in the meter reading when you point your meter to a different part of the scene that has more light or dark areas. Most meters average things out. Even a spot meter averages the reading for the spot it's seeing. That's why it's important to be smarter than your meter; be aware of what you are pointing it at and compensate if you think the average of what the meter is seeing shouldn't be in Zone V (or whatever).
And yes, your meter gives you f-stop and shutter speed combinations based on how much light it sees. That's a separate discussion from the fact that all lenses set at f/8 will give the same exposure for a given object regardless of how much or little it fills the frame. Lenses don't average the light, they focus it. The film has the same dark and light areas as the scene. The total amount of light hitting the film changes depending on the proportion of light to dark areas in the scene being photographed, but the rendering of any one evenly bright object stays the same when you zoom in or out; it just changes size. Same when you move closer or farther; that object gets bigger or smaller, but the density on film at a given exposure remains the same.
So, if you need f/8 for a close-up photo of an apple, then if you back up to photograph the whole tree, or change lenses from long to wide or whatever, you'll still need f/8 to get the same exposure of that apple. It doesn't matter if it takes up 90% or 10% of the image.
Best,
Doremus
My hyper-confused self and the f-stop
Quote:
Originally Posted by
BennoLF
This is fallacious from the start because a 90mm sees a cone 4x as wide as a 360mm in BOTH dimensions. That has it seeing an area of that surface 16x more than the 360mm lens.
Let’s just say both lenses are at f8 and the surface is still lit at 150 units of light per unit of area. The cameras are right next to each other, the same distance from the wall, and the wall isn’t particularly reflective such that the small change in position would change perceived brightness. The 90mm takes in a whopping 2400 units of light (150*16) total while the 360mm only takes in 150 units of light (1*150) total.
That 16x difference in area and in light units seems crazy. They can’t both expose that surface the same on film, can they?
Let’s figure it out. We set both lenses at f8 so we must find the area of the entrance pupil to see how much light actually gets in and focused onto the film.
The diameter of the entrance pupil of the 90mm is 11.25mm
Area is pi*r^2
A[90mm]=pi*(5.625^2)
A[90mm]=99.40mm^2
The diameter of the 360mm’s entrance pupil is 45mm
A[360mm]=pi*r^2
A[360mm]=pi*(22.5^2)
A[360mm]=1590.43mm^2
Check our logic and math by finding the proportion between these aperture areas. It should be inversely proportional to the areas of that surface being photographed, or volumes of field of view, depending on which one is easier to think about.
1590.43/99.40=16.0000
The proportion between entrance pupils is 16x, the very same as the proportion of area of that wall and between pyramidal volumes of field of view between lenses.
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Re: My hyper-confused self and the f-stop
Quote:
Originally Posted by
C. D. Keth
This is fallacious from the start because a 90mm sees a cone 4x as wide as a 360mm in BOTH dimensions. That has it seeing an area of that surface 16x more than the 360mm lens.
Oh gracious wow. Speaking of kicking yourself geez.
Quote:
Originally Posted by
Doremus Scudder
Benno,
You're right about the total amount of light fooling an averaging meter. And, it does make a difference in the meter reading when you point your meter to a different part of the scene that has more light or dark areas. Most meters average things out. Even a spot meter averages the reading for the spot it's seeing. That's why it's important to be smarter than your meter; be aware of what you are pointing it at and compensate if you think the average of what the meter is seeing shouldn't be in Zone V (or whatever).
And yes, your meter gives you f-stop and shutter speed combinations based on how much light it sees. That's a separate discussion from the fact that all lenses set at f/8 will give the same exposure for a given object regardless of how much or little it fills the frame. Lenses don't average the light, they focus it. The film has the same dark and light areas as the scene. The total amount of light hitting the film changes depending on the proportion of light to dark areas in the scene being photographed, but the rendering of any one evenly bright object stays the same when you zoom in or out; it just changes size. Same when you move closer or farther; that object gets bigger or smaller, but the density on film at a given exposure remains the same.
So, if you need f/8 for a close-up photo of an apple, then if you back up to photograph the whole tree, or change lenses from long to wide or whatever, you'll still need f/8 to get the same exposure of that apple. It doesn't matter if it takes up 90% or 10% of the image.
Best,
Doremus
AHAH!!! I finally get it!! It's the size changing that had me throwed. I was worried about luminous density from the subject but not on the film, I was thinking only of the total amount on the film. Now that it's clicked all the times you guys said "we don't care about total amount" make sense now.
Now that it makes sense it makes the f/stop system so much more cool to me. Having the amount of light diminish at the same rate the area of a given subject does so the density stays the same is so clever. Man that took too long.
Thanks so much for all y'all's help! I apologize again for any and all cardiac issue this thread has caused. Hopefully I won't try to think past the practical anymore.
Re: My hyper-confused self and the f-stop
It’s all good. Sometimes you have to kick your brain in circles before it clicks. Now don’t go thinking too hard about depth of field. If it looks sharp on the groundglass it’ll be sharp on film.[emoji6]
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