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 | | From: | Repeating Rifle | | Subject: | Scheimflug condition again | | Date: | Mon, 10 Jan 2005 21:29:31 GMT |
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 | If you already know that object planes get imaged into image planes, the
condition can made to appear "obvious."
Every object point on the intersection between the object plane and the lens
plane coincides with its image point. Light emanates from such a point with
spherical wavefronts. Thus, to be in focus, the film plane must pass through
the same intersection.
Bill
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| | From: | AES/newspost | | Subject: | Re: Scheimflug condition again | | Date: | Mon, 10 Jan 2005 14:24:45 -0800 |
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| In article ,
Repeating Rifle wrote:
> If you already know that object planes get imaged into image planes, the
> condition can made to appear "obvious."
>
> Every object point on the intersection between the object plane and the lens
> plane coincides with its image point. Light emanates from such a point with
> spherical wavefronts. Thus, to be in focus, the film plane must pass through
> the same intersection.
>
> Bill
>
CLEVER!
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| | From: | Phil Hobbs | | Subject: | Re: Scheimflug condition again | | Date: | Tue, 11 Jan 2005 12:24:55 -0500 |
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| Repeating Rifle wrote:
>If you already know that object planes get imaged into image planes, the
>condition can made to appear "obvious."
>
>Every object point on the intersection between the object plane and the lens
>plane coincides with its image point. Light emanates from such a point with
>spherical wavefronts. Thus, to be in focus, the film plane must pass through
>the same intersection.
>
>Bill
>
>
>
I don't think this works, unfortunately.
The object and image points are on opposite sides of the axis, so
premise #1 isn't true.
Even on axis, having the object and image points of a lens coincide
requires a focal length of zero, because so*si=f**2.
And the wavefronts are irrelevant--this is all geometric optics, and a
focus is a focus.
Cheers,
Phil Hobbs
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| | From: | AES/newspost | | Subject: | Re: Scheimflug condition again | | Date: | Tue, 11 Jan 2005 20:51:37 -0800 |
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| In article <10u82tnkg0jkn74@corp.supernews.com>,
Phil Hobbs wrote:
> Repeating Rifle wrote:
>
> >If you already know that object planes get imaged into image planes, the
> >condition can made to appear "obvious."
> >
> >Every object point on the intersection between the object plane and the lens
> >plane coincides with its image point. Light emanates from such a point with
> >spherical wavefronts. Thus, to be in focus, the film plane must pass through
> >the same intersection.
> >
> >Bill
> >
> >
> >
> I don't think this works, unfortunately.
>
> The object and image points are on opposite sides of the axis, so
> premise #1 isn't true.
>
> Even on axis, having the object and image points of a lens coincide
> requires a focal length of zero, because so*si=f**2.
>
> And the wavefronts are irrelevant--this is all geometric optics, and a
> focus is a focus.
>
> Cheers,
>
> Phil Hobbs
Phil, I read this as saying that you should imagine a lens (a
hypothetical very thin, sheet-like lens) that's so big transversely that
it extends out all the way to the point (or line, actually) where the
tilted object plane intersects it.
Any object points on the object plane in the immediate vicinity of that
intersection are therefore essentially right on the lens surface (on one
side), and so the associated image points must also be right on the lens
surface (on the other side).
Assuming that an object plane is imaged into an image _plane_ (not some
curved surface), doesn't that imply that the tilted image plane has to
intersect the lens plane at that same point (or line)?
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| | From: | Phil Hobbs | | Subject: | Re: Scheimflug condition again | | Date: | Wed, 12 Jan 2005 12:06:27 -0500 |
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| AES/newspost wrote:
>In article <10u82tnkg0jkn74@corp.supernews.com>,
> Phil Hobbs wrote:
>
>
>
>>Repeating Rifle wrote:
>>
>>
>>
>>>If you already know that object planes get imaged into image planes, the
>>>condition can made to appear "obvious."
>>>
>>>Every object point on the intersection between the object plane and the lens
>>>plane coincides with its image point. Light emanates from such a point with
>>>spherical wavefronts. Thus, to be in focus, the film plane must pass through
>>>the same intersection.
>>>
>>>Bill
>>>
>>>
>>>
>>>
>>>
>>I don't think this works, unfortunately.
>>
>>The object and image points are on opposite sides of the axis, so
>>premise #1 isn't true.
>>
>>Even on axis, having the object and image points of a lens coincide
>>requires a focal length of zero, because so*si=f**2.
>>
>>And the wavefronts are irrelevant--this is all geometric optics, and a
>>focus is a focus.
>>
>>Cheers,
>>
>>Phil Hobbs
>>
>>
>
>Phil, I read this as saying that you should imagine a lens (a
>hypothetical very thin, sheet-like lens) that's so big transversely that
>it extends out all the way to the point (or line, actually) where the
>tilted object plane intersects it.
>
>Any object points on the object plane in the immediate vicinity of that
>intersection are therefore essentially right on the lens surface (on one
>side), and so the associated image points must also be right on the lens
>surface (on the other side).
>
>Assuming that an object plane is imaged into an image _plane_ (not some
>curved surface), doesn't that imply that the tilted image plane has to
>intersect the lens plane at that same point (or line)?
>
>
Tony,
I suppose so, but points in that neighbourhood cast only a virtual
image, which makes the argument not especially compelling to me--how are
you going to put your film there? A tilted plane is imaged on to
another tilted plane only in a small neighbourhood, even if we're
talking about infinite aperture and first-order imaging. The image goes
from real to virtual when d_o=f, becoming arbitrarily far away when d_o
is near f, and reappearing on the other side of the lens. This focal
surface is not a plane, which makes the "aha" in this argument illusory,
I think. Why on the face of it should this plane coincide with the
virtual image plane near the axis? (It does, of course, but it isn't
instantly obvious.) In my earlier post, I was thinking of real images
only, and taking the planar object-planar image mapping as an axiom,
which is what I thought Bill invited us to do.
The Scheimpflug condition can be looked on as a consequence of the fact
that if the transverse magnification from an object point to its image
point is M, the longitudinal magnification is M**2. Consider such a
point at a point z = -d_o with respect to the lens centre, on the
optical axis. If you draw any (tilted) object plane through it, the
image plane cuts the axis at d_i = M*d_o by similar triangles. The
slope of the image plane is -1/M times the slope of the object plane,
because a point (dx, -d_o+dz) gets imaged to (-M*dx, d_i-M**2*dz).
Because the plane has 1/M times the slope, and is M times further away,
it intercepts the z=0 (lens) plane at the same x coordinate. Axial
symmetry means that the lines of intersection have to be parallel, so
therefore they're collinear--which is the Scheimpflug condition. This
has the advantage of using only paraxial ideas and real images, which
stays a good deal closer to the application.
Cheers,
Phil
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| | From: | AES/newspost | | Subject: | Re: Scheimflug condition again | | Date: | Wed, 12 Jan 2005 11:55:25 -0800 |
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| In article <10uam78602ar404@corp.supernews.com>,
Phil Hobbs wrote:
> >Phil, I read this as saying that you should imagine a lens (a
> >hypothetical very thin, sheet-like lens) that's so big transversely that
> >it extends out all the way to the point (or line, actually) where the
> >tilted object plane intersects it.
> >
> >Any object points on the object plane in the immediate vicinity of that
> >intersection are therefore essentially right on the lens surface (on one
> >side), and so the associated image points must also be right on the lens
> >surface (on the other side).
> >
> >Assuming that an object plane is imaged into an image _plane_ (not some
> >curved surface), doesn't that imply that the tilted image plane has to
> >intersect the lens plane at that same point (or line)?
> >
> >
>
> Tony,
>
> I suppose so, but points in that neighbourhood cast only a virtual
> image, which makes the argument not especially compelling to me--how are
> you going to put your film there? A tilted plane is imaged on to
> another tilted plane only in a small neighbourhood, even if we're
> talking about infinite aperture and first-order imaging. The image goes
> from real to virtual when d_o=f, becoming arbitrarily far away when d_o
> is near f, and reappearing on the other side of the lens. This focal
> surface is not a plane, which makes the "aha" in this argument illusory,
> I think. Why on the face of it should this plane coincide with the
> virtual image plane near the axis? (It does, of course, but it isn't
> instantly obvious.) In my earlier post, I was thinking of real images
> only, and taking the planar object-planar image mapping as an axiom,
> which is what I thought Bill invited us to do.
>
> The Scheimpflug condition can be looked on as a consequence of the fact
> that if the transverse magnification from an object point to its image
> point is M, the longitudinal magnification is M**2. Consider such a
> point at a point z = -d_o with respect to the lens centre, on the
> optical axis. If you draw any (tilted) object plane through it, the
> image plane cuts the axis at d_i = M*d_o by similar triangles. The
> slope of the image plane is -1/M times the slope of the object plane,
> because a point (dx, -d_o+dz) gets imaged to (-M*dx, d_i-M**2*dz).
>
> Because the plane has 1/M times the slope, and is M times further away,
> it intercepts the z=0 (lens) plane at the same x coordinate. Axial
> symmetry means that the lines of intersection have to be parallel, so
> therefore they're collinear--which is the Scheimpflug condition. This
> has the advantage of using only paraxial ideas and real images, which
> stays a good deal closer to the application.
>
> Cheers,
>
> Phil
Yes, you're right also -- the "geometric arguments only" approach gets
trickier the closer you look at it.
--AES
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| | From: | Repeating Rifle | | Subject: | Re: Scheimflug condition again | | Date: | Wed, 12 Jan 2005 04:57:14 GMT |
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| in article 10u82tnkg0jkn74@corp.supernews.com, Phil Hobbs at
pcdh@SpamMeSenseless.us.ibm.com wrote on 1/11/05 9:24 AM:
> Repeating Rifle wrote:
>
>> If you already know that object planes get imaged into image planes, the
>> condition can made to appear "obvious."
>>
>> Every object point on the intersection between the object plane and the lens
>> plane coincides with its image point. Light emanates from such a point with
>> spherical wavefronts. Thus, to be in focus, the film plane must pass through
>> the same intersection.
>>
>> Bill
>>
>>
>>
> I don't think this works, unfortunately.
>
> The object and image points are on opposite sides of the axis, so
> premise #1 isn't true.
>
> Even on axis, having the object and image points of a lens coincide
> requires a focal length of zero, because so*si=f**2.
>
> And the wavefronts are irrelevant--this is all geometric optics, and a
> focus is a focus.
>
> Cheers,
>
> Phil Hobbs
Put your thinking cap on again and go through the material CAREFULLY. For
example, you last equation is the newtonian lens formula and not the
gaussian formula. The distances s are measured from the focal point and not
the lens center. Thus, at the lens s = f.
Bill
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| | From: | redbelly | | Subject: | Re: Scheimflug condition again | | Date: | 11 Jan 2005 21:54:12 -0800 |
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| Thanks, I was having trouble with the explanation too.
AES wrote:
> the associated image points must also be right on the lens
> surface (on the other side)
Of course, the (virtual) image is actually on the same side of the lens
as the object, but that doesn't really matter since we are taking the
limit as the object approaches the lens surface anyway.
Mark
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