Why do lower f-stop numbers mean larger apertures? [duplicate]
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2
down vote
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This question already has an answer here:
What does f-stop mean?
4 answers
I thought this would have been asked before but it seems not.
I don't understand how f-stops are numbered. From what I read, the lower the f-stop number, the larger the aperture. That is, f/2 is larger than f/22.
This seems odd to me. The maximum aperture width is (at least theoretically) infinite, while the minimum is 0 (i.e. the lens is completely closed and not letting in any light).
The way the numbering system is set up, it allows for infinitely tiny apertures but a limit on how large they can be (i.e. I guess the largest is f/0)
Can someone clarify why the aperture is numbered in this way?
aperture f-stop
marked as duplicate by scottbb, xiota, Philip Kendall, Rafael, Hueco 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
2
down vote
favorite
This question already has an answer here:
What does f-stop mean?
4 answers
I thought this would have been asked before but it seems not.
I don't understand how f-stops are numbered. From what I read, the lower the f-stop number, the larger the aperture. That is, f/2 is larger than f/22.
This seems odd to me. The maximum aperture width is (at least theoretically) infinite, while the minimum is 0 (i.e. the lens is completely closed and not letting in any light).
The way the numbering system is set up, it allows for infinitely tiny apertures but a limit on how large they can be (i.e. I guess the largest is f/0)
Can someone clarify why the aperture is numbered in this way?
aperture f-stop
marked as duplicate by scottbb, xiota, Philip Kendall, Rafael, Hueco 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Also see photo.stackexchange.com/questions/12072/…
– Tetsujin
7 hours ago
There may be limits on how large an aperture can be, but practical considerations make lenses faster than f/1 very, very, very rare. The fastest I have ever heard of is f/0.7.
– Jim MacKenzie
6 hours ago
There are many measuring systems where a larger numerical value means a smaller physical size.
– whatsisname
34 mins ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question already has an answer here:
What does f-stop mean?
4 answers
I thought this would have been asked before but it seems not.
I don't understand how f-stops are numbered. From what I read, the lower the f-stop number, the larger the aperture. That is, f/2 is larger than f/22.
This seems odd to me. The maximum aperture width is (at least theoretically) infinite, while the minimum is 0 (i.e. the lens is completely closed and not letting in any light).
The way the numbering system is set up, it allows for infinitely tiny apertures but a limit on how large they can be (i.e. I guess the largest is f/0)
Can someone clarify why the aperture is numbered in this way?
aperture f-stop
This question already has an answer here:
What does f-stop mean?
4 answers
I thought this would have been asked before but it seems not.
I don't understand how f-stops are numbered. From what I read, the lower the f-stop number, the larger the aperture. That is, f/2 is larger than f/22.
This seems odd to me. The maximum aperture width is (at least theoretically) infinite, while the minimum is 0 (i.e. the lens is completely closed and not letting in any light).
The way the numbering system is set up, it allows for infinitely tiny apertures but a limit on how large they can be (i.e. I guess the largest is f/0)
Can someone clarify why the aperture is numbered in this way?
This question already has an answer here:
What does f-stop mean?
4 answers
aperture f-stop
aperture f-stop
asked 7 hours ago
CodyBugstein
2742416
2742416
marked as duplicate by scottbb, xiota, Philip Kendall, Rafael, Hueco 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by scottbb, xiota, Philip Kendall, Rafael, Hueco 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Also see photo.stackexchange.com/questions/12072/…
– Tetsujin
7 hours ago
There may be limits on how large an aperture can be, but practical considerations make lenses faster than f/1 very, very, very rare. The fastest I have ever heard of is f/0.7.
– Jim MacKenzie
6 hours ago
There are many measuring systems where a larger numerical value means a smaller physical size.
– whatsisname
34 mins ago
add a comment |
Also see photo.stackexchange.com/questions/12072/…
– Tetsujin
7 hours ago
There may be limits on how large an aperture can be, but practical considerations make lenses faster than f/1 very, very, very rare. The fastest I have ever heard of is f/0.7.
– Jim MacKenzie
6 hours ago
There are many measuring systems where a larger numerical value means a smaller physical size.
– whatsisname
34 mins ago
Also see photo.stackexchange.com/questions/12072/…
– Tetsujin
7 hours ago
Also see photo.stackexchange.com/questions/12072/…
– Tetsujin
7 hours ago
There may be limits on how large an aperture can be, but practical considerations make lenses faster than f/1 very, very, very rare. The fastest I have ever heard of is f/0.7.
– Jim MacKenzie
6 hours ago
There may be limits on how large an aperture can be, but practical considerations make lenses faster than f/1 very, very, very rare. The fastest I have ever heard of is f/0.7.
– Jim MacKenzie
6 hours ago
There are many measuring systems where a larger numerical value means a smaller physical size.
– whatsisname
34 mins ago
There are many measuring systems where a larger numerical value means a smaller physical size.
– whatsisname
34 mins ago
add a comment |
5 Answers
5
active
oldest
votes
up vote
5
down vote
The f stop is a comparison between the physical size of the focal length and the aperture.
If the aperture is 25mm and the focal length is 100mm, then you'd have f/4 because the aperture value is 1/4 the focal length.
So, the f stop is the fractional representation of this comparison. As with fractions, 1/2 is bigger than 1/22.
add a comment |
up vote
3
down vote
/ is division.
A number divided by 22 (f/22) is smaller than the same number divided by 2 (f/2).
As simple as that.
Now, WHY?
f is the focal length. Describing the aperture as a fraction of the focal length has one advantage: it immediately reveals the image brightness.
Example:
20mm aperture can be...
- Telephoto lens, 200mm f/10 - pretty dark
- Wide angle lens, 30mm f/1.5 - pretty bright
1
Aha didn't realize that we were dividingf
. So what isf
? What doesf/1
equal?
– CodyBugstein
7 hours ago
Aha So aperture off/6
means something very different on a 30mm lens and a 200mm lens?
– CodyBugstein
7 hours ago
f/1 would be aperture equal to the focal length (which means quite big aperture and bright lens but otherwise not a special case of any kind. it is always possible to push a little harder and make lens a bit bigger and more expensive)
– szulat
7 hours ago
1
f/6 means the physical diameter would be different but they produce the same brightness.
– szulat
6 hours ago
So the maximum aperture is always equal to the length of the lens? In other words, longer lenses must be fatter too?
– CodyBugstein
6 hours ago
|
show 1 more comment
up vote
2
down vote
It's no accident that f-stops are written with a "division" slash.
You say, "f/2 is larger than f/22". Here, "f" stands for the focal length, and the result of the expression gives the physical size of the aperture opening. So, let's take a 50mm lens... At f/2, the aperture opening measures 25mm. At f/22, the aperture opening measures ~2.3mm. Obviously an opening measuring 25mm is going to let in more light than one that measures ~2.3mm.
You say, "I guess the largest is f/0". No. You can't divide by zero. Or in other words, there is no lens such that the ratio of the focal length to the size of the aperture is f:∞.
I always say... photography is the art for mathematicians :-P
– osullic
3 hours ago
I have literally been told, you're about the science and I'm more about the art, when trying to teach darkroom principals. Le sigh.
– Hueco
2 hours ago
add a comment |
up vote
0
down vote
Because the lens operates much like a funnel in that it gathers light, the greater the working diameter of the lens, the brighter the projected image. How bright this image will be is dependent on the brightness of the scene and the magnification realized by the lens. The longer the focal length, the more the lens magnifies. The deed of magnifying to produce an image, takes its toll on image brightness. In other words, the longer the focal length, the more the lens magnifies. This higher magnification result is a larger but dimmer image of objects.
Another way to say this, image brightness intertwines the working diameter and the focal length. Because these two factors are so interwoven, gauging image brightness is demanding. We are forced to fall back on a mathematical ratio that will take the chaos out of figuring out image brightness. This is true because a ratio is dimensionless. If I tell you the ratio of boys to girls in a 6th grade class is 3 boys for every 4 girls, I have given you a ratio that works independent of the number of students. For example if the class consist of 28 kids, then 12 boys to 16 girls is the breakdown (ratio is dimensionless).
For the camera lens: if the working diameter is 4 inches and the focal length is 4 inches, then the focal ratio (f-number) = 4 ÷ 4 = 1 (written as f/1 (f/1 is produces a very bright image). If the working diameter is 2 inches and the focal length is 4 inches, then the f-number is 4 ÷ 2 = 2 (written as f/2.).
The splendor of using a ratio is, any lens operating at the same f-number as another lens, yields the same image brightness regardless of the dimensions (diameter or focal length), for an identical scene. It’s complicated; but the f-number system actually takes away the chaos.
add a comment |
up vote
0
down vote
The "why" is that f/stop number is focal length / aperture diameter.
So a 100 mm lens at f/4 has an effective aperture of 25 mm. This is not exactly the physical diameter opening, but is the effective diameter, specifically the front entrance pupil, as seen by the magnification of the front lens elements.
Because of this definition, as aperture diameter becomes larger (and exposure increases), f/stop number becomes smaller. f/2.8 is wide, and f/22 is narrow.
It may seem backwards at first, but we get over it easily, and the huge advantage of the f/stop number system is that it applies for any lens, of any size and focal length and aperture, so then we all know what we mean when we say f/8. f/8 is f/8 exposure on any lens. This is a plus. It wasn't that way in the early beginning. And of course, light meters meter with the same meanings.
My site has more about this at https://www.scantips.com/lights/fstop.html
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The f stop is a comparison between the physical size of the focal length and the aperture.
If the aperture is 25mm and the focal length is 100mm, then you'd have f/4 because the aperture value is 1/4 the focal length.
So, the f stop is the fractional representation of this comparison. As with fractions, 1/2 is bigger than 1/22.
add a comment |
up vote
5
down vote
The f stop is a comparison between the physical size of the focal length and the aperture.
If the aperture is 25mm and the focal length is 100mm, then you'd have f/4 because the aperture value is 1/4 the focal length.
So, the f stop is the fractional representation of this comparison. As with fractions, 1/2 is bigger than 1/22.
add a comment |
up vote
5
down vote
up vote
5
down vote
The f stop is a comparison between the physical size of the focal length and the aperture.
If the aperture is 25mm and the focal length is 100mm, then you'd have f/4 because the aperture value is 1/4 the focal length.
So, the f stop is the fractional representation of this comparison. As with fractions, 1/2 is bigger than 1/22.
The f stop is a comparison between the physical size of the focal length and the aperture.
If the aperture is 25mm and the focal length is 100mm, then you'd have f/4 because the aperture value is 1/4 the focal length.
So, the f stop is the fractional representation of this comparison. As with fractions, 1/2 is bigger than 1/22.
edited 2 hours ago
answered 7 hours ago
Hueco
9,80732346
9,80732346
add a comment |
add a comment |
up vote
3
down vote
/ is division.
A number divided by 22 (f/22) is smaller than the same number divided by 2 (f/2).
As simple as that.
Now, WHY?
f is the focal length. Describing the aperture as a fraction of the focal length has one advantage: it immediately reveals the image brightness.
Example:
20mm aperture can be...
- Telephoto lens, 200mm f/10 - pretty dark
- Wide angle lens, 30mm f/1.5 - pretty bright
1
Aha didn't realize that we were dividingf
. So what isf
? What doesf/1
equal?
– CodyBugstein
7 hours ago
Aha So aperture off/6
means something very different on a 30mm lens and a 200mm lens?
– CodyBugstein
7 hours ago
f/1 would be aperture equal to the focal length (which means quite big aperture and bright lens but otherwise not a special case of any kind. it is always possible to push a little harder and make lens a bit bigger and more expensive)
– szulat
7 hours ago
1
f/6 means the physical diameter would be different but they produce the same brightness.
– szulat
6 hours ago
So the maximum aperture is always equal to the length of the lens? In other words, longer lenses must be fatter too?
– CodyBugstein
6 hours ago
|
show 1 more comment
up vote
3
down vote
/ is division.
A number divided by 22 (f/22) is smaller than the same number divided by 2 (f/2).
As simple as that.
Now, WHY?
f is the focal length. Describing the aperture as a fraction of the focal length has one advantage: it immediately reveals the image brightness.
Example:
20mm aperture can be...
- Telephoto lens, 200mm f/10 - pretty dark
- Wide angle lens, 30mm f/1.5 - pretty bright
1
Aha didn't realize that we were dividingf
. So what isf
? What doesf/1
equal?
– CodyBugstein
7 hours ago
Aha So aperture off/6
means something very different on a 30mm lens and a 200mm lens?
– CodyBugstein
7 hours ago
f/1 would be aperture equal to the focal length (which means quite big aperture and bright lens but otherwise not a special case of any kind. it is always possible to push a little harder and make lens a bit bigger and more expensive)
– szulat
7 hours ago
1
f/6 means the physical diameter would be different but they produce the same brightness.
– szulat
6 hours ago
So the maximum aperture is always equal to the length of the lens? In other words, longer lenses must be fatter too?
– CodyBugstein
6 hours ago
|
show 1 more comment
up vote
3
down vote
up vote
3
down vote
/ is division.
A number divided by 22 (f/22) is smaller than the same number divided by 2 (f/2).
As simple as that.
Now, WHY?
f is the focal length. Describing the aperture as a fraction of the focal length has one advantage: it immediately reveals the image brightness.
Example:
20mm aperture can be...
- Telephoto lens, 200mm f/10 - pretty dark
- Wide angle lens, 30mm f/1.5 - pretty bright
/ is division.
A number divided by 22 (f/22) is smaller than the same number divided by 2 (f/2).
As simple as that.
Now, WHY?
f is the focal length. Describing the aperture as a fraction of the focal length has one advantage: it immediately reveals the image brightness.
Example:
20mm aperture can be...
- Telephoto lens, 200mm f/10 - pretty dark
- Wide angle lens, 30mm f/1.5 - pretty bright
edited 7 hours ago
answered 7 hours ago
szulat
3,58711126
3,58711126
1
Aha didn't realize that we were dividingf
. So what isf
? What doesf/1
equal?
– CodyBugstein
7 hours ago
Aha So aperture off/6
means something very different on a 30mm lens and a 200mm lens?
– CodyBugstein
7 hours ago
f/1 would be aperture equal to the focal length (which means quite big aperture and bright lens but otherwise not a special case of any kind. it is always possible to push a little harder and make lens a bit bigger and more expensive)
– szulat
7 hours ago
1
f/6 means the physical diameter would be different but they produce the same brightness.
– szulat
6 hours ago
So the maximum aperture is always equal to the length of the lens? In other words, longer lenses must be fatter too?
– CodyBugstein
6 hours ago
|
show 1 more comment
1
Aha didn't realize that we were dividingf
. So what isf
? What doesf/1
equal?
– CodyBugstein
7 hours ago
Aha So aperture off/6
means something very different on a 30mm lens and a 200mm lens?
– CodyBugstein
7 hours ago
f/1 would be aperture equal to the focal length (which means quite big aperture and bright lens but otherwise not a special case of any kind. it is always possible to push a little harder and make lens a bit bigger and more expensive)
– szulat
7 hours ago
1
f/6 means the physical diameter would be different but they produce the same brightness.
– szulat
6 hours ago
So the maximum aperture is always equal to the length of the lens? In other words, longer lenses must be fatter too?
– CodyBugstein
6 hours ago
1
1
Aha didn't realize that we were dividing
f
. So what is f
? What does f/1
equal?– CodyBugstein
7 hours ago
Aha didn't realize that we were dividing
f
. So what is f
? What does f/1
equal?– CodyBugstein
7 hours ago
Aha So aperture of
f/6
means something very different on a 30mm lens and a 200mm lens?– CodyBugstein
7 hours ago
Aha So aperture of
f/6
means something very different on a 30mm lens and a 200mm lens?– CodyBugstein
7 hours ago
f/1 would be aperture equal to the focal length (which means quite big aperture and bright lens but otherwise not a special case of any kind. it is always possible to push a little harder and make lens a bit bigger and more expensive)
– szulat
7 hours ago
f/1 would be aperture equal to the focal length (which means quite big aperture and bright lens but otherwise not a special case of any kind. it is always possible to push a little harder and make lens a bit bigger and more expensive)
– szulat
7 hours ago
1
1
f/6 means the physical diameter would be different but they produce the same brightness.
– szulat
6 hours ago
f/6 means the physical diameter would be different but they produce the same brightness.
– szulat
6 hours ago
So the maximum aperture is always equal to the length of the lens? In other words, longer lenses must be fatter too?
– CodyBugstein
6 hours ago
So the maximum aperture is always equal to the length of the lens? In other words, longer lenses must be fatter too?
– CodyBugstein
6 hours ago
|
show 1 more comment
up vote
2
down vote
It's no accident that f-stops are written with a "division" slash.
You say, "f/2 is larger than f/22". Here, "f" stands for the focal length, and the result of the expression gives the physical size of the aperture opening. So, let's take a 50mm lens... At f/2, the aperture opening measures 25mm. At f/22, the aperture opening measures ~2.3mm. Obviously an opening measuring 25mm is going to let in more light than one that measures ~2.3mm.
You say, "I guess the largest is f/0". No. You can't divide by zero. Or in other words, there is no lens such that the ratio of the focal length to the size of the aperture is f:∞.
I always say... photography is the art for mathematicians :-P
– osullic
3 hours ago
I have literally been told, you're about the science and I'm more about the art, when trying to teach darkroom principals. Le sigh.
– Hueco
2 hours ago
add a comment |
up vote
2
down vote
It's no accident that f-stops are written with a "division" slash.
You say, "f/2 is larger than f/22". Here, "f" stands for the focal length, and the result of the expression gives the physical size of the aperture opening. So, let's take a 50mm lens... At f/2, the aperture opening measures 25mm. At f/22, the aperture opening measures ~2.3mm. Obviously an opening measuring 25mm is going to let in more light than one that measures ~2.3mm.
You say, "I guess the largest is f/0". No. You can't divide by zero. Or in other words, there is no lens such that the ratio of the focal length to the size of the aperture is f:∞.
I always say... photography is the art for mathematicians :-P
– osullic
3 hours ago
I have literally been told, you're about the science and I'm more about the art, when trying to teach darkroom principals. Le sigh.
– Hueco
2 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
It's no accident that f-stops are written with a "division" slash.
You say, "f/2 is larger than f/22". Here, "f" stands for the focal length, and the result of the expression gives the physical size of the aperture opening. So, let's take a 50mm lens... At f/2, the aperture opening measures 25mm. At f/22, the aperture opening measures ~2.3mm. Obviously an opening measuring 25mm is going to let in more light than one that measures ~2.3mm.
You say, "I guess the largest is f/0". No. You can't divide by zero. Or in other words, there is no lens such that the ratio of the focal length to the size of the aperture is f:∞.
It's no accident that f-stops are written with a "division" slash.
You say, "f/2 is larger than f/22". Here, "f" stands for the focal length, and the result of the expression gives the physical size of the aperture opening. So, let's take a 50mm lens... At f/2, the aperture opening measures 25mm. At f/22, the aperture opening measures ~2.3mm. Obviously an opening measuring 25mm is going to let in more light than one that measures ~2.3mm.
You say, "I guess the largest is f/0". No. You can't divide by zero. Or in other words, there is no lens such that the ratio of the focal length to the size of the aperture is f:∞.
edited 4 hours ago
answered 4 hours ago
osullic
5,53311021
5,53311021
I always say... photography is the art for mathematicians :-P
– osullic
3 hours ago
I have literally been told, you're about the science and I'm more about the art, when trying to teach darkroom principals. Le sigh.
– Hueco
2 hours ago
add a comment |
I always say... photography is the art for mathematicians :-P
– osullic
3 hours ago
I have literally been told, you're about the science and I'm more about the art, when trying to teach darkroom principals. Le sigh.
– Hueco
2 hours ago
I always say... photography is the art for mathematicians :-P
– osullic
3 hours ago
I always say... photography is the art for mathematicians :-P
– osullic
3 hours ago
I have literally been told, you're about the science and I'm more about the art, when trying to teach darkroom principals. Le sigh.
– Hueco
2 hours ago
I have literally been told, you're about the science and I'm more about the art, when trying to teach darkroom principals. Le sigh.
– Hueco
2 hours ago
add a comment |
up vote
0
down vote
Because the lens operates much like a funnel in that it gathers light, the greater the working diameter of the lens, the brighter the projected image. How bright this image will be is dependent on the brightness of the scene and the magnification realized by the lens. The longer the focal length, the more the lens magnifies. The deed of magnifying to produce an image, takes its toll on image brightness. In other words, the longer the focal length, the more the lens magnifies. This higher magnification result is a larger but dimmer image of objects.
Another way to say this, image brightness intertwines the working diameter and the focal length. Because these two factors are so interwoven, gauging image brightness is demanding. We are forced to fall back on a mathematical ratio that will take the chaos out of figuring out image brightness. This is true because a ratio is dimensionless. If I tell you the ratio of boys to girls in a 6th grade class is 3 boys for every 4 girls, I have given you a ratio that works independent of the number of students. For example if the class consist of 28 kids, then 12 boys to 16 girls is the breakdown (ratio is dimensionless).
For the camera lens: if the working diameter is 4 inches and the focal length is 4 inches, then the focal ratio (f-number) = 4 ÷ 4 = 1 (written as f/1 (f/1 is produces a very bright image). If the working diameter is 2 inches and the focal length is 4 inches, then the f-number is 4 ÷ 2 = 2 (written as f/2.).
The splendor of using a ratio is, any lens operating at the same f-number as another lens, yields the same image brightness regardless of the dimensions (diameter or focal length), for an identical scene. It’s complicated; but the f-number system actually takes away the chaos.
add a comment |
up vote
0
down vote
Because the lens operates much like a funnel in that it gathers light, the greater the working diameter of the lens, the brighter the projected image. How bright this image will be is dependent on the brightness of the scene and the magnification realized by the lens. The longer the focal length, the more the lens magnifies. The deed of magnifying to produce an image, takes its toll on image brightness. In other words, the longer the focal length, the more the lens magnifies. This higher magnification result is a larger but dimmer image of objects.
Another way to say this, image brightness intertwines the working diameter and the focal length. Because these two factors are so interwoven, gauging image brightness is demanding. We are forced to fall back on a mathematical ratio that will take the chaos out of figuring out image brightness. This is true because a ratio is dimensionless. If I tell you the ratio of boys to girls in a 6th grade class is 3 boys for every 4 girls, I have given you a ratio that works independent of the number of students. For example if the class consist of 28 kids, then 12 boys to 16 girls is the breakdown (ratio is dimensionless).
For the camera lens: if the working diameter is 4 inches and the focal length is 4 inches, then the focal ratio (f-number) = 4 ÷ 4 = 1 (written as f/1 (f/1 is produces a very bright image). If the working diameter is 2 inches and the focal length is 4 inches, then the f-number is 4 ÷ 2 = 2 (written as f/2.).
The splendor of using a ratio is, any lens operating at the same f-number as another lens, yields the same image brightness regardless of the dimensions (diameter or focal length), for an identical scene. It’s complicated; but the f-number system actually takes away the chaos.
add a comment |
up vote
0
down vote
up vote
0
down vote
Because the lens operates much like a funnel in that it gathers light, the greater the working diameter of the lens, the brighter the projected image. How bright this image will be is dependent on the brightness of the scene and the magnification realized by the lens. The longer the focal length, the more the lens magnifies. The deed of magnifying to produce an image, takes its toll on image brightness. In other words, the longer the focal length, the more the lens magnifies. This higher magnification result is a larger but dimmer image of objects.
Another way to say this, image brightness intertwines the working diameter and the focal length. Because these two factors are so interwoven, gauging image brightness is demanding. We are forced to fall back on a mathematical ratio that will take the chaos out of figuring out image brightness. This is true because a ratio is dimensionless. If I tell you the ratio of boys to girls in a 6th grade class is 3 boys for every 4 girls, I have given you a ratio that works independent of the number of students. For example if the class consist of 28 kids, then 12 boys to 16 girls is the breakdown (ratio is dimensionless).
For the camera lens: if the working diameter is 4 inches and the focal length is 4 inches, then the focal ratio (f-number) = 4 ÷ 4 = 1 (written as f/1 (f/1 is produces a very bright image). If the working diameter is 2 inches and the focal length is 4 inches, then the f-number is 4 ÷ 2 = 2 (written as f/2.).
The splendor of using a ratio is, any lens operating at the same f-number as another lens, yields the same image brightness regardless of the dimensions (diameter or focal length), for an identical scene. It’s complicated; but the f-number system actually takes away the chaos.
Because the lens operates much like a funnel in that it gathers light, the greater the working diameter of the lens, the brighter the projected image. How bright this image will be is dependent on the brightness of the scene and the magnification realized by the lens. The longer the focal length, the more the lens magnifies. The deed of magnifying to produce an image, takes its toll on image brightness. In other words, the longer the focal length, the more the lens magnifies. This higher magnification result is a larger but dimmer image of objects.
Another way to say this, image brightness intertwines the working diameter and the focal length. Because these two factors are so interwoven, gauging image brightness is demanding. We are forced to fall back on a mathematical ratio that will take the chaos out of figuring out image brightness. This is true because a ratio is dimensionless. If I tell you the ratio of boys to girls in a 6th grade class is 3 boys for every 4 girls, I have given you a ratio that works independent of the number of students. For example if the class consist of 28 kids, then 12 boys to 16 girls is the breakdown (ratio is dimensionless).
For the camera lens: if the working diameter is 4 inches and the focal length is 4 inches, then the focal ratio (f-number) = 4 ÷ 4 = 1 (written as f/1 (f/1 is produces a very bright image). If the working diameter is 2 inches and the focal length is 4 inches, then the f-number is 4 ÷ 2 = 2 (written as f/2.).
The splendor of using a ratio is, any lens operating at the same f-number as another lens, yields the same image brightness regardless of the dimensions (diameter or focal length), for an identical scene. It’s complicated; but the f-number system actually takes away the chaos.
answered 4 hours ago
Alan Marcus
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The "why" is that f/stop number is focal length / aperture diameter.
So a 100 mm lens at f/4 has an effective aperture of 25 mm. This is not exactly the physical diameter opening, but is the effective diameter, specifically the front entrance pupil, as seen by the magnification of the front lens elements.
Because of this definition, as aperture diameter becomes larger (and exposure increases), f/stop number becomes smaller. f/2.8 is wide, and f/22 is narrow.
It may seem backwards at first, but we get over it easily, and the huge advantage of the f/stop number system is that it applies for any lens, of any size and focal length and aperture, so then we all know what we mean when we say f/8. f/8 is f/8 exposure on any lens. This is a plus. It wasn't that way in the early beginning. And of course, light meters meter with the same meanings.
My site has more about this at https://www.scantips.com/lights/fstop.html
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The "why" is that f/stop number is focal length / aperture diameter.
So a 100 mm lens at f/4 has an effective aperture of 25 mm. This is not exactly the physical diameter opening, but is the effective diameter, specifically the front entrance pupil, as seen by the magnification of the front lens elements.
Because of this definition, as aperture diameter becomes larger (and exposure increases), f/stop number becomes smaller. f/2.8 is wide, and f/22 is narrow.
It may seem backwards at first, but we get over it easily, and the huge advantage of the f/stop number system is that it applies for any lens, of any size and focal length and aperture, so then we all know what we mean when we say f/8. f/8 is f/8 exposure on any lens. This is a plus. It wasn't that way in the early beginning. And of course, light meters meter with the same meanings.
My site has more about this at https://www.scantips.com/lights/fstop.html
add a comment |
up vote
0
down vote
up vote
0
down vote
The "why" is that f/stop number is focal length / aperture diameter.
So a 100 mm lens at f/4 has an effective aperture of 25 mm. This is not exactly the physical diameter opening, but is the effective diameter, specifically the front entrance pupil, as seen by the magnification of the front lens elements.
Because of this definition, as aperture diameter becomes larger (and exposure increases), f/stop number becomes smaller. f/2.8 is wide, and f/22 is narrow.
It may seem backwards at first, but we get over it easily, and the huge advantage of the f/stop number system is that it applies for any lens, of any size and focal length and aperture, so then we all know what we mean when we say f/8. f/8 is f/8 exposure on any lens. This is a plus. It wasn't that way in the early beginning. And of course, light meters meter with the same meanings.
My site has more about this at https://www.scantips.com/lights/fstop.html
The "why" is that f/stop number is focal length / aperture diameter.
So a 100 mm lens at f/4 has an effective aperture of 25 mm. This is not exactly the physical diameter opening, but is the effective diameter, specifically the front entrance pupil, as seen by the magnification of the front lens elements.
Because of this definition, as aperture diameter becomes larger (and exposure increases), f/stop number becomes smaller. f/2.8 is wide, and f/22 is narrow.
It may seem backwards at first, but we get over it easily, and the huge advantage of the f/stop number system is that it applies for any lens, of any size and focal length and aperture, so then we all know what we mean when we say f/8. f/8 is f/8 exposure on any lens. This is a plus. It wasn't that way in the early beginning. And of course, light meters meter with the same meanings.
My site has more about this at https://www.scantips.com/lights/fstop.html
edited 2 hours ago
answered 3 hours ago
WayneF
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Also see photo.stackexchange.com/questions/12072/…
– Tetsujin
7 hours ago
There may be limits on how large an aperture can be, but practical considerations make lenses faster than f/1 very, very, very rare. The fastest I have ever heard of is f/0.7.
– Jim MacKenzie
6 hours ago
There are many measuring systems where a larger numerical value means a smaller physical size.
– whatsisname
34 mins ago