Talk:Lenticular optics

Duckweed

What the heck is a duckweed? The term is used in this article, and the same author used the same word in the lenticular lens article, but doesn't explain it. I suspect this is some literal translation from French that doesn't work in English, or perhaps its some technical term the author didn't think it necessary to define.--Srleffler (talk) 22:41, 25 December 2007 (UTC)

Plan of editing

This article needs a lot of work. Besides basic cleanup and wikification, it also needs major correction of English (the original author is French). I plan also to convert the worked examples into clear expositions of the math. This is necessary both for clarity and to comply with Wikipedia:What Wikipedia is not. Wikipedia is not a textbook, and it is not generally appropriate to give long worked examples here. (There are exceptions.)--Srleffler (talk) 06:33, 23 February 2008 (UTC)

Flawed argument

The calculation of the focal length of a lenticular lens is incorrect. You can't find the focal length of a thick lens by tracing a single off-axis ray, because of spherical aberration. The focal length and the location of the focal plane are easily calculated from the lensmaker's equation etc. This error gets me wondering, though: is this page original research? The main reference for the lenticular calculations is the page creator's own website.

Material removed

A priori, one could think that the focal distance is with the tangent of the back of a lenticular sheet, there where is the impression. Is this well the case?

On the basis of what was previously exposed, it is possible determine the focal distance from a lenticular network, it is possible to make the following calculations:

Data input:

• Lenticular Network of 75,45 LPI is pitch of 336,65µ
• Ray of the lenticular 190,5µ (the microscope with sweeping gives 196µ)
• Thickness of 457µ
• Index of the resin Lenstar 1,557

Data calculated:

For this calculation, we are going from a light beam parallel to the axis of a lenticule. This beam has an incident angle I of 15°. The distance that beam to the axis is equal to a half a rope c arc either equal to the sinuses (I) * radius r of the lenticule. Either c = 2 * (sin (15 °) * 190.5) = 98.61 µ. The arrow f is equal to the radius r of the lenticule minus the square root of the square of the radius squared least half of the rope either r-f = √ (r² - (c / 2)²) Either f = 190.5 µ - √ ((190.5 * 190.5) - ((98.61 / 2)²) = 6.49 µ. The focal length is equal to more arrow f the tangent of the sum of the corner opposite of the angle of incidence I more the angle of refraction R multiplied by the half rope c. The angle of refraction R on the basis of the formula sin (i2) = sin (i1) n1 / n2, n1 is the index of mid-air (1003), n2 is the index in the mid-PET ( 1557). R = 9.6 ° Either F = f + (tan ((90 ° -15 °) +9.6 °) * (c / 2)) = 527.85 µ The manufacturer gives a focal length of 452 µ theoretical to 508 µ It is here that the focus of this network is farther than the back of the network of 527 µ - 457 µ (thickness) ≥ 70 µ. In other words this means that an observer in the axis of the lenticule does not a line but a tape.

What is the width of this band ? The rope of the arc of the lenticule is equal to the Pitch networks. The boom, as seen above, is equal to the radius r of the lenticule minus the square root of the square of the radius squared least half of the pitch is p r-f = √ (r² - (p / 2)²) Either f = 190.5 µ - √ ((190.5 * 190.5) - ((336.65 / 2) * (336.65 / 2)) = 101.3 µ. The band width is equal to a 2 (((p / 2) * (F-e)) / (F-f)). Either b = 2 * (((336.65 / 2) * (527.85-457)) / (527.85 - 101.3) = 55.92 µ. This is interesting because it shows that the maximum number of images that can be seen separately through a lenticule is p / 55.92 = 6 images. In calculating resurfacing with an incident angle of 30° I, the focal length is 506.11 µ and the band observation is 40.84 µ or 8 images. These calculations are based for an observer to infinity, it is clear that over the observer is closer tape expands and the number of images possible in a lenticule decreases.

Defocus in above example

The example doesn't address why the focal plane doesn't appear to coincide with the back of the lens. I presume this comes from an actual example. I can think of several reasons why the focal plane might be displaced, including things like manufacturing tolerances, compensating for spherical aberration, optimizing for some aspect of human vision, or just simple mismeasurement of the lens. Are the lenses actually circular cylinders? I would have expected aspheric surfaces to be used. If you're going to mold lenses out of plastic, they might as well be aspheric cylinders.

I deleted the analysis of the resolution limit imposed by the defocus because that is just not the right way to tackle that problem, and coming up with a new analysis would certainly violate WP:OR. --Srleffler (talk) 05:00, 26 February 2008 (UTC)

Focal length

I consider removing the focal length section. The calculation of BFD is wrong. If the focus is inside the lenticule, the formula is different. If the focus is in the air (outside), one should use thick lens formula, which is different from the formula displayed in the article. —Preceding unsigned comment added by Itsikw (talk • contribs) 20:17, 16 July 2009 (UTC) Well, after I learn how to enter equations and other staff, maybe I will post the correct formulas... —Preceding unsigned comment added by Itsikw (talk • contribs) 20:20, 16 July 2009 (UTC)

I just looked at it, and I think the equations are correct. Perhaps the formalism used here is different from the one you are used to? "Focal length" here is the reciprocal of the optical power of the lens, and is the same for a thick plano-convex lens as for a thin one. The thickness-dependent term drops out. The position of the focal planes still depends on the thickness, however, because the position of the principal planes depends on the lens's thickness. The formula given for BFD looks correct to me, and I believe it should work regardless of whether the focal plane is inside or outside the lens. Take another look and let me know what you think. If you have other equations to insert, I can help with formatting them.--Srleffler (talk) 04:35, 20 July 2009 (UTC)

Yes, I did not formulate my thoughts correctly. You are right, the equations are correct, only not complete. I made the change, hope that you will approve it. Yitzhak Weissman 20:06, 23 July 2009 (UTC) —Preceding unsigned comment added by Itsikw (talk • contribs)

I think there is a subtle problem with what you have done. Your calculation for the case where the focal plane is inside the lens treats the lens as if it were infinitely thick, such that the light never leaves it. Your calculation gives the position of the rear focal plane in the optical space of the lens material, not as viewed from the image space behind the lens. Refraction at the flat surface shifts the focal plane position as viewed from outside the lens.

The formula for BFD previously given presumes air in the optical space behind the lens. I'll make the same assumption for the case where the focal plane lies inside the lens. As you wrote:

$F'={nr \over n-1}=nF$

As viewed in the optical space inside the lens, the distance from the focal plane to the flat surface is

$e-F'=e-nF\,$

To get the position of the focal plane as viewed from behind the lens, you divide by n. In the image space, then, the focal plane is in front of the plane surface of the lens by

${1 \over n}\left(e-nF\right)={e \over n}-F$

But this is just -1 times the value of the back focal length given by the previous formula:

$BFD=F-{e \over n}$

In other words, this formula for back focal length works regardless whether the focal plane is inside the lens or outside. Negative values give the correct location of the focal plane inside the lens as viewed from outside the lens, which is the consistent way to describe the performance of the lens.

This does gloss over the fact that there isn't in fact air behind the lens, but rather an opaque strip of backing material containing an image. Any treatment of defocus at that plane, however, should treat the error in position of the focal plane as being in the same optical space, regardless whether the focal plane is in front of the backing or behind it.

Really, this is the wrong way to describe the problem altogether. One should treat the backing as an object, and calculate the position of the image formed by the lens. --Srleffler (talk) 03:14, 24 July 2009 (UTC)

Yes, Srleffler, you are right, I overlooked the fact that both cases are described by the same formula, and the only difference is the sign. Just a note, it is not generally true that the printed image is in contact with the back of the lens. This is the case when the image is printed on the lens. However, there are techniques in which the image is printed on another material (like paper), and bonded or laminated to the lens. In this case there is an additional layer representing the adhesive thickness (which is not air, of course). Yitzhak Weissman 16:02, 24 July 2009 (UTC)

Delete article?

I am leaning toward proposing deletion of this article. The derivations are mostly original research, and are not structured very well, taking naive optical models and abusing them to get simple answers (see above for examples). If the material were viable, it should be merged into the article on lenticular printing, not hanging here as a separate article. --Srleffler (talk) 03:23, 24 July 2009 (UTC)
Yes, this makes sense.Yitzhak Weissman 16:02, 24 July 2009 (UTC)