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Small update

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Formatted the article and added a small section on relative brightness. This article may need a picture showing the lens on a camera... I would add one except I need my camera to photograph the lens and the lens is firmly attached to the camera... I tried a mirror but I'm just not steady enough. Also an equation relating the f-number to the focal length and the objective lens to the magnification would be a good addition except I'm not sure of the standard symbols for each. -- Bob-505 00:22, 8 June 2006 (UTC)[reply]

I took out the bit on relative brightness, as it was manifestly untrue for camera lenses, and was covered sort of already in the section on direct-observation instruments. I'm not sure what you have in mind with camera lenses; the exit pupil size and position are not functions of focal length and f-number, but more complicated. Dicklyon 07:20, 1 August 2006 (UTC)[reply]

Diagram

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Perhaps someone can fire up Inkscape and draw a diagram? Shinobu 23:15, 6 May 2007 (UTC)[reply]

Added diagram and removed reqdiagram tag. Egmason (talk) 03:43, 22 June 2012 (UTC)[reply]
I removed the diagram and restored the tag. The diagram does not correctly illustrate the exit pupil. The exit pupil is an image of the aperture stop. It has a definite location in space; it is not merely a diameter of something (collimated rays??) emerging from the back of the lens. --Srleffler (talk) 04:52, 22 June 2012 (UTC)[reply]
I agree. I looked around for examples of illustrations, and most are not very good. Some show the exit pupil as the image of the objective lens, which is OK for simple lenses, but it would be nice to show it as the image of the aperture stop more generally. Dicklyon (talk) 05:57, 22 June 2012 (UTC)[reply]
<Sigh> Ok I get it. Well, I get that I don't get it. I will look for better pictures because I don't really understand where the exit pupil is located in space. Is it the focal point of the lens, is it the focal plane of the lens (are they different?) or is it located at the imaging plane - the film or sensor? Once someone can set me straight or I can find a good reference I can understand, then I will update the picture. I also added the picture of the binoculars, but if I get what you are saying, there is no 'exit pupil' without an eye or an imaging plane to receive it - it is a 'virtual' construct? Egmason (talk) 05:34, 26 June 2012 (UTC)[reply]
The focal point is by definition in the focal plane. The imaging plane (where the film or sensor is) is not in the focal plane unless the lens is focused at infinity. The exit pupil is not typically in either the focal plane or the imaging plane. I don't know how to determine where the exit pupil of a lens will be without doing a ray trace or a Gaussian analysis of the lens.
The exit pupil can be a "real" image, in the same sense that a picture projected on a screen is. An eye or image sensor is not necessary.
Regarding image location: not all images formed by a lens fall in the imaging plane. If you focus a camera lens to take a picture of something, you are adjusting the lens so that the image from that object is in the camera's imaging plane, which is where the CCD sensor is. Objects that are more distant from the camera will form images that fall in front of the sensor, while objects closer to the camera will form images that would be behind the sensor if the sensor weren't there to block the light. We say that these images that don't fall in the imaging plane are out of focus. Because the images are in the wrong place, the sensor sees a blurred image. In the case of the exit pupil, this blurred image has an important and unintuitive effect on the formation of other images by the lens. --Srleffler (talk) 05:29, 27 June 2012 (UTC)[reply]
I like the binoculars image by the way. When you look into the rear face of a lens, what you see in there really is the exit pupil. For microscopes, the exit pupil is designed to be in space several millimeters above each eyepiece. The optimum viewing position is to place the pupil of your eye at the exit pupil of the microscope. The optics in your eye then produce the best image on your retina.--Srleffler (talk) 05:29, 27 June 2012 (UTC)[reply]
I believe the exit pupil is always a virtual, not real, image. In the lead photo, as it says, it's that bright spot inside the lens, which has a location in 3-space as you can see by looking at it. There's not necessarily anything real at that location in space; it might even be infinitely far away, in the case of a lens that is image-side telecentric. I'm less sure about things like binoculars; probably it's the same: the exit pupil is where it looks like it is. Here's good info [1]; [2]. Dicklyon (talk) 06:05, 27 June 2012 (UTC)[reply]
Exit pupil can be a real or a virtual image, depending on the geometry. Consider a system consisting only of an aperture stop followed by a thin lens (like figure 12.3a in Introduction to Geometrical Optics, which you link to above). If the aperture stop is more than one focal length in front of the lens, the exit pupil is a real image. As I said above, I believe that microscope eyepieces typically have a real exit pupil, several mm from the output lens. The distance from the face of the eyepiece to the exit pupil is the "eye relief" of the eyepiece. From The Eye and Visual Optical Instruments: "The higher the power of the eyepiece, the closer is the ... exit pupil to its back focal point". That's a real image. I agree that the pupil is where it looks like it is. It's an image like any other. --Srleffler (talk) 02:11, 28 June 2012 (UTC)[reply]
Yes, I see what you mean; seems right. Dicklyon (talk) 05:09, 28 June 2012 (UTC)[reply]
Is the diagram shown in [3] ok?? If yes, I can make something similar to that. Roshan (talk) 10:01, 10 July 2012 (UTC)[reply]
That looks non-standard to me. The position of the pupil does not appear to be an image of any sort. Dicklyon (talk) 14:57, 10 July 2012 (UTC)[reply]
I don't really know anything about this, but from what I gather from the article, isn't Figure 3 from the link above appropriate? If not, if you can give me an idea of how the diagram should look like(even a hand-drawing on a tissue paper would suffice) or atleast what elements should be present in it, I can make one. Roshan (talk) 21:00, 10 July 2012 (UTC)[reply]
Figure 3 in the linked page doesn't illustrate the important point that we would need for this article. The text identifies the rays as marginal rays, and two pairs of them cross at the exit pupil the way marginal rays should, but if these are marginal rays, then the aperture stop of the system is somewhere off the bottom of the diagram. The marginal rays that cross on each side of the exit pupil should also cross at each side of the aperture stop. In figure 3 those rays only cross at the exit pupil; they don't meet anywhere else. This is probably because the diagram only shows the eyepiece; presumably the aperture stop of the system is in the objective lens assembly. The key point the illustration for the WP article needs to show is that the exit pupil is an image of the aperture stop. This means that all rays that pass through a given point in the plane of the stop meet again (cross) at a corresponding point in the exit pupil.
This kind of diagram is very hard to do unless the illustrator also has a sound technical understanding of the subject. --Srleffler (talk) 02:23, 11 July 2012 (UTC)[reply]
Isn't the field diaphragm in the picture the aperture?? Roshan (talk) 07:28, 11 July 2012 (UTC)[reply]
If it is, then the rays are drawn wrong. The exit pupil shown is not an image of the field diaphragm.--Srleffler (talk) 16:40, 11 July 2012 (UTC)[reply]

Revised diagram

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The new image is better, but still isn't accurate. There is really no substitute for knowing how to do a simple ray trace here. Given the two rays shown, it is possible to determine the focal length of the lens (in arbitrary units, of course), and then to determine where the pupil is and how big it is.

On the plus side, it does suggest that the exit pupil is an image, and that it has a definite location in space. It's unfortunate that it suggests that the exit pupil is located at the aperture; this is not typically the case.--Srleffler (talk) 05:01, 21 November 2013 (UTC)[reply]

Created in Inkscape to illustrate the concept of the Exit Pupil. svg file.
Grumble grumble. Have just updated the picture again, with reference to several telescope diagrams. Diagram is simplified with simple lenses and rays refracting at the centre of the lens (probably don't need to point that out). I hope this is right now. I might try to work on getting a better photograph of some of this. Egmason (talk) 11:16, 22 May 2014 (UTC)[reply]
Sorry, but it is still not right. The thing you have marked "exit pupil" is an image of the objective lens, not an image of the aperture. Look at the rays. Rays that start together at the objective come back together at the plane marked "exit pupil". Rays that are together at the aperture appear to leave the lens parallel to one another, indicating that the exit pupil is at infinity. The diagram would be correct if you removed the aperture. In that case, the objective lens would be the aperture, and the exit pupil would be as indicated.
On a correct diagram, rays that are together at the aperture will be together again at the exit pupil. That is the definition of what the exit pupil is.--Srleffler (talk) 04:13, 23 May 2014 (UTC)[reply]
Diagram edited to remove purported aperture. After going back to The Internet, I see that I have confused the aperture concept with the exit pupil concept. I hope that this is acceptable now. I will check back and hopefully finally wipe Exit Pupil off the Reqdiagram list (which has blown out from around 600 to 730 in the time I have been watching it). Egmason (talk) 08:50, 1 June 2014 (UTC) This is based on several diagrams on internet and books, mainly astronomical and microscope sites. Know any free ray trace software? Egmason (talk) 00:44, 7 June 2014 (UTC)[reply]
No, I don't. All of the free entries at List of ray tracing software seem to be for graphics rather than physics.--Srleffler (talk) 01:55, 7 June 2014 (UTC)[reply]
No-one seems to have objected in the last three years, I have now removed the "Optical Diagram requested" tag. Egmason (talk) 07:42, 2 February 2018 (UTC)[reply]

Merge

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I propose that Ramsden disc be merged with this article. There is no reason for an archaic term for a thing to have an article of its own, separate from the article on the thing itself. This is especially true for technical topics.--Srleffler (talk) 01:59, 20 March 2008 (UTC)[reply]

Since there was no objection, I went ahead with the merge. I think this article is much improved—there was some good material in the Ramsden disc article.--Srleffler (talk) 04:19, 12 May 2008 (UTC)[reply]

Reference

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I have read the phrase "Image sensors often have a limited range of angles over which they will efficiently accept light" in this and several other Wikipedia entries but have never been able to find a scientific paper (or image sensor data sheet) that supports it. This needs to be supported by a proper reference or references. GromitRoberts (talk) 20:26, 23 September 2009 (UTC)[reply]

This book has some good stuff on it. Here is another. Dicklyon (talk) 20:52, 23 September 2009 (UTC)[reply]
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Small exit pupil

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The article states that "If the disc is larger than the eye's pupil, light will be lost instead of entering the eye; if smaller, the view will be vignetted." The first part is true - if the exit pupil is bigger than the eye pupil, light will hit the iris and not enter the eye. But I don't believe the second part; if I use a x250 eyepiece on my 250mm aperture telescope and consequently the exit pupil diameter is a tiny 1.0mm, there is no vignetting: the full field of view is visible, with no shading effects. Am I using the wrong definition of "vignetting"? Funast (talk) 12:50, 25 October 2020 (UTC)[reply]