Aperature and Depth of field Physics question

[Hulk]

Next week I'm going to be teaching optics to my physics class.

I know that as aperature is increased less light is let into the lens and depth of field is increased.

Can someone give my a good explanation as to WHY this happens? It would be nice to be able to describe it using a ray diagram.

[Dr Mordrid]

Book definition;

Depth of Field

The amount of distance between the nearest and farthest objects that appear in acceptably sharp focus in a photograph. Depth of field depends on the lens opening, the focal length of the lens, and the distance from the lens to the subject.

Generally speaking....

A smaller apature (larger F-stop) = more depth of field = near & far objects are both in focus.

A larger apature (smaller F-stop) = less depth of field = near & far objects are NOT both in focus.

Of course it isn't quite that simple as the focal length of the lens, distance to the primary subject and apature all take part in producing the effect. Here are the relationships;

Focal length of the lens

DOF is strongly inverse to the focal length of the lens. This means that a photograph taken with a 28mm lens will exhibit much greater depth of field than one taken with a 100mm lens.

Distance from camera to subject

DOF is directly proportional to the camera-subject distance. The further away the camera is from the subject in primary focus the greater depth of field you will see in the photograph.

Size of the apature (f-stop)

The diameter of the apature is inversely proportional to DOF. A physically larger apature (small f-stop) will result in reduced DOF. A physically smaller apature (large f-stop) will result in an incresed DOF.

In real terms the effect of the apature is actually smaller than that of the focal length of the lens when photographing subjects at moderate to long distances. It's influence is greater when doing closeups.

This page might be useful;

http://www.dof.pcraft.com/dof-frames.cgi

Here is a DOF calculator you can use to produce some graphs;

http://www.johndesq.com/formulas/dof2.htm

Dr. Mordrid



[This message has been edited by Dr Mordrid (edited 19 February 2001).]

[Brian Ellis]

Hulk

I'll try to explain it slightly differently. Imagine a large diameter lens and a small diameter one with the same focal length. Draw them on a piece of paper, with rays from an object at a fixed distance going through the periphery of the lens. The angle of the rays reaching the focal plane is much larger with the big lens than the small one. If you move the hypothetical film from the focal plane slightly closer to the two lens, the blurring will be greater for a given movement with the big lens, as the section of the cone of light will be bigger. Simplistic explanation.

Not counting diffraction effects, you can get an infinite depth of field with an infinitely small aperture, as the cone of light becomes a line. Of course, you don't even need a lens, then, as I'm describing a pin-hole camera. The effect of refraction in a lens is negligible under such conditions. This ideal is, unfortunately, spoilt by diffraction, which limits the minimum size of the aperture and the best "pinhole" is made from the thinnest opaque medium, such as alu foil, and should be a clean-cut hole (not punched with a needle) with a diameter not smaller than about 0.5 mm. This gives the best compromise between blurring due the hole diameter and that due to diffraction at the hole edges. Diffraction can be likened, in laymen's terms, to photons ("packets of light") "tripping up" as they stumble through the hole and being diverted from the straight and narrow.

With a real life lens, let's take a well-designed 35 mm camera fixed focal length lens of 50 mm, as an example, the image sharpness, even when correctly focussed, varies with aperture. Let us assume it has a max aperture of f1.6 to f2. At full aperture, the lens designer has had to adopt some compromises, possibly due to the fact that the glasses available do not have the absolutely ideal range of refractive index for the range of light wavelengths (colours) "seen" by the film or due to manufacturing tolerances etc. This will usually limit the best sharpness to the centre of the image. As the aperture is stopped down, so the effect of these compromises decreases, but that horrible diffraction increases. This is why such a lens has a minimum aperture of f16 or, occasionally, f22: it is not a mechanical limitation of the iris diaphragm, but one where the lens designer does not wish to show that the resolution becomes poor. The optimum sharpness, over the whole field, of such a lens is usually obtained at somewhere between f5.6 and f11. With zoom lenses, the equations become much more complex and many more compromises are needed to be made, by the designer.

Hope this helps



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Brian (the terrible)

[wizz]

Hulk, this is what i do for a living, but Doc and Brian have explained the process well so there's not much for me to say

Tony

[Hulk]

Dr Mordrid and Brian -

Thanks a lot for the info.

I just want to be sure that I'm prepared to for all of the questions I know the kids are going to ask. This is an AP class and I like to go as far into a subject as they would desire.

I stopped by my local lens maker yesterday and they were nice enough to give me some large diverging/converging lenses for demos. Our school doesn't have much in the way of optics.

I do have a few questions if you guys don't mind if I pick your brains.

Here's my sequence for class:

I'm going to start with some basic definitions and then proceed to ray diagrams for concave and convex mirrors.

After that I'll do the ray diagrams for concave and convex lenses.

Next is the compound microscope. No problem there either. The ray diagram is pretty straightforward.

I do have a question about the human eye. Most texts show the image forming on the retina AT the focal point. Wouldn't the focal point be only a POINT? When demonstrating lenses, NO image is formed AT the focal point. Is the image on the retina formed just AFTER the focal point? Hopefully someone can clear this up.

Also for the astronomical refracting telescope. Most texts show only parallel rays entering the objective lens. Again there would be no image at the focal point. The thin lens equation says that the image will move to the focal point as the object moves to infinity. Now, most texts also show the focal point of the eyepiece at the focal point of the objective. Again, I though no image was formed at the focal point.

Some texts show nonparalled rays entering the objective. In this way they can show an image. My question is where do these nonparallel rays come from if the telescope is pointing directly at a star?

I have a feeling I'm missing something simple here because all of my questions are centered on the same concept.

Another thing. As the object moves to infinity and the object to the focal point with zero height, how do you project in onto the film (or CCD) of a camera?

Hopefully someone can help me out here.
I've gone through my college physics book and a lot of internet site but haven't really cleared this up in my mind yet.

Thanks for any help.

[Hulk]

Brian -

I think the difraction you speak of for small aperatures is best described using the wave (not particle)theory of light but I could be wrong.

The interference pattern produced is a result of the path difference the light takes from the center of the hole from other locations of the hole. When these path differences equal the multiples of the lights wavelength constructive interference occurs and a bright fringe is seen, multiples of one-half wavelength produce destructive interference or dark regions

[KRSESQ]

Hulk:

It is helpful to remember that light never enters a lens at only one angle.

Imagine a line drawn through the center of the lens from the object to the focal point on the film (or CCD). Light rays travelling parallel to this line will focus on the center of the image plane.

Light originating from any point (for example, 45 degrees) above our imaginary centerline will be brought into focus at a point on the film 45 degrees below the line. Likewise, light originating below the line will be brought into focus an equal distance above it.

This is why the image formed on a photographic negative, or the retina of the eye, is inverted and flipped left for right.

Whether the image at the periphery of the lens stays in focus is a function of the quality of the lens. A cheap lens will have a degree of focus "fall-off" at the edges, whereas an expensive lens will keep the peripheral image sharp.

Hope this helps clarify.

Kevin

[This message has been edited by KRSESQ (edited 19 February 2001).]

[This message has been edited by KRSESQ (edited 19 February 2001).]

[Brian Ellis]

Hulk

Diffraction producing an image with true interference patterns require monochromatic light (or nearly so). With a photo-type image illuminated by white light, the interference patterns merge to form a loss of resolution.

You are quite right about using waves, rather than particles, if you are going into interference. I tried to give a simplistic explanation your kids can understand.

I'm not sure what you mean by "point". In optics, there is no sich animal as a true geometrical point source or image: everything is of finite size. If the image of a distant star, through a telescope, is less than a 1/2 wavelength across, you won't see it. If it is bigger, you will.

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Brian (the terrible)

[Brent]

An interesting I experiment I recall doing at college was using a moderate wide angle and tele lens at say f4 on a tripod focused at the same distance at an object, using a fine grain film, and then enlarging the central portion of the wide angle shot to match the full frame print of the tele shot. I think the depth of field and perspective was identical (within things like the different resolution and distortion of the two lens designs). Was to prove that subect to lens distance was an important consideration I think (my course was a long time ago).