Optics (Parallax and Luminosity)

[tuoni]

Does anyone know the optical algorithms for calculating (or at least estimating) parallax and luminosity? So far I have only found something for parallax, and not quite sure it is correct. For luminosity I haven't really found anything useable. This is all in regards to rifle scopes and similar smaller lenticular systems. Luminosity should preferably be in denary ratio, 1.0 = no change, <1.0 = less luminous, >1.0 = more luminous. What I do know is that the greater magnification the greater luminosity loss. As far as I know, luminosity should depend primarily on objective diameter and magnification.

The algorithm I found for parallax:

$I = (F^{-1} - O^{-1})^{-1}$

I := distance from image to lens
O := distance from object to lens
F := focal length
​

 

[Andy Resnick]

I'm confused about applying parallax to a monocular system- AFAIK, parallax is the difference in apparent direction of a source from two (or more) separated detectors.

Luminosity is a property of the source- how much energy it radiates.

Can you provide more details or context?

 

[tuoni]

As I mentioned earlier, it is for optical scopes used in rifles. Parallax occurs when the object image is formed either in front of or behind the reticle image, thus when you move your head horizontally you will see the reticle move. Or something like that, the documents I've read have been quite messy, and it takes a while to sort it all out and understand the terminology.

Perhaps luminosity was a bad choice of word, how about: apparent brightness, light transfer ratio, or something similar. Basically how bright the ocular image is compared to how the naked eye would see it. The greater the magnification the lower the brightness, and the greater the objective diameter the greater the brightness.

 

[Andy Resnick]

I'm still confused about the use of "parallax", but I understand what you mean by luminosity now.

In the design of eyepieces, a critical feature is the location and size of the exit pupil- the aperture through which all light must pass upon exiting the optical system. For a well-designed eyepiece, the exit pupil coincides both in diameter and in location with the entrance pupil of the eye- the hole through which all light must enter before passing through the optical system. If the exit pupil is too large, the eye does not collect all available light.

A simple demonstration (for me) is to look through binoculars both with and without my glasses on. With my glasses on, the exit pupils do not match up properly and the scenery looks dim. Taking my glasses off (and re-focusing the bionculars) places the exit pupil coincident with my entrance pupil and I recover a significant amount of light.

The entrance pupil of a dark-adapted eye is about 7mm in diamter. it's normally 5 mm, although everyone is different.

So- a larger objective diameter = larger entrance pupil (of the scope), so more light is funneled through the exit pupil, which means your image is brighter. Higher magnification, holding the objective diameter constant, means the exit pupil must decrease in size, so the image appears dimmer to your eye.

Does that help?

 

[tuoni]

Here is a picture that illustrates the problem.

http://www.6mmbr.citymaker.com/i/Products/Parallax_Diagramx400.gif
(http://www.6mmbr.com/parallax.html <-- original article)

Fixed scopes with low magnification are usually preset to be parallax-free to 100 m or yrds — and doesn't have any parallax adjustment — but with higher magnification or variable scopes, parallax adjustment becomes a necessity. So if your scope is adjusted to 100 m, if the object is at a distance of 50 m or 200 m you will need to make parallax adjustment. If it is closer the image will form behind the reticle, and if it is farther the image will form in front of the reticle, the end result being that the primary image and reticle do not align, which is the source of parallax.

Thus you need to move some part of the lenticular system, maybe it's the erector cell, to adjust for this, which is usually quite small.

For luminosity you pretty much solved the problem ^_^ Thank you!

 

[Andy Resnick]

Thanks for the links- that helps.

In your application, parallax arises from the difference between two image planes arising from the reticle and the object. I would have thought that one of the two would be defocused, but I guess if the depth of focus (for your eye) is large enough both can appear to be in focus.

So the algorithm you originally posted makes sort-of sense- it's simply the thin lens law. My question would be where the image plane of the object is with respect to the reticle. If they are in different locations, you have parallax.

The adjustment knobs, I couldn't find out what exactly they do- does it move an optical element, does it move the reticle?

 

[tuoni]

I have attached a picture I made to illustrate how it works, don't mind any faults I may have introduced, as I am not an expert in optics ^_^. The descriptions (although edited) are taken from http://www.snipercountry.com/Articles/Parallax.asp

x. Object — the object (target) that you are looking (shooting) at.
y. Oculus — the eye.

1. Objective — the front lens is called the objective, and it forms the first image of the object we are looking at.
2. Erector group — because the first image is upside down/wrong way around, we (as shooters) can't use it, so we flip it around with a simple optical group called the erector group. This group gives us a new image that is right way around, called the second image plane. But this group has another very important job. Moving this group causes this second image plane to move, so micrometer spindles are put against the group, to get elevation and windage adjustments.
3. Zoom group — this group of lenses can change the size of the second image plane and then form a new (third) image plane behind it.
4. Ocular — this optical group is like a jewelers' loupe. It is (or should be) a super fine magnifier. Its only job in the whole world, is to focus on the reticle.
​

a. First image plane — the objective focuses the light to make an image of the subject, just like a camera lens. This image is upside down, and right/left reversed.
b. Second image plane — this is the second real image plane in the scope. In a fixed or variable power scope with a magnifying reticle, the reticle would be placed in this image plane.
c. Third image plane — in variable power scopes, this is the plane that the reticle is placed in. By being here, it allows the image to change size, but the reticle to stay the same size.
​

"In the scope that is set for infinity range, the object forms an image (upside down, right/left reversed) behind the objective (the first image plane), the erector cell "sees" that image, and flips it over and makes it right way around in a NEW image plane (the Second image plane). The zoom group adjusts the size of this image plane, and makes a NEW image plane (the Third image plane) that is the desired size. There is a reticle placed in this last image plane, and the eyepiece focuses on the reticle AND the image at the same time. When things are good, that's how the scope works!"

So to summarise, the knobs adjust elevation and windage by moving the erector lens/cell. The reticle can be placed in either the second (magnifying) or third (non-magnifying) real image plane. Thus, you probably adjust parallax by moving the reticle or something in the ocular to align the image planes.

 

[Andy Resnick]

That's a nice website- very informative. Does it make more sense to you now?