Talk:Lenticular optics

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[edit] 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)

[edit] 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)

[edit] 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.

[edit] 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:

Image:sq3d-distance-focale.gif

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.

Image:sq3d-bande-focale.gif

[edit] 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)