F-stops Explained

[thunderkyss]

If I set the 105mm lens' aperture to, let’s say 10mm diameter, I would have an f-number of roughly f10.5. But if I set the 28mm to the same 10mm diameter, it would be set to f2.8! To get f10.5 on a 28mm lens, the aperture would need a diameter of 2.67 mm.

So even though the areas created by the aperture blades in lenses of different focal lengths are different…… optically they create the same f-number…. And ultimately the same exposure! That’s why f-numbers are used instead of the areas created by the aperture blades…….. it makes it just that much easier for us to work with. Otherwise, converting aperture areas from one lens to create an identical exposure in another lens would REALLY give you a headache!

So what the manufacturers do when they put seemingly archaic numbers like 2.8, 5.6 and 11 on the lenses is really just taking the math out of the equation for us! So f8 on one lens gives us the same exposure aperture on any other lens! Whether it's a $50 point-and-shoot, a vintage 8x10 view camera or a 5-digit Hassy dream setup.... f8 is f8.

So a diameter of 2.67mm on a 28 millimeter lens lets in just as much light as a 10mm diameter on a 105mm lens?

How is that possible?

 

[IByte]

Sparks I love ya man lol, but I'm going to come back to this after a few pints.

 

[cwcaesar]

The very definition of f-stop is focal length divided by diameter, or F/D.

Very well explained OP! :hail:

 

[Helen B]

If I set the 105mm lens' aperture to, let’s say 10mm diameter, I would have an f-number of roughly f10.5. But if I set the 28mm to the same 10mm diameter, it would be set to f2.8! To get f10.5 on a 28mm lens, the aperture would need a diameter of 2.67 mm.

So even though the areas created by the aperture blades in lenses of different focal lengths are different…… optically they create the same f-number…. And ultimately the same exposure! That’s why f-numbers are used instead of the areas created by the aperture blades…….. it makes it just that much easier for us to work with. Otherwise, converting aperture areas from one lens to create an identical exposure in another lens would REALLY give you a headache!

So what the manufacturers do when they put seemingly archaic numbers like 2.8, 5.6 and 11 on the lenses is really just taking the math out of the equation for us! So f8 on one lens gives us the same exposure aperture on any other lens! Whether it's a $50 point-and-shoot, a vintage 8x10 view camera or a 5-digit Hassy dream setup.... f8 is f8.

So a diameter of 2.67mm on a 28 millimeter lens lets in just as much light as a 10mm diameter on a 105mm lens?

How is that possible?

Obviously it lets in less light. The image of an object, however, is smaller and that exactly compensates for the lesser amount of light - less light spread over a smaller area, to give the same light/area (in simple terms).

 

[thunderkyss]

You have succeeded in in making simple things way more complicated than they have to be...

Each stop is a function of the square root of 2... the square root of 2 is roughly 1.4... √2*2=2, √2*3=4.2 (rounded to 4 on lenses), √2*4=5.6 (rounded), √2*5=7.07 (rounded to 8 on lenses), etc...

The reason why is because that is how you figure out the area of a circle - we're not interested in the diameter of the aperture, just the area.

If the square root of 2 is roughly 1.4, how is √2*2=2; wouldn't it be 2.8?

Why do we round 4.2 to 4, but don't round 2.8 to 3?

How is f4 on 50mm lens equivalent to f4 on a 300mm lens? Or why?

Your explanation makes it easy to figure out the number, but doesn't explain how you get that number or what that number represents. It's nice to know that the ratio of the area of the aperture opening to the focal length of the lens produces a quotient that is common between lenses & that the f-stop is that quotient.

 

[pixmedic]

You have succeeded in in making simple things way more complicated than they have to be...

Each stop is a function of the square root of 2... the square root of 2 is roughly 1.4... √2*2=2, √2*3=4.2 (rounded to 4 on lenses), √2*4=5.6 (rounded), √2*5=7.07 (rounded to 8 on lenses), etc...

The reason why is because that is how you figure out the area of a circle - we're not interested in the diameter of the aperture, just the area.

If the square root of 2 is roughly 1.4, how is √2*2=2; wouldn't it be 2.8?

Why do we round 4.2 to 4, but don't round 2.8 to 3?

How is f4 on 50mm lens equivalent to f4 on a 300mm lens? Or why?

Your explanation makes it easy to figure out the number, but doesn't explain how you get that number or what that number represents. It's nice to know that the ratio of the area of the aperture opening to the focal length of the lens produces a quotient that is common between lenses & that the f-stop is that quotient.

its magic. just turn the dial and let the magic work.

 

[thunderkyss]

Each stop is a function of the square root of 2... the square root of 2 is roughly 1.4... √2*2=2, √2*3=4.2 (rounded to 4 on lenses), √2*4=5.6 (rounded), √2*5=7.07 (rounded to 8 on lenses), etc...

If the square root of 2 is roughly 1.4, how is √2*2=2; wouldn't it be 2.8?

its magic. just turn the dial and let the magic work.

No offense, I know you're shooting for levity, but this is the very definition of "dumbing down"

The OP obviously wanted to go beyond the basic, "A full stop up is twice as much light & a full stop down is half as much." & while it is true, that is all that a beginner "needs" to know, there are some people who would like to know why a full stop up lets in twice as much light. The simple answer of doubling the area doesn't fully explain as it is the ratio of the focal length to the aperture diameter.

& I appreciate it.

 

[thunderkyss]

So a diameter of 2.67mm on a 28 millimeter lens lets in just as much light as a 10mm diameter on a 105mm lens?

How is that possible?

Obviously it lets in less light. The image of an object, however, is smaller and that exactly compensates for the lesser amount of light - less light spread over a smaller area, to give the same light/area (in simple terms).

Thank you. I think the OP would greatly benefit if that info was included.

 

[JayJeep]

You have succeeded in in making simple things way more complicated than they have to be...

Each stop is a function of the square root of 2... the square root of 2 is roughly 1.4... √2*2=2, √2*3=4.2 (rounded to 4 on lenses), √2*4=5.6 (rounded), √2*5=7.07 (rounded to 8 on lenses), etc...

The reason why is because that is how you figure out the area of a circle - we're not interested in the diameter of the aperture, just the area.

√2*2=2 is 2.8 and why would 7.07 get rounded to 8. and √2*6=8.48 do we round that to 8 or 9 or 11.

You have it wrong I think. It should be the √2^2=2, √2^3=2.8 and √2^6=8. You take it to the next power not multiply it by the next integer.

 

[Helen B]

You have succeeded in in making simple things way more complicated than they have to be...

Each stop is a function of the square root of 2... the square root of 2 is roughly 1.4... √2*2=2, √2*3=4.2 (rounded to 4 on lenses), √2*4=5.6 (rounded), √2*5=7.07 (rounded to 8 on lenses), etc...

The reason why is because that is how you figure out the area of a circle - we're not interested in the diameter of the aperture, just the area.

If the square root of 2 is roughly 1.4, how is √2*2=2; wouldn't it be 2.8?

Why do we round 4.2 to 4, but don't round 2.8 to 3?

How is f4 on 50mm lens equivalent to f4 on a 300mm lens? Or why?

Your explanation makes it easy to figure out the number, but doesn't explain how you get that number or what that number represents. It's nice to know that the ratio of the area of the aperture opening to the focal length of the lens produces a quotient that is common between lenses & that the f-stop is that quotient.

They are mostly rounded to two significant figures. The 0.5 - 1 - 2 - 4 - 8 series do not need to be rounded because they are exact numbers, 0.5 being the lowest f-number theoretically possible. The others usually get rounded to a nominal f-number with no more than two significant figures, such as 1.4, 2.8. 5.6, 11 etc.

The sequence is (√2)^x, where x is a positive or negative integer equal or greater than -2, or zero.

F/4 on a 50 mm lens is equivalent to f/4 on a 300 mm lens in terms of image brightness. As I mentioned earlier, the 300 mm lets more light in from an object, but it forms a larger image, so the brightness is the same. How much detail do you want on this? How good are you at maths?

 

[ClickAddict]

Both Sparky and O|||||||O have the same math. Just one is longer form.
Sparky's math Breaking it down:

focal lens = m
diameter = d
aperture = f
He states m/d = f........................... (His 200mm lense to 25 diameter = f8 example)
which means m=df ...........................(multiply both sides by d)
and d=m/f .....................................(divide both sides by f)

opening area = A
A = pr^2 .............................(This is a known calculation for area of circle)
p=3.14 ..................................(rounded)
r = radius

r= 1/2 d .....................................(By definition radius = 1/2 diameter)

r= 1/2X(m/f)................................ (replacing d with m/f as per above)
r= m/2f........................................ (result of 1/2 X m/f)

A1 ...................area of lens at a given aperture
A2 .................area of same lense at a different aperture
r1............... radius of lense to get Area = A1
r2.................... radius of lense to get Area = A2

if A1 = 1/2 of A2 then ............................(looking for next stop which is half the area)

A1 = A2/2
pr1^2=(pr2^2)/2................................... (A1 = pr1^2 as per area formula, same for A2))
p(m1/2f1)^2 = p(m2/2f2)^2/2 ............................(replacing r with m/2f)
(m1/2f1)^2 = (m2/2f2)^2/2 ...........................divide both sides by p
2(m1/2f1)^2 = (m2/2f2)^2 ........................multiply both sides by2
2 = ((m2/2f2)^2)/(m1/2f1)^2...................... divedi both sides by
2 = (m2^2/(2f2)^2)/(m1^2)/(2f1)^2 ......................(square the fractions)
2= (m2^2)(2f1)^2)/((2f2)^2)(m1^2) ......................(a/b)/(c/d) = a/b X d/c = ad/bc
2= (m1^2)(2f1)^2)/((2f2)^2)(m1^2) .....................since were keeping m the same and only changing f, m1=m2)
2=(2f1)^2)/((2f2)^2) ...................................(xy/zx = y/x)
2= 4f1^2 / 4f2^2 ........................................(ab)^2 = (a^2)(b^2)
2=f1^2 / f2^2................................ (4A/4B = A/B)
sqrt(2) = f1/f2 ..................................sqrt = square root)
1.4 = f1/f2

so to get the half amount of light as per sparkys math AND O||||||O you need to have that ration

Essentially many of the variables eliminate themselves to get to the 1.4 value

 

[bhop]

 

[JayJeep]

Both Sparky and O|||||||O have the same math. Just one is longer form.
Sparky's math Breaking it down:

focal lens = m
diameter = d
aperture = f
He states m/d = f........................... (His 200mm lense to 25 diameter = f8 example)
which means m=df ...........................(multiply both sides by d)
and d=m/f .....................................(divide both sides by f)

opening area = A
A = pr^2 .............................(This is a known calculation for area of circle)
p=3.14 ..................................(rounded)
r = radius

r= 1/2 d .....................................(By definition radius = 1/2 diameter)

r= 1/2X(m/f)................................ (replacing d with m/f as per above)
r= m/2f........................................ (result of 1/2 X m/f)

A1 ...................area of lens at a given aperture
A2 .................area of same lense at a different aperture
r1............... radius of lense to get Area = A1
r2.................... radius of lense to get Area = A2

if A1 = 1/2 of A2 then ............................(looking for next stop which is half the area)

A1 = A2/2
pr1^2=(pr2^2)/2................................... (A1 = pr1^2 as per area formula, same for A2))
p(m1/2f1)^2 = p(m2/2f2)^2/2 ............................(replacing r with m/2f)
(m1/2f1)^2 = (m2/2f2)^2/2 ...........................divide both sides by p
2(m1/2f1)^2 = (m2/2f2)^2 ........................multiply both sides by2
2 = ((m2/2f2)^2)/(m1/2f1)^2...................... divedi both sides by
2 = (m2^2/(2f2)^2)/(m1^2)/(2f1)^2 ......................(square the fractions)
2= (m2^2)(2f1)^2)/((2f2)^2)(m1^2) ......................(a/b)/(c/d) = a/b X d/c = ad/bc
2= (m1^2)(2f1)^2)/((2f2)^2)(m1^2) .....................since were keeping m the same and only changing f, m1=m2)
2=(2f1)^2)/((2f2)^2) ...................................(xy/zx = y/x)
2= 4f1^2 / 4f2^2 ........................................(ab)^2 = (a^2)(b^2)
2=f1^2 / f2^2................................ (4A/4B = A/B)
sqrt(2) = f1/f2 ..................................sqrt = square root)
1.4 = f1/f2

so to get the half amount of light as per sparkys math AND O||||||O you need to have that ration

Essentially many of the variables eliminate themselves to get to the 1.4 value

Except O||||||O is wrong. He says to multiply 1.4*(integer) to get 2, 2.8, 4, 5.6, 8, 11. You don't multiply you take it to the power of the integers.

sqrt(2)^0=1, sqrt(2)^1=1.41, sqrt(2)^2=2, sqrt(2)^3=2.82, sprt(2)^4=4

If I used his math of rounding 7.07 to 8 and 8.48 to 11 in my line of work I would most likely be fired and endangering lives.

The problem I see is that he insisted that he was right and the OP was wrong to post a complicated explanation for something I know has been asked.

"where do they get f/5.6 as double the light of f/8" and "where do those numbers come from"

 

[jwbryson1]

How so? That is the foundation of the whole thing.

I really don't think you're qualified to speak for everyone here on TPF.

If anybody on TPF disputes that the square root of 2 is 1.414213562..., speak up now.

Also, if anybody on TPF disputes that the area of a circle = pi*r^2, speak up now.

I won't dispute those but I will dispute the following:

Each stop is a function of the square root of 2... the square root of 2 is roughly 1.4... √2*2=2, √2*3=4.2 (rounded to 4 on lenses), √2*4=5.6 (rounded), √2*5=7.07 (rounded to 8 on lenses), etc...

 

[ratssass]

i like turtles