Is there a theoretical limit to aperture?

For a lens in air the theoretical limit for the aperture is f/0.5. The fastest lens built that you can read about in the normal open literature seems to be a Zeiss lens with a maximum aperture of f/0.7.

Joe

Knoxis said:

...there is technically no theoretical limit to how big aperture can get. You could get an F0.01 lens if the money and resources were available.

Click to expand...

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Jack

Jack Hogan said:

Knoxis said:

...there is technically no theoretical limit to how big aperture can get. You could get an F0.01 lens if the money and resources were available.

Click to expand...

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

This does not take the index of refraction into account and does not explain why we cannot have a lens faster than 1/2 in f/D sense. Like a 50mm (single element) lens with 1m diameter. I believe the answer is that such a lens would not be able to focus rays in an acceptable way but I have not seen a good exposition. For a single lens element that should be doable but I am not sure about a multiple element one.

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

It appears that lens-system F-Ratio is only approximately equal to 1 / ( ( 2 ) * ( Numerical Aperture) ).

When the using the stated (as opposed to the approximate) identity for F-Ratio relative to (image-side) Numerical Aperture (for a single symmetrical lens), I recently came up with something rather different.

.

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

NA = sin (theta) = sin (arctangent ( D / ( 2*L ) ) )

where:

D is the (virtual) circular aperture-opening diameter;

L is the (actual) focal-length;

F is the lens-system F-Ratio value.

.

I rearranged to solve for F-Ratio (F) as a function of Numerical Aperture (NA), and calculated the % error (of the derived F-Ratio value, relative to the derived approximated value)..

Using the inverse trigonometric identity for arcsin(x) (appearing first below):


Source: https://en.wikipedia.org/wiki/Inver...ips_among_the_inverse_trigonometric_functions

F = 1 / ( 2 * Tangent ( 2 * Arctangent ( NA / ( 1 + ( 1 - NA^2 )^(1/2) ) ) )

where:

F is F-Ratio;

NA is (image-space) Numerical Aperture.

The % error (at any particular F-Ratio value, relative to the value derived from the approximation) is a negative valued error (-2.021% at F=2.450, -1.000% at F=3.54439, and -0.501% at F=4.975).

A simple formula for (a close approximation of) that (F-Ratio/Approx.) error [when F>2.0] is:

E[%] ~ ( -1 ) * ( 3.54439 / F )^(2) ... [in units of percentage change, %]

where:

E[%] is the (negative signed) percentage error (of the F-Ratio divided by approximated F-Ratio);

F is the lens-system F-Ratio.

.

For cases where a thin-lens-system is not focused at infinity, the (Effective) F-Ratio is:

F = 1 / ( 2 * Tangent ( 2 * Arctangent ( NA / ( 1 + ( 1 - NA^2 )^(1/2) ) ) ) * (1 - M/P)

where:

F is F-Ratio;

NA is (image-space) Numerical Aperture;

M is Image Magnification [ sensor-size / object-size ];

P is Pupil Magnification [exit-pupil / entrance-pupil ].

.

An interesting result of using the (exact) identity for converting (image-side) NA to F-Ratio:

(Using the approximate NA to F conversion formula), the maximum possible Numerical Aperture value of unity limits (such a thin-lens system) F-Ratio to (appear to be limited to) a minimum value of F=0.5.

Using the precise formula listed above, the minimum value of F-Ratio is (actually) lower (F=0.1767767).

Notes: The result directly above is calculated for the case of a thin-lens focused at infinity. This result (of course) says nothing in particular (whatsoever) about the realizability of manufacturing such a lens design.

.

From: http://www.dpreview.com/forums/post/57028944

DM

J A C S said:

Jack Hogan said:

Knoxis said:

...there is technically no theoretical limit to how big aperture can get. You could get an F0.01 lens if the money and resources were available.

Click to expand...

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

This does not take the index of refraction into account and does not explain why we cannot have a lens faster than 1/2 in f/D sense. Like a 50mm (single element) lens with 1m diameter. I believe the answer is that such a lens would not be able to focus rays in an acceptable way but I have not seen a good exposition. For a single lens element that should be doable but I am not sure about a multiple element one.

Click to expand...

Hi, Jack, Nakamura correctly writes down the formula for F#, but as JACS points out doesn't say where it comes from. It comes from a well corrected lens satisfying the Abbe sine condition, thereby making the principal surface of the lens curved (spherical).

Most of the time, I see lenses being drawn with flat principal surfaces, but as you go faster, the curvature of the principal surface begins to show up (brian used to write about this back in the day). So you can easily draw the situation as follows: draw a horizontal line as your optical axis, and mark the image focus point on it on the right. Then take a compass, set it to your focal length f, and centered on the focus point, sweep out an arc above and below the axis. Now, draw a line parallel to the axis and a distance r away from the axis. r is = d/2, the radius of your entrance pupil. Now, mark the point where this line intersects the arc. So, the arc from the axis to this point is half of the principal surface. From this point, draw a ray to the focus point, which is the marginal ray and defines theta`, it is also of length f because it is on the circle. From this layout you can immediately see that r/f = sin(theta`).

Now, how big can this curve get? Well, if you keep increasing r, you see that the maximum size it can go is a forward facing hemisphere with r = f, and so f/d, (which still holds by the way) has its minimum value of F# = f/d = f/(2f) = 0.5!

I wrote about this more here (with a reference on where you can see a curved principal surface):


Chris

cpw said:

J A C S said:

Jack Hogan said:

Knoxis said:

...there is technically no theoretical limit to how big aperture can get. You could get an F0.01 lens if the money and resources were available.

Click to expand...

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

This does not take the index of refraction into account and does not explain why we cannot have a lens faster than 1/2 in f/D sense. Like a 50mm (single element) lens with 1m diameter. I believe the answer is that such a lens would not be able to focus rays in an acceptable way but I have not seen a good exposition. For a single lens element that should be doable but I am not sure about a multiple element one.

Click to expand...

Hi, Jack, Nakamura correctly writes down the formula for F#, but as JACS points out doesn't say where it comes from. It comes from a well corrected lens satisfying the Abbe sine condition, thereby making the principal surface of the lens curved (spherical).

Most of the time, I see lenses being drawn with flat principal surfaces, but as you go faster, the curvature of the principal surface begins to show up (brian used to write about this back in the day). So you can easily draw the situation as follows: draw a horizontal line as your optical axis, and mark the image focus point on it on the right. Then take a compass, set it to your focal length f, and centered on the focus point, sweep out an arc above and below the axis. Now, draw a line parallel to the axis and a distance r away from the axis. r is = d/2, the radius of your entrance pupil. Now, mark the point where this line intersects the arc. So, the arc from the axis to this point is half of the principal surface. From this point, draw a ray to the focus point, which is the marginal ray and defines theta`, it is also of length f because it is on the circle. From this layout you can immediately see that r/f = sin(theta`).

Now, how big can this curve get? Well, if you keep increasing r, you see that the maximum size it can go is a forward facing hemisphere with r = f, and so f/d, (which still holds by the way) has its minimum value of F# = f/d = f/(2f) = 0.5!

I wrote about this more here (with a reference on where you can see a curved principal surface):

http://www.dpreview.com/forums/post/39973191

Click to expand...

Good explanation Chris. And yes, I (and I assume Nakamura) was referring to lenses designed for reasonable image quality in typical photographic applications. So n is assumed to be air.

Jack

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

It appears that lens-system F-Ratio is only approximately equal to 1 / ( ( 2 ) * ( Numerical Aperture) )...

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

Click to expand...

Hi DM,

I see, the approximation of an approximation ;-)

I never liked that article, it's unclear and it starts with a circuitous fallacy: it defines N as f/D, which we know is only an approximation. If they defined N correctly, they would get N = 1/(2NA) exactly.

Jack

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

It appears that lens-system F-Ratio is only approximately equal to 1 / ( ( 2 ) * ( Numerical Aperture) )...

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

Click to expand...

Hi DM,

I see, the approximation of an approximation ;-)

I never liked that article, it's unclear and it starts with a circuitous fallacy: it defines N as f/D, which we know is only an approximation. If they defined N correctly, they would get N = 1/(2NA) exactly.

Click to expand...

Interesting. Thank you for your thoughts presented. So, (image-side) Numerical Aperture equals what ?


Source: https://upload.wikimedia.org/wikipe...g/568px-Numerical_aperture_for_a_lens.svg.png

What is the full and correct closed-form identity (or other math) that defines Numerical Aperture precisely ?

.

What is specifically wrong with the statement on the Wiki page that you seem to be saying is incorrect (?):

... the angular aperture of a lens (or an imaging mirror) is expressed by the f-number, written f/# or N, which is defined as the ratio of the focal length f to the diameter of the entrance pupil D ...

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

A note on this subject a long while ago (if memory serves me) by Bob indicated that the issue is with the index of refraction of modern glass. He intimated that new novel materials might overcome the problem.

--
Charles Darwin: "ignorance more frequently begets confidence than does knowledge."
tony
http://www.tphoto.ca

A delightful "world's fastest lens" made by Zeiss as a joke for fun. And another version.

--
Charles Darwin: "ignorance more frequently begets confidence than does knowledge."
tony
http://www.tphoto.ca

Detail Man said:

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

It appears that lens-system F-Ratio is only approximately equal to 1 / ( ( 2 ) * ( Numerical Aperture) )...

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

Click to expand...

Hi DM,

I see, the approximation of an approximation ;-)

I never liked that article, it's unclear and it starts with a circuitous fallacy: it defines N as f/D, which we know is only an approximation. If they defined N correctly, they would get N = 1/(2NA) exactly.

Click to expand...

Interesting. Thank you for your thoughts presented. So, (image-side) Numerical Aperture equals what ?

Source: https://upload.wikimedia.org/wikipe...g/568px-Numerical_aperture_for_a_lens.svg.png

What is the full and correct closed-form identity (or other math) that defines Numerical Aperture precisely ?

.

What is specifically wrong with the statement on the Wiki page that you seem to be saying is incorrect (?):

... the angular aperture of a lens (or an imaging mirror) is expressed by the f-number, written f/# or N, which is defined as the ratio of the focal length f to the diameter of the entrance pupil D ...

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

Click to expand...

It's not my expertise but I understand that in optics the opening angle theta' (theta in the backwards ideal thin lens picture above) shows up in many complex formulas. N has been defined as 1/[2sin(theta')]* because it simplifies notation in a number of practical applications, likewise NA=sin(theta')*. If one wants to convert some advanced optics theory into language familiar to photographers it is useful to know the original definition.

For instance an optical scientist might say that the aperture diameter above is

D = r * 2sin(theta')

However most photographers normally do not have at their disposal r or theta'. Fortunately we can translate the optical formula above into photographese by expressing it in terms of N, exactly:

D = r / N

When theta' is small r is approximately equal to f, so D is approximately equal to f / N. Photographers understand and usually know their f's and N's - and theta's are usually quite small - so all is good.

Many formulas need translating like that to be understood by photographers, starting from the one for exposure.

Jack

* Yes, in air

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

It appears that lens-system F-Ratio is only approximately equal to 1 / ( ( 2 ) * ( Numerical Aperture) )...

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

Click to expand...

Hi DM,

I see, the approximation of an approximation ;-)

I never liked that article, it's unclear and it starts with a circuitous fallacy: it defines N as f/D, which we know is only an approximation. If they defined N correctly, they would get N = 1/(2NA) exactly.

Click to expand...

Interesting. Thank you for your thoughts presented. So, (image-side) Numerical Aperture equals what ?

Source: https://upload.wikimedia.org/wikipe...g/568px-Numerical_aperture_for_a_lens.svg.png

What is the full and correct closed-form identity (or other math) that defines Numerical Aperture precisely ?

.

What is specifically wrong with the statement on the Wiki page that you seem to be saying is incorrect (?):

... the angular aperture of a lens (or an imaging mirror) is expressed by the f-number, written f/# or N, which is defined as the ratio of the focal length f to the diameter of the entrance pupil D ...

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

Click to expand...

It's not my expertise but I understand that in optics the opening angle theta' (theta in the backwards ideal thin lens picture above) shows up in many complex formulas. N has been defined as 1/[2sin(theta')]* because it simplifies notation in a number of practical applications, likewise NA=sin(theta')*.

* Yes, in air

Click to expand...

Are you saying that N (as stated by the mathematical identity given in your quoted text directly above) is - or is not - in that form fully and accurately defined ? Not sure how to interpret your statements meaning.

Jack Hogan said:

If one wants to convert some advanced optics theory into language familiar to photographers it is useful to know the original definition.

Click to expand...

Right. That it was we are trying get towards as we speak now.

Jack Hogan said:

For instance an optical scientist might say that the aperture diameter above is

D = r * 2sin(theta')

Click to expand...

Are they correct ?

Jack Hogan said:

However most photographers normally do not have at their disposal r or theta'.

Click to expand...

Not sure what that has to do with our getting to the bottom of the (non-approximated) relationships.

Jack Hogan said:

Fortunately we can translate the optical formula above into photographese by expressing it in terms of N, exactly:

D = r / N

Click to expand...

"Fortunately" implies that an approximation is (reasonably) able to be taken in your above statement ?

Jack Hogan said:

When theta' is small r is approximately equal to f, so D is approximately equal to f / N. Photographers understand and usually know their f's and N's - and theta's are usually quite small - so all is good.

Many formulas need translating like that to be understood by photographers, starting from the one for exposure.

Click to expand...

Yet, but without (precise, even if laborious) "first translations" from which to justifiably base approximations being made thereof we have no sense of errors between actual and approximated. Let us be "laborious".

Detail Man said:

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

Click to expand...

Click to expand...

Click to expand...

It's not my expertise but I understand that in optics the opening angle theta' (theta in the backwards ideal thin lens picture above) shows up in many complex formulas. N has been defined as 1/[2sin(theta')]* because it simplifies notation in a number of practical applications, likewise NA=sin(theta')*.

* Yes, in air

Click to expand...

Are you saying that N (as stated by the mathematical identity given in your quoted text directly above) is ... in that form fully and accurately defined ? Not sure how to interpret your statements meaning.

Click to expand...

Yes

Detail Man said:

Detail Man said:

For instance an optical scientist might say that the aperture diameter above is

D = r * 2sin(theta')

Click to expand...

Are they correct ?

Click to expand...

If geometry is not an opinion ...

Detail Man said:

Detail Man said:

Fortunately we can translate the optical formula above into photographese by expressing it in terms of N, exactly:

D = r / N

Click to expand...

"Fortunately" implies that an approximation is (reasonably) able to be taken in your above statement ?

Click to expand...

No approximation there, that's an exact substitution. N and NA are defined precisely that way on purpose to allow for simplification of use in the field.

Jack

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

Click to expand...

Click to expand...

Click to expand...

It's not my expertise but I understand that in optics the opening angle theta' (theta in the backwards ideal thin lens picture above) shows up in many complex formulas. N has been defined as 1/[2sin(theta')]* because it simplifies notation in a number of practical applications, likewise NA=sin(theta')*.

* Yes, in air

Click to expand...

Are you saying that N (as stated by the mathematical identity given in your quoted text directly above) is ... in that form fully and accurately defined ? Not sure how to interpret your statements meaning.

Click to expand...

Yes

Jack Hogan said:

Jack Hogan said:

For instance an optical scientist might say that the aperture diameter above is

D = r * 2sin(theta')

Click to expand...

Are they correct ?

Click to expand...

If geometry is not an opinion ...

Jack Hogan said:

Jack Hogan said:

Fortunately we can translate the optical formula above into photographese by expressing it in terms of N, exactly:

D = r / N

Click to expand...

"Fortunately" implies that an approximation is (reasonably) able to be taken in your above statement ?

Click to expand...

No approximation there, that's an exact substitution. N and NA are defined precisely that way on purpose to allow for simplification of use in the field.

Click to expand...

Have for some reason not followed your of thinking along the way considering the sum of your statements

From another route (so to speak), have found the following (unrelated to utilization of Numerical Aperture):

Jack Hogan said:

I found this derivation (unrelated to the utilization of "Numerical Aperture"):

Click to expand...

Source: "The Manual of Photography, Ninth Edition, Page 64


Source: "The Manual of Photography, Ninth Edition, Page 65

"E" (as it is defined on Page 64) is restated (within the right-hand column of the displayed text on Page 65)

.

DM

Detail Man said:

Detail Man said:

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

Click to expand...

Click to expand...

I found this derivation (unrelated to the utilization of "Numerical Aperture"):


Source: "The Manual of Photography, Ninth Edition, Page 65

"E" (as it is defined on Page 64) is restated (within the right-hand column of the displayed text on Page 65)

Click to expand...

Right, this manual being for Photography it has already being 'translated'. Here is "E" from the Optical Engineering text mentioned by cpw earlier:


From page 241 of Modern Optical Engineering by Warren J. Smith

Feast your eyes on equations (6) and (13) of your reference.

Jack

Jack Hogan said:

Detail Man said:

Jack Hogan said:

Jack Hogan said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

Click to expand...

Click to expand...

I found this derivation (unrelated to the utilization of "Numerical Aperture"):


Source: "The Manual of Photography, Ninth Edition, Page 65

"E" (as it is defined on Page 64) is restated (within the right-hand column of the displayed text on Page 65)

Click to expand...

Right, this manual being for Photography it has already being 'translated'. Here is "E" from the Optical Engineering text mentioned by cpw earlier:


From page 241 of Modern Optical Engineering by Warren J. Smith

Feast your eyes on equations (6) and (13) of your reference.

Click to expand...

Some additional related thoughts that I have had between our postings and replies:

For a single, symmetrical flat-lens (where Image and Pupil Magnifications are unity), it seems that "Effective F-Ratio" (as here discussed) is a quantity that exists which is distinct from "Numerical Aperture". However, it appears that the values (of F-Ratio, and approximate F-Ratio extrapolated from approximate "Numerical Aperture") approach the same numerical value as the value of F-Ratio increases.

The "approximation" identified below refers to an approximated value of F-Ratio derived utilizing the approximations sometimes made when equating F-Ratio and (image-space) "Numerical Aperture".

The % error (at any particular F-Ratio value, relative to the value derived from the approximation) is a negative valued error (-2.021% at F=2.450, -1.000% at F=3.54439, and -0.501% at F=4.975).

A simple formula for (a close approximation of) that (F-Ratio/Approximation) error [when F>2.0] is:

E[%] ~ ( -1 ) * ( 3.54439 / F )^(2) ... [in units of percentage change, %]

where:

E[%] is the (negative signed) percentage error (of the F-Ratio divided by approximated F-Ratio);

F is the lens-system F-Ratio.

From: http://www.dpreview.com/forums/post/57344381

.

DM

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

It appears that lens-system F-Ratio is only approximately equal to 1 / ( ( 2 ) * ( Numerical Aperture) ).

When the using the stated (as opposed to the approximate) identity for F-Ratio relative to (image-side) Numerical Aperture (for a single symmetrical lens), I recently came up with something rather different.

.

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

NA = sin (theta) = sin (arctangent ( D / ( 2*L ) ) )

...

Click to expand...

Hi Detail Man,

The approximation occurs when you have a flat principal plane in the derivation, and that leads you and Wikipedia to include arctangent in the derivation right off the bat (the use of the flat principal plane is the approximation, if you follow it, it takes you down the path of inaccurate conclusions).

But then Wikipedia goes on, and Kingslake says it best in his quote about using sin(theta) in the much simpler derivations. This goes along with what I was saying about the curved principal surface above, so that NA = sin(theta`) = 1/ 2*F# is all self-consistent.

Chris

tony field said:

A note on this subject a long while ago (if memory serves me) by Bob indicated that the issue is with the index of refraction of modern glass. He intimated that new novel materials might overcome the problem.

Click to expand...

I believe the 0.5 limit is a hard limit set by the fact that sin(x) is never great than 1 (for real x). Approaching the limit in a real lens may well be a question of the available high refractive index glass. I guess diffractive optics can approach the limit closer but at the cost of other problems.

Joe

cpw said:

Detail Man said:

Jack Hogan said:

Joe is correct, the physical lower limit on f-number (N) is 0.5. The reason is that f/D is only an approximation valid when the opening angle theta' is small. The actual definition of f-number in air is

N = 1/[2sin(theta')]

from which it becomes obvious that N can never be less than 0.5, as better explained by Nakamura .

Click to expand...

It appears that lens-system F-Ratio is only approximately equal to 1 / ( ( 2 ) * ( Numerical Aperture) ).

When the using the stated (as opposed to the approximate) identity for F-Ratio relative to (image-side) Numerical Aperture (for a single symmetrical lens), I recently came up with something rather different.

.

From: https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number

NA = sin (theta) = sin (arctangent ( D / ( 2*L ) ) )

Click to expand...

Hi Detail Man,

The approximation occurs when you have a flat principal plane in the derivation, and that leads you and Wikipedia to include arctangent in the derivation right off the bat (the use of the flat principal plane is the approximation, if you follow it, it takes you down the path of inaccurate conclusions).

But then Wikipedia goes on, and Kingslake says it best in his quote about using sin(theta) in the much simpler derivations. This goes along with what I was saying about the curved principal surface above, so that NA = sin(theta`) = 1/ 2*F# is all self-consistent.

Click to expand...

Thanks for replying. Am trying to understand what you are intending to mean. A single, symmetrical flat-lens does not have a flat (merged front/back) principal plane (as would then apply to cases of all other lens-systems). Or, a single, symmetrical flat-lens does have a flat (merged front/back) principal plane ?