PDA

View Full Version : Achromatic diopters and 35mm adapters?



Coco Bermudez
03-16-2006, 08:57 AM
Anyone has attempted to use an achromatic diopter with their 35mm solution? Does it beat out using just a couple plain macros? I was thinking of using the Canon 500D or even the Century Optics achromatic diopter as a solution to better improve the image size of the GG as I focus in on the 35mm adapter i'm building.
Any ideas?

PaPa
03-16-2006, 09:00 AM
someone please tell me what acrhomatic, and diopters is, and also what they mean together :) please

ovjamaica
03-16-2006, 09:30 AM
I'm glad you started this topic, as I was just about to discuss this in the giant Edweirdo thread. But I think it has greater application here.

An achromatic lens is a lens made up of two types of glass. These two types of glass are cemented together. What they do is alter the way certain wavelengths pass through the lens. This reduces chromatic aberration because chromatic aberration is caused by the "drifting" of certain wavelengths of light. This is usually seen around bright edges in an images. I hope that this helps to explain it a bit better.

The reason I was going to bring this up is because there is a cheap way for anyone who wants to try an achromat to do so. I found this auction on eBay (http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=7598581609). The guy is selling one hundred 51mm x 175mm at $4 for the achromat and $5 for shipping. Shipping is the same if you buy 1 or 2. I am going to get one and try it out in my setup. From what little I know about the focal length of achromats, a 175mm focal length will be pretty powerful. The 51mm refers to the diameter of the lens and the 175mm is the focal length of the lens (at least from what I understand about these). They are Zeiss lenses too so I'm sure they'll be decent quality. Anyway, that's just a thought on this matter. I'll post my results when I get this and the rest of my parts in. Still waiting on that freaking Nikon screen.

ovjamaica
03-16-2006, 12:09 PM
Sorry to double post, but are there any optics experts out there that can relate the focal length of an achromatic lens to a magnification power? For example, an achromat with a focal length of 150mm magnifies an image by what power? +5, +7, +10? I've been trying to research this but so far I am mostly finding out interesting things about telescopes. :D Thanks in advance.

j
03-16-2006, 12:50 PM
Perhaps Bill Turner?

LHejza
03-16-2006, 02:11 PM
The formula for this is:

FL = 1000/Diopters

Why? Because 1000 cm = 1 meter and 1 diopter will focus at 1 Meter (1000cm) with the lens at infinity.

So a 2 diopter lens focuses at 500 cm or (1.2 Meter) 10 diopter at 100 cm (.1 meter)

if you know the focal length and want a diopter equivalent then use

Diopters = 1000/FL

Make sense?

Leo

LHejza
03-16-2006, 02:13 PM
Just for the picky. THere is a small effect on the actual focal length (focus distance) based on the focal length of the camera lens the diopter is attached to, but for most uses the difference is insiginificant.

ovjamaica
03-16-2006, 02:50 PM
Ok, so let's use an example. I just ordered an achromatic lens with a focal length of 175mm. So if I use your calculation concerning diopters I take 1000/175, right? This gives me 5.71, approximately. So what does that number mean? Is that that magnification level or something else? I really appreciate your calculations, I just want to use them for information I can understand. Thanks a bunch.

j
03-16-2006, 03:01 PM
So in order to get a +10 diopter, I should look for a 1000mm focal length achromat? That's going to be hard to find in the 58mm size I am shopping for, but at least I know what to look for...

Thx.

LHejza
03-16-2006, 06:10 PM
No, For a +10 diopter you would look for 1000/10 or 100 mm focal length.

The 5.71 is the Diopter rating of the lens or it's "strength" if you will. Eyeglass lenses are measured in diopters with larger numbers meaning stronger magnifications. the + and - in front of the number tell if it's a positive (magnifying) or negative (reducing) lens.

Does that help?