Diffraction determined by entrance pupil or exit pupil?

Started 2 months ago | Questions
(unknown member) New Member • Posts: 15
Diffraction determined by entrance pupil or exit pupil?

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

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bclaff Forum Pro • Posts: 10,328
Re: Diffraction determined by entrance pupil or exit pupil?

Crazybird wrote:

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

It originates at the physical diaphragm but the size on the image plane is determined by the exit pupil.

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

Jack Hogan Veteran Member • Posts: 7,470
Either but Exit
2

Crazybird wrote:

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

Entrance and Exit pupils are virtual planes - and images of each other related by pupil magnification, so either will work. Likewise diffraction can equally be thought of as occurring at one or the other, treating the rest geometrically.

However many references find it easier to think of it as occurring at the exit pupil, since we can just Fourier Transform the field within it to get to the aberrated/diffracted PSF and MTF on the sensing plane. In that case diffraction depends on the half opening angle drawn from the center of the imaging plane to the edge of the Exit pupil.

Then f-number N = 1/2sin of that, and the radius of the Airy Disc is 1.22lambdaN.

Jack

PS Finding out the exact size and location of the exit pupil is not trivial, Bill's tool is an excellent step in the right direction.

bclaff Forum Pro • Posts: 10,328
Re: Either but Exit
2

Jack Hogan wrote:

...

Entrance and Exit pupils are virtual planes - and images of each other related by pupil magnification, so either will work. ...

A couple small points, since this is the PS&T forum.

We generally treat them as planes but they are in fact normally curved.

We generally only consider them viewed on-axis; off-axis they often move and rotate.

-- hide signature --

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

Jack Hogan Veteran Member • Posts: 7,470
Pupils: Planes vs Curves
1

bclaff wrote:

Jack Hogan wrote:

...

Entrance and Exit pupils are virtual planes - and images of each other related by pupil magnification, so either will work. ...

A couple small points, since this is the PS&T forum.

We generally treat them as planes but they are in fact normally curved.

We generally only consider them viewed on-axis; off-axis they often move and rotate.

Yeah, I've been thinking about that, I believe this is the source of many incorrect assumptions and conclusions, at least on my part. My current understanding, which comes in large part from Goodman, is that whether they are plane or curved depends on the definition one uses, which depends on the intended purpose.

When dealing with diffraction in Fourier Optics one takes the complex-lens as black-box approach, where the Entrance and Exit pupils represent the terminal properties of a complex lens simplified down to just those two elements. The diffracting aperture is assumed to lie in a plane and the field within it is what is propagated to the image plane. Any deviations from the ideal (including aberrations) are assumed to be phasor differences built into that field. So in this case I believe the pupils are planes.

On the other hand if one takes an entirely geometric approach, I understand how one could come to a different conclusion.

Happy to learn otherwise if my understanding is incorrect.

Jack

bclaff Forum Pro • Posts: 10,328
Re: Pupils: Planes vs Curves
1

Jack Hogan wrote:

bclaff wrote:

Jack Hogan wrote:

...

Entrance and Exit pupils are virtual planes - and images of each other related by pupil magnification, so either will work. ...

A couple small points, since this is the PS&T forum.

We generally treat them as planes but they are in fact normally curved.

We generally only consider them viewed on-axis; off-axis they often move and rotate.

Yeah, I've been thinking about that, I believe this is the source of many incorrect assumptions and conclusions, at least on my part. My current understanding, which comes in large part from Goodman, is that whether they are plane or curved depends on the definition one uses, which depends on the intended purpose.

When dealing with diffraction in Fourier Optics one takes the complex-lens as black-box approach, where the Entrance and Exit pupils represent the terminal properties of a complex lens simplified down to just those two elements. The diffracting aperture is assumed to lie in a plane and the field within it is what is propagated to the image plane. Any deviations from the ideal (including aberrations) are assumed to be phasor differences built into that field. So in this case I believe the pupils are planes.

On the other hand if one takes an entirely geometric approach, I understand how one could come to a different conclusion.

Happy to learn otherwise if my understanding is incorrect.

I think how you treat them depends on your purpose.

Certainly, with (geometrical) ray tracing then are not planes.
So, in my mind, in actuality they are curved but we treat them as planes for good reasons.

-- hide signature --

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

OP (unknown member) New Member • Posts: 15
Pupil magnification

bclaff wrote:

Crazybird wrote:

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

It originates at the physical diaphragm but the size on the image plane is determined by the exit pupil.

Does this mean that a pupil magnification of 0.5 leads to an angular diameter of the Airy disc that is twice as large or is it more complicated?

I just looked at https://www.photonstophotos.net/GeneralTopics/Lenses/Pupil_Magnification_versus_Angle_of_View.htm , so the exit pupil can be nearly 6 times as large as the entrance pupil? But for lenses with a small angle of view (I guess it's not the angle of coverage) the exit pupil seems to be smaller than the entrance pupil or only slightly larger. Camera manufacturers usually don't reveal the exit pupil?

I wonder how smartphones behave? Most smartphone main cameras have a diagonal field of view of approximately 79°. Is it reasonable to expect a pupil magnification of 1?

AiryDiscus Senior Member • Posts: 2,119
Re: Either but Exit

Jack Hogan wrote:

Crazybird wrote:

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

Entrance and Exit pupils are virtual planes - and images of each other related by pupil magnification, so either will work. Likewise diffraction can equally be thought of as occurring at one or the other, treating the rest geometrically.

However many references find it easier to think of it as occurring at the exit pupil, since we can just Fourier Transform the field within it to get to the aberrated/diffracted PSF and MTF on the sensing plane. In that case diffraction depends on the half opening angle drawn from the center of the imaging plane to the edge of the Exit pupil.

Then f-number N = 1/2sin of that, and the radius of the Airy Disc is 1.22lambdaN.

Jack

PS Finding out the exact size and location of the exit pupil is not trivial, Bill's tool is an excellent step in the right direction.

In geometric optics, we refer to real and virtual images.  Light focus to a real image, and diverges to a virtual one.  In this (domain consistent) language, there is nothing peculiar about the EP or XP being real images.

AiryDiscus Senior Member • Posts: 2,119
Re: Diffraction determined by entrance pupil or exit pupil?

Crazybird wrote:

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

You will not get the wrong answer if you use the entrance pupil.

The manufactures very rarely tell you anything about it, though.

-

For those waiting in the wings to jump at this and say I am wrong, please go ahead and rationalize P =/= 1 with the airy radius being ~= 1.22 * F# * lambda, with F# == EFL/EPD.

AiryDiscus Senior Member • Posts: 2,119
Re: Pupil magnification

Crazybird wrote:

bclaff wrote:

Crazybird wrote:

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

It originates at the physical diaphragm but the size on the image plane is determined by the exit pupil.

Does this mean that a pupil magnification of 0.5 leads to an angular diameter of the Airy disc that is twice as large or is it more complicated?

I just looked at https://www.photonstophotos.net/GeneralTopics/Lenses/Pupil_Magnification_versus_Angle_of_View.htm , so the exit pupil can be nearly 6 times as large as the entrance pupil? But for lenses with a small angle of view (I guess it's not the angle of coverage) the exit pupil seems to be smaller than the entrance pupil or only slightly larger. Camera manufacturers usually don't reveal the exit pupil?

Imagine you have a 600mm F/4 lens.  Your entrance pupil by definition is 150 mm in diameter.  You also have an exit pupil; it must be viewable through the bayonet, which for a given range of reasonable positions for the exit pupil constrains its size.

I wonder how smartphones behave? Most smartphone main cameras have a diagonal field of view of approximately 79°. Is it reasonable to expect a pupil magnification of 1?

No, because the stop in a cell phone lens is virtual, and inside the front element (generally).  This means that the exit pupil is basically formed by placing an object near a magnifying glass, while the entrance pupil is formed -- conceptually -- by placing an object on top of a lens and looking at it (thus, unchanged in size).

AiryDiscus Senior Member • Posts: 2,119
Re: Pupils: Planes vs Curves

Jack Hogan wrote:

bclaff wrote:

Jack Hogan wrote:

...

Entrance and Exit pupils are virtual planes - and images of each other related by pupil magnification, so either will work. ...

A couple small points, since this is the PS&T forum.

We generally treat them as planes but they are in fact normally curved.

We generally only consider them viewed on-axis; off-axis they often move and rotate.

Yeah, I've been thinking about that, I believe this is the source of many incorrect assumptions and conclusions, at least on my part. My current understanding, which comes in large part from Goodman, is that whether they are plane or curved depends on the definition one uses, which depends on the intended purpose.

No, their meaning is never ambiguous.  The entrance and exit pupils are clearly and strictly defined as the images of the aperture stop, viewed through the front or rear member of a lens.

A 'member' is all optical elements in front of or behind the stop.

When dealing with diffraction in Fourier Optics one takes the complex-lens as black-box approach, where the Entrance and Exit pupils represent the terminal properties of a complex lens simplified down to just those two elements.

No, this is not what the pupil is in physical optics.  Physical optics inherits all elements, assumptions, and definitions from geometric optics.  Superset, not alternative set.

The diffracting aperture is assumed to lie in a plane and the field within it is what is propagated to the image plane. Any deviations from the ideal (including aberrations) are assumed to be phasor differences built into that field. So in this case I believe the pupils are planes.

No, the aperture is assumed to lie at the instantaneous transition between collimated and focusing space.  Your exact interpretation of where this is depends on the assumptions you make.  If you assume your system is described by paraxial optics of slow F/#, then this is (approximately) a plane.  If the F/# is fast, then it is almost certainly the surface of a paraboloid.

For the typical black box treatment, we use a pupil function which is not precisely the same as a pupil.  The pupil function exists in a plane and has amplitude and phase such that it is equivalent to the pupil, with all of its complexity.  It cannot capture pupil aberrations, for example.

OP (unknown member) New Member • Posts: 15
Entrance pupil for comparing different cameras?

AiryDiscus wrote:

Crazybird wrote:

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

You will not get the wrong answer if you use the entrance pupil.

That's actually the reason why I am asking. Does this mean that if I know that a smartphone tele lens has an entrance pupil of 5mm, it could resolve (regarding diffraction) approximately the same (theoretical) maximum amount of details from distant objects as a full frame camera with 5mm entrance pupil? This was my understanding, but since I heard that the exit pupil plays a role, I don't know anymore whether that's really the case.

According to https://www.dpreview.com/articles/2666934640/what-is-equivalence-and-why-should-i-care/2 , "diffraction will have the same impact on two images shot at equivalent apertures" and they are talking about the total light equivalent f-number (which should lead to the same entrance pupil for a given field of view, no matter which sensor size)

AiryDiscus Senior Member • Posts: 2,119
Re: Entrance pupil for comparing different cameras?
1

Crazybird wrote:

AiryDiscus wrote:

Crazybird wrote:

Which diameter determines the diffraction (angular radius of the airy disc)? Diameter of the entrance pupil? Diameter of the exit pupil? A combination of both? Something else? Camera manufacturers only reveal the entrance pupil? Is it ok to use the entrance pupil diameter in most cases?

When I focus on something close, then I need to multiply the f number with (1+magnification/P) for the size of the airy disc (micrometer) ?

You will not get the wrong answer if you use the entrance pupil.

That's actually the reason why I am asking. Does this mean that if I know that a smartphone tele lens has an entrance pupil of 5mm, it could resolve (regarding diffraction) approximately the same (theoretical) maximum amount of details from distant objects as a full frame camera with 5mm entrance pupil? This was my understanding, but since I heard that the exit pupil plays a role, I don't know anymore whether that's really the case.

In a system without pupil aberrations, the quantity that matters is the product of the exit pupil and the pupil magnification.  Conveniently, this is the entrance pupil size.

For any lens you may use which you care about diffraction as a user of, the apertures you ask the questions about are small enough that there are no appreciable pupil aberrations.

According to https://www.dpreview.com/articles/2666934640/what-is-equivalence-and-why-should-i-care/2 , "diffraction will have the same impact on two images shot at equivalent apertures" and they are talking about the total light equivalent f-number (which should lead to the same entrance pupil for a given field of view, no matter which sensor size)

I think equivalence is a fantastic way of making simple things more complicated.

The airy radius of any entrance pupil is 1.22 lambda / D.  This means the angular resolution limit of any two 5 mm pupils is the same.

If the Q of the systems -- (wavelength x F#) / pixel pitch -- are the same, then they have equal sampling.  This means that the M4/3 pixel pitch must be 1/2 that of the FF.

I bolded a sentence in your yellow quote above.  that product will give you the working F/#, assuming "F/#" in the equation is the infinite F/#.  The working F/# will always give you the right answer.  Some macro lenses, like Nikon's, report working F/#s directly.  If you perform this calculation on that, you will get the wrong answer.

Jack Hogan Veteran Member • Posts: 7,470
Re: Pupils: Planes vs Curves
1

AiryDiscus wrote:

Jack Hogan wrote:

bclaff wrote:

Jack Hogan wrote:

...

Entrance and Exit pupils are virtual planes - and images of each other related by pupil magnification, so either will work. ...

A couple small points, since this is the PS&T forum.

We generally treat them as planes but they are in fact normally curved.

We generally only consider them viewed on-axis; off-axis they often move and rotate.

Yeah, I've been thinking about that, I believe this is the source of many incorrect assumptions and conclusions, at least on my part. My current understanding, which comes in large part from Goodman, is that whether they are plane or curved depends on the definition one uses, which depends on the intended purpose.

No, their meaning is never ambiguous. The entrance and exit pupils are clearly and strictly defined as the images of the aperture stop, viewed through the front or rear member of a lens.

That may very well be for you and your cohorts.  As for me, I've been confused aplenty by many pictures in supposedly respected papers and texts that show incorrectly located vectors representing distance to the imaging and/or focal plane.  For a lens focused at infinity the arrow jumps arbitrarily from gaussian sphere to principal plane to exit pupil at the whim of the writer as if they were all the same, though they clearly aren't.

A 'member' is all optical elements in front of or behind the stop.

When dealing with diffraction in Fourier Optics one takes the complex-lens as black-box approach, where the Entrance and Exit pupils represent the terminal properties of a complex lens simplified down to just those two elements.

No, this is not what the pupil is in physical optics. Physical optics inherits all elements, assumptions, and definitions from geometric optics. Superset, not alternative set.

That may very well be, but In Fourier optics the source I mentioned is pretty clear, referring to the terms in italics above (Goodman ch. 6):

... the "terminals" of this black box consist of the planes containing the entrance and the exit pupils.

The diffracting aperture is assumed to lie in a plane and the field within it is what is propagated to the image plane. Any deviations from the ideal (including aberrations) are assumed to be phasor differences built into that field. So in this case I believe the pupils are planes.

No, the aperture is assumed to lie at the instantaneous transition between collimated and focusing space. Your exact interpretation of where this is depends on the assumptions you make. If you assume your system is described by paraxial optics of slow F/#, then this is (approximately) a plane. If the F/# is fast, then it is almost certainly the surface of a paraboloid.

As I mentioned earlier, that may very well be if you take a different perspective.  But Goodman says clearly above that that's not the case in Fourier Optics.  The assumption there is that the exit pupil is a plane containing a flat phase shifting plate representing all abnormalities.  The pupil and plate have only x and y dimensions, not z.  If another interpretation of Goodman's words is possible, I'd be happy to hear it.

For the typical black box treatment, we use a pupil function which is not precisely the same as a pupil. The pupil function exists in a plane and has amplitude and phase such that it is equivalent to the pupil, with all of its complexity. It cannot capture pupil aberrations, for example.

The way I read it, Goodman makes different assumptions for his purposes and therefore uses the terms differently.  In the meantime I have a skill-testing question for you or anyone interested:

Let's say you have such an exit pupil and related function, a lens of focal length f and f-number N, focused at infinity.  You use Fourier Optics to propagate the field at the exit pupil to the focal plane.  How far precisely is that from the exit pupil?  From the principal plane?

Be precise, show your assumptions and your work.

Jack

AiryDiscus Senior Member • Posts: 2,119
Re: Pupils: Planes vs Curves

Jack Hogan wrote:

AiryDiscus wrote:

Jack Hogan wrote:

bclaff wrote:

Jack Hogan wrote:

...

Entrance and Exit pupils are virtual planes - and images of each other related by pupil magnification, so either will work. ...

A couple small points, since this is the PS&T forum.

We generally treat them as planes but they are in fact normally curved.

We generally only consider them viewed on-axis; off-axis they often move and rotate.

Yeah, I've been thinking about that, I believe this is the source of many incorrect assumptions and conclusions, at least on my part. My current understanding, which comes in large part from Goodman, is that whether they are plane or curved depends on the definition one uses, which depends on the intended purpose.

No, their meaning is never ambiguous. The entrance and exit pupils are clearly and strictly defined as the images of the aperture stop, viewed through the front or rear member of a lens.

That may very well be for you and your cohorts. As for me, I've been confused aplenty by many pictures in supposedly respected papers and texts that show incorrectly located vectors representing distance to the imaging and/or focal plane. For a lens focused at infinity the arrow jumps arbitrarily from gaussian sphere to principal plane to exit pupil at the whim of the writer as if they were all the same, though they clearly aren't.

Would you like to link something?

A 'member' is all optical elements in front of or behind the stop.

When dealing with diffraction in Fourier Optics one takes the complex-lens as black-box approach, where the Entrance and Exit pupils represent the terminal properties of a complex lens simplified down to just those two elements.

No, this is not what the pupil is in physical optics. Physical optics inherits all elements, assumptions, and definitions from geometric optics. Superset, not alternative set.

That may very well be, but In Fourier optics the source I mentioned is pretty clear, referring to the terms in italics above (Goodman ch. 6):

... the "terminals" of this black box consist of the planes containing the entrance and the exit pupils.

This text does not appear in chapter six of the current edition of the book. I will assume you're using an illegally acquired copy of the second edition, which is easy to find online.

If you look in the appendix of that edition, which you are directed to do mere words away from what you quoted, you find a definition that is not in any material way different to what I described;

The entrance pupil of the optical system is defined as the image of the most severely
limiting aperture, when viewed from the object space, looking through any optical
elements that may precede the physical aperture. The exit pupil of the system is also defined as the image of the physical aperture, but this time looking from the image
space through any optical elements that may lie between that aperture and the image
plane.

So, "in Fourier optics the source you mentioned" is pretty clear, indeed, and it is in discord with your representation of what it says.

The diffracting aperture is assumed to lie in a plane and the field within it is what is propagated to the image plane. Any deviations from the ideal (including aberrations) are assumed to be phasor differences built into that field. So in this case I believe the pupils are planes.

No, the aperture is assumed to lie at the instantaneous transition between collimated and focusing space. Your exact interpretation of where this is depends on the assumptions you make. If you assume your system is described by paraxial optics of slow F/#, then this is (approximately) a plane. If the F/# is fast, then it is almost certainly the surface of a paraboloid.

As I mentioned earlier, that may very well be if you take a different perspective. But Goodman says clearly above that that's not the case in Fourier Optics.

See above.

The assumption there is that the exit pupil is a plane containing a flat phase shifting plate representing all abnormalities.

You put too many buzzwords in there to make sense.  If you're saying the pupil is a binary object that may contain some phase change, that is not true either.  Nonbinary pupils are covered in some depth in the good book, and are 'widely' used for coronagraphy.

The pupil and plate have only x and y dimensions, not z. If another interpretation of Goodman's words is possible, I'd be happy to hear it.

The pupil function P (!= pupil, again) must exist in a plane.  The pupil must not.  See the paragraph I quoted.

For the typical black box treatment, we use a pupil function which is not precisely the same as a pupil. The pupil function exists in a plane and has amplitude and phase such that it is equivalent to the pupil, with all of its complexity. It cannot capture pupil aberrations, for example.

The way I read it, Goodman makes different assumptions for his purposes and therefore uses the terms differently.

No, he does not.

In the meantime I have a skill-testing question for you or anyone interested:

Let's say you have such an exit pupil and related function, a lens of focal length f and f-number N, focused at infinity. You use Fourier Optics to propagate the field at the exit pupil to the focal plane. How far precisely is that from the exit pupil? From the principal plane?

Be precise, show your assumptions and your work.

Depends on your assumptions and approach.

In the case that you wish to use a plane to plane (free space) propagation to move from the XP to the image, you would use whatever the distance is between the pupil and the image.  The principle plane is not necessary to be considered.  There is not really any work to show for this.  You are asking to move from one plane to another, so you must propagate the distance between them.  You must apply a quadratic phase at the XP with this approach.

In the case that you wish to invoke the Fourier transforming property of a lens and use what I would call a focusing propagation, simply taking an FT, then the distance is one focal length.

There is, again, no work to show (this is the fourier transforming property of a lens), unless you would like me to recreate proofs that have been known for what is now a very long time.

I encourage you to read the whole book, and not skim it around formulas you want to lift without context.  Or take a course in the topic.

Jack Hogan Veteran Member • Posts: 7,470
Diffraction, Fourier and Focal Planes
1

AiryDiscus wrote:

Jack Hogan wrote:

AiryDiscus wrote:

This posting pattern is tiresome, destructive and wasteful.  A thoughtful contribution is made:

Jack Hogan wrote:

When dealing with diffraction in Fourier Optics one takes the complex-lens as black-box approach, where the Entrance and Exit pupils represent the terminal properties of a complex lens simplified down to just those two elements.

Someone on an ego trip doesn't take the time to understand what is being said or the context, yet questions the premise with a peremptory, often off topic, edict:

No, this is not what the pupil is in physical optics. Physical optics inherits all elements, assumptions, and definitions from geometric optics. Superset, not alternative set.

The response is easily rebutted and the conversation brought back to the original topic

That may very well be, but In Fourier optics the source I mentioned is pretty clear, referring to the terms in italics above (Goodman ch. 6):

... the "terminals" of this black box consist of the planes containing the entrance and the exit pupils.

Caught, to save face the message and the messenger need to be thrown into question:

This text does not appear in chapter six of the current edition of the book. I will assume you're using an illegally acquired copy of the second edition, which is easy to find online.

To what purpose did I just waste time to show that I knew what I knew and that the sources I said supported what I said, supported it? For the record, my source is Introduction to Fourier Optics third edition, which I have owned and read for several years. It says exactly what I said it says. Same with Goodman’s phase shifting plate analogy, which may be ‘buzzwords’ to some but to me have always been useful in visualizing the math. From my own personal library of real books:

Diffraction and Fourier Optics, Introduction to Fourier Optics 3rd edition, Goodman ch 6

Enough wasteful diversions, back to substance. Where the rubber meets the road with regards to this sub-thread is the skill-testing question that was answered incorrectly below:

In the meantime I have a skill-testing question for you or anyone interested:

Let's say you have such an exit pupil and related function, a lens of focal length f and f-number N, focused at infinity. You use Fourier Optics to propagate the field at the exit pupil to the focal plane. How far precisely is that from the exit pupil? From the principal plane?

In the case that you wish to invoke the Fourier transforming property of a lens and use what I would call a focusing propagation, simply taking an FT, then the distance is one focal length.

Goodman would beg to differ. Vectors are once again being placed wantonly and without thought, just like some of the others I lamented about earlier , QED. Where is focal length measured from?

There is, again, no work to show (this is the fourier transforming property of a lens), unless you would like me to recreate proofs that have been known for what is now a very long time.

I encourage you to read the whole book, and not skim it around formulas you want to lift without context. Or take a course in the topic.

Prompt: using the variables at hand and any others that may be needed,

zi = f + ...

Of course the context is photography.

Jack

Be precise, show your assumptions and your work

AiryDiscus Senior Member • Posts: 2,119
Re: Diffraction, Fourier and Focal Planes
1

Jack Hogan wrote:

AiryDiscus wrote:

Jack Hogan wrote:

AiryDiscus wrote:

This posting pattern is tiresome, destructive and wasteful. A thoughtful contribution is made:

Jack Hogan wrote:

When dealing with diffraction in Fourier Optics one takes the complex-lens as black-box approach, where the Entrance and Exit pupils represent the terminal properties of a complex lens simplified down to just those two elements.

Someone on an ego trip doesn't take the time to understand what is being said or the context, yet questions the premise with a peremptory, often off topic, edict:

No, this is not what the pupil is in physical optics. Physical optics inherits all elements, assumptions, and definitions from geometric optics. Superset, not alternative set.

The response is easily rebutted and the conversation brought back to the original topic

That may very well be, but In Fourier optics the source I mentioned is pretty clear, referring to the terms in italics above (Goodman ch. 6):

... the "terminals" of this black box consist of the planes containing the entrance and the exit pupils.

Caught, to save face the message and the messenger need to be thrown into question:

This text does not appear in chapter six of the current edition of the book. I will assume you're using an illegally acquired copy of the second edition, which is easy to find online.

To what purpose did I just waste time to show that I knew what I knew and that the sources I said supported what I said, supported it?

You posted a photo that shows what I said it does. Literally words away from what you're quoting is a reference to an appendix that defines the pupils as I quoted them. Here, again, in bold

The entrance pupil of the optical system is defined as the image of the most severely limiting aperture, when viewed from the object space, looking through any optical elements that may precede the physical aperture. The exit pupil of the system is also defined as the image of the physical aperture, but this time looking from the image space through any optical elements that may lie between that aperture and the image plane.

Attributed to Goodman, at the reference in his book that he makes in the image you posted.

If you would like to show how you know that all systems which image stops do so in a way that is totally planar, please go ahead.  You might want to show there are no aberrations at all while you're at it and not limit your proof to just field curvature and astigmatism.

For the record, my source is Introduction to Fourier Optics third edition, which I have owned and read for several years. It says exactly what I said it says. Same with Goodman’s phase shifting plate analogy, which may be ‘buzzwords’ to some but to me have always been useful in visualizing the math. From my own personal library of real books:

Diffraction and Fourier Optics, Introduction to Fourier Optics 3rd edition, Goodman ch 6

Enough wasteful diversions, back to substance. Where the rubber meets the road with regards to this sub-thread is the skill-testing question that was answered incorrectly below:

In the meantime I have a skill-testing question for you or anyone interested:

Let's say you have such an exit pupil and related function, a lens of focal length f and f-number N, focused at infinity. You use Fourier Optics to propagate the field at the exit pupil to the focal plane. How far precisely is that from the exit pupil? From the principal plane?

In the case that you wish to invoke the Fourier transforming property of a lens and use what I would call a focusing propagation, simply taking an FT, then the distance is one focal length.

Goodman would beg to differ. Vectors are once again being placed wantonly and without thought, just like some of the others I lamented about earlier , QED. Where is focal length measured from?

Here's some quotes from Goodman, on the section "Fourier transforming properties of lenses:"

Case of field 'at' the lens:

To find the distribution Uf(u, v ) in the back focal plane of the lens, the Fresnel diffraction formula, Eq. (4-17), is applied. Thus, putting z = f ,[...]

Case of field 'in front of' the lens:

Thus the amplitude and phase of the light at coordinates (u,v)are again related to the amplitude and phase of the input spectrum at frequencies (u/lambdaf ,v/lambdaf). Note that a quadratic phase factor again precedes the transform integral, but that it vanishes for the very special case d = f . Evidently when the input is placed in the front focal plane of the lens, the phase curvature disappears, leaving an exact Fourier transform relation!

There is, again, no work to show (this is the fourier transforming property of a lens), unless you would like me to recreate proofs that have been known for what is now a very long time.

I encourage you to read the whole book, and not skim it around formulas you want to lift without context. Or take a course in the topic.

Prompt: using the variables at hand and any others that may be needed,

zi = f + ...

No definition of variables, vague open ended prompt.  No thanks.

Jack Hogan Veteran Member • Posts: 7,470
Re: Diffraction, Fourier and Focal Planes
1

AiryDiscus wrote:

Jack Hogan wrote:

AiryDiscus wrote:

Jack Hogan wrote:

Enough wasteful diversions, back to substance. Where the rubber meets the road with regards to this sub-thread is the skill-testing question that was answered incorrectly below:

In the meantime I have a skill-testing question for you or anyone interested:

Let's say you have such an exit pupil and related function, a lens of focal length f and f-number N, focused at infinity. You use Fourier Optics to propagate the field at the exit pupil to the focal plane. How far precisely is that from the exit pupil? From the principal plane?

In the case that you wish to invoke the Fourier transforming property of a lens and use what I would call a focusing propagation, simply taking an FT, then the distance is one focal length.

Goodman would beg to differ. Vectors are once again being placed wantonly and without thought, just like some of the others I lamented about earlier , QED. Where is focal length measured from?

Here's some quotes from Goodman, on the section "Fourier transforming properties of lenses:"

Case of field 'at' the lens:

To find the distribution Uf(u, v ) in the back focal plane of the lens, the Fresnel diffraction formula, Eq. (4-17), is applied. Thus, putting z = f ,[...]

Case of field 'in front of' the lens:

Thus the amplitude and phase of the light at coordinates (u,v)are again related to the amplitude and phase of the input spectrum at frequencies (u/lambdaf ,v/lambdaf). Note that a quadratic phase factor again precedes the transform integral, but that it vanishes for the very special case d = f . Evidently when the input is placed in the front focal plane of the lens, the phase curvature disappears, leaving an exact Fourier transform relation!

I asked a pretty basic question: where is focal length measured from?  Anybody?

There is, again, no work to show (this is the fourier transforming property of a lens), unless you would like me to recreate proofs that have been known for what is now a very long time.

I encourage you to read the whole book, and not skim it around formulas you want to lift without context. Or take a course in the topic.

Prompt: using the variables at hand and any others that may be needed,

zi = f + ...

No definition of variables, vague open ended prompt. No thanks.

You really do not understand what I am asking, do you? What's the difference between  z1 and zi?

Jack

AiryDiscus Senior Member • Posts: 2,119
Re: Diffraction, Fourier and Focal Planes

Jack Hogan wrote:

AiryDiscus wrote:

Jack Hogan wrote:

AiryDiscus wrote:

Jack Hogan wrote:

Enough wasteful diversions, back to substance. Where the rubber meets the road with regards to this sub-thread is the skill-testing question that was answered incorrectly below:

In the meantime I have a skill-testing question for you or anyone interested:

Let's say you have such an exit pupil and related function, a lens of focal length f and f-number N, focused at infinity. You use Fourier Optics to propagate the field at the exit pupil to the focal plane. How far precisely is that from the exit pupil? From the principal plane?

In the case that you wish to invoke the Fourier transforming property of a lens and use what I would call a focusing propagation, simply taking an FT, then the distance is one focal length.

Goodman would beg to differ. Vectors are once again being placed wantonly and without thought, just like some of the others I lamented about earlier , QED. Where is focal length measured from?

Here's some quotes from Goodman, on the section "Fourier transforming properties of lenses:"

Case of field 'at' the lens:

To find the distribution Uf(u, v ) in the back focal plane of the lens, the Fresnel diffraction formula, Eq. (4-17), is applied. Thus, putting z = f ,[...]

Case of field 'in front of' the lens:

Thus the amplitude and phase of the light at coordinates (u,v)are again related to the amplitude and phase of the input spectrum at frequencies (u/lambdaf ,v/lambdaf). Note that a quadratic phase factor again precedes the transform integral, but that it vanishes for the very special case d = f . Evidently when the input is placed in the front focal plane of the lens, the phase curvature disappears, leaving an exact Fourier transform relation!

So where is focal length measured from?

I suggest you to look in the other parts of the appendix of that book you have.  If you want a different book, Welford is excellent.

'measure' implies a certain amount of rendering to reality.  The focal length is the distance from the rear principle plane to the image plane.  The language of "plane" is a bit casual, but these are all paraxial quantities anyway.

Here's a toy example that might help you in your crusade, which I'm sure likely exists in the good book.

A stop, one focal length from a thin lens of focal length and an object at infinity.  The exit pupil is at -inf, and the object at -f.

There are several truthy priors, like the size of the airy disk, or where the image is in relation to the lens.

There is, again, no work to show (this is the fourier transforming property of a lens), unless you would like me to recreate proofs that have been known for what is now a very long time.

I encourage you to read the whole book, and not skim it around formulas you want to lift without context. Or take a course in the topic.

Prompt: using the variables at hand and any others that may be needed,

zi = f + ...

No definition of variables, vague open ended prompt. No thanks.

You really do not understand what I am asking, do you? What's the difference between z1 and zi?

Jack

You haven't defined anything  What is even z1 an zi?  Where is z0?

Jack Hogan Veteran Member • Posts: 7,470
Re: Diffraction, Fourier and Focal Planes

AiryDiscus wrote:

Jack Hogan wrote:

AiryDiscus wrote:

Jack Hogan wrote:

AiryDiscus wrote:

Jack Hogan wrote:

Enough wasteful diversions, back to substance. Where the rubber meets the road with regards to this sub-thread is the skill-testing question that was answered incorrectly below:

In the meantime I have a skill-testing question for you or anyone interested:

Let's say you have such an exit pupil and related function, a lens of focal length f and f-number N, focused at infinity. You use Fourier Optics to propagate the field at the exit pupil to the focal plane. How far precisely is that from the exit pupil? From the principal plane?

In the case that you wish to invoke the Fourier transforming property of a lens and use what I would call a focusing propagation, simply taking an FT, then the distance is one focal length.

Goodman would beg to differ. Vectors are once again being placed wantonly and without thought, just like some of the others I lamented about earlier , QED. Where is focal length measured from?

Here's some quotes from Goodman, on the section "Fourier transforming properties of lenses:"

Case of field 'at' the lens:

To find the distribution Uf(u, v ) in the back focal plane of the lens, the Fresnel diffraction formula, Eq. (4-17), is applied. Thus, putting z = f ,[...]

Case of field 'in front of' the lens:

Thus the amplitude and phase of the light at coordinates (u,v)are again related to the amplitude and phase of the input spectrum at frequencies (u/lambdaf ,v/lambdaf). Note that a quadratic phase factor again precedes the transform integral, but that it vanishes for the very special case d = f . Evidently when the input is placed in the front focal plane of the lens, the phase curvature disappears, leaving an exact Fourier transform relation!

So where is focal length measured from?

I suggest you to look in the other parts of the appendix of that book you have. If you want a different book, Welford is excellent.

'measure' implies a certain amount of rendering to reality. The focal length is the distance from the rear principle plane to the image plane. The language of "plane" is a bit casual, but these are all paraxial quantities anyway.

Here's a toy example that might help you in your crusade, which I'm sure likely exists in the good book.

A stop, one focal length from a thin lens of focal length and an object at infinity. The exit pupil is at -inf, and the object at -f.

There are several truthy priors, like the size of the airy disk, or where the image is in relation to the lens.

There is, again, no work to show (this is the fourier transforming property of a lens), unless you would like me to recreate proofs that have been known for what is now a very long time.

I encourage you to read the whole book, and not skim it around formulas you want to lift without context. Or take a course in the topic.

Prompt: using the variables at hand and any others that may be needed,

zi = f + ...

No definition of variables, vague open ended prompt. No thanks.

You really do not understand what I am asking, do you? What's the difference between z1 and zi?

Jack

You haven't defined anything What is even z1 an zi? Where is z0?

You have the same book I have. But let me help you: In Goodman z1 is measured from the principal plane, zi from the exit pupil.

So where is focal length measured from?

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