Dalo

Just now, Strange said:

With omnidirectional light sources (multiple paths) or with lasers aimed towards the centre of the lens they will not disappear.

I do not understand where you get those omnidirectional light sources from. We are talking about 5 beams just like in the video.

The problem with this explanation is that we have two principles from the same light source:

1) a collimated beam that we can follow from the source, through the lens and onto a screen; in the video we see that it leaves a visible trace on the white board.

2) random rays that are invisible to the eye and also to the white board, but still show on the screen.

The same light source is therefore a source of random and collimated beams.

56 minutes ago, Strange said:

So I guess the question now is: do you understand why this doesn't affect the field of view?

Yes. And what I would like to understand, is once again how both facts can be true at the same time:

1) only three beams reach the screen;

2) 5 lamps are projected on the screen.

On 11/28/2017 at 3:38 PM, Dalo said:

The simplest setup would look like this on an optic rail, from left to right:
1) a laser ray box (at least 3, preferably 5 beams or more)

what is unclear about this?

Yes, and my description was very clear on that. As were my explanations afterwards. That was the reason why I stopped reacting because it seemed that everybody was just avoiding the subject and demanding that I deal with their own examples instead of with mine.

1 minute ago, Strange said:

I think it is easier to consider a single light source but whatever.

I think that you have not read my posts very carefully but analyzed your own examples. One single source would be useless for my experiment.

This is how the setup, as far as the light source is concerned, would look like.

14 minutes ago, Strange said:

Filters? Where did that come from? Are you completely changing the subject again?

Nope, you should read my posts, is all.

Your explanation is lacking for the following reason.

Draw lines, as you suggested, from the light sources through the lens, in such a way that, for instance, 3 beams get through the diaphragm, while two are blocked.

You have now an image of three light sources. The image will of course be darker than one with five passing through the lens.

Use a gray filter, or if the light sources can be dimmed, reduce the intensity of the beams to show the lamps themselves. That is what one does when one wants to take a picture of the sun.

The question now is. How many lamps will appear on the screen/picture? Three or five?

If it is three lamps, the same number as the beams which have been let through, then all is well that ends well and I will shut up.

If it is five lamps then it becomes interesting.

edit: the dimming of the beams is more for us, because otherwise  we would not see the lamps because of the intensity of the light.

Just now, Strange said:

See the second post in this thread. Feel free to ask questions about it, if you think it is not sufficiently clear.

That is not the problem. The issue is how a beam can be blocked while we still can get an image of the light source (with the proper filters). The combination is a problem, not that one or the other happens.

The experiment proposed here is really nothing new. It just makes things very explicit. We are all familiar with the fact that pictures of a light source get darker the smaller the aperture is. That can be explained by the fact that less rays from the source go through the diaphragm.
There is, apparently, no reason to expect a different result from a laser ray box (or a normal lamp shining through slits). Smaller apertures will give darker images. 
The only, big, difference is that the different sources are perfectly delimited. If a light source is blocked, we know why the image gets darker.
But then how come we are still able to get an image of the blocked source?
It would seem that the reflection of the light source follows another path than the beam. 
Or maybe not. Maybe it is easily explainable by the laws of optics and em waves.
Maybe I am making the wrong assumption by thinking that an image of the blocked source would be projected on the screen or the sensors.

That is what the experiment aims at making clear. Others apparently do not need such an assurance. I feel I do.

On 11/28/2017 at 3:38 PM, Dalo said:

edit 2: by putting gray filters in front of the beams we would decrease their intensity and make it possible for us to get a picture of the lamps themselves, as we would of the sun. My prediction is that the number of bright dots representing the lamps would remain unchanged, whatever the position of the diaphragm and its aperture.  In contrast with the number of beams that would still be able to reach the screen.

If this expectation is confirmed it would mean that the projection of the image of the lamps (of the object or scene) would follow different rules than the path of the beams. Otherwise, if the expectation is falsified, nothing has to be changed in optical theory.

As long as the experiment has not been done all anybody can do is speculate on the outcome.

4 hours ago, Dalo said:

The simplest setup would look like this on an optic rail, from left to right:
1) a laser ray box (at least 3, preferably 5 beams or more)

It does not need to be lasers. A strong lamp and a cover with slits would work as well.

I will only react to posts concerning actual experiments, and not ideas about how or what they should look like.

1 minute ago, swansont said:

No, it's not. The number of rays going to a particular point on the image only affects the brightness.

I recall a talk from an AAPT conference I went to in grad school, on teaching methods in optics. Students had been asked what would happen to an image if part of the lens was blocked. The misconception that part of the image would be missing was common. All that does is make the image dimmer.  

Blocking the rays reduces the amount of light getting to the image. But light from any point on the object will will pass through any point you pick on the lens. The same concept applies to the aperture area — there will always be light going through the center of it, from all points on the object, to form the image.

Millions of photographers have done so. Changing the aperture does not change the field of view — that's a function of the focal length of the lens. It's why a wide-angle lens is always a wide-angle lens. It does not become a "regular" lens by changing the aperture, or vice-versa. There wouldn't be much of a point to having these different lenses if the aperture had this effect.

I completely agree with you as far as it concerns illuminated objects or scenes. The experiment I propose is to see whether the same principle is applicable to beams of light. Strange seems to agree that some beams would be blocked by the aperture.

Those are two different situations and it would be very interesting to see if the difference between them is real, and how we could interpret this difference. 

But first I would like to attest, in a non-amateurish way, whether there is indeed a difference.

1 minute ago, Strange said:

Why would they? Why not just take some photographs? Or draw a diagram? Or learn some basic optics?

If nobody does it, I will just have to keep saving for the proper equipment and do it myself. It might take a while though.

4 minutes ago, Strange said:

This, equally, obviously, has no effect on the field of view. 

yes and no. It depends what you understand with field of view in different situations. I have already agreed  with the situation in which a slide is used, or only the images of the lamps are projected. You agree that some beams will be blocked. Which means that we will see, for instance, 3 instead of 5 beams. That is a kind of diminished field or angle of view, but I will not make a point of this.

More importantly, I hope someone will do the experiment and then we can discuss its interpretation.

The simplest setup would look like this on an optic rail, from left to right:
1) a laser ray box (at least 3, preferably 5 beams or more)
2) a convex lens
3) a variable diaphragm
4) a screen

The first objective would be to position the screen, and/or the other elements, in such a way as to get a sharp image of all beams.

a) Determination of the position of the focal length (where all or most of the rays meet before they diverge again) relative to the diaphragm. (different from the focal plane on the screen)
b) Effects of different positions of the diaphragm on the image.
c) Effects of different apertures on the image.

Expected result: the number of beams reaching the screen will depend on the aperture and position of the diaphragm.

That might seem obvious until we compare it to what would happen if we used a single light source and a slide. In such a case we would expect the field of view to remain unchanged and only the brightness of the image to be affected.
 

edit: I am ignoring the depth of field in all this discussion.

edit 2: by putting gray filters in front of the beams we would decrease their intensity and make it possible for us to get a picture of the lamps themselves, as we would of the sun. My prediction is that the number of bright dots representing the lamps would remain unchanged, whatever the position of the diaphragm and its aperture.  In contrast with the number of beams that would still be able to reach the screen.

1 minute ago, Strange said:

The diaphragm makes no difference to that.

None of the image will be blocked.

 

I really do not want to start a discussion which will only frustrate both of us. 

I prefer to wait for someone with, once again, practical information, or lab gear making it possible to answer my questions. Maybe then it will become apparent that I have asked the wrong thing and should approach the problem differently. But I will only know that in a practical situation.

I thank you once again and advise you not to waste your time.

3 minutes ago, Strange said:

Is this "too theoretical" because it is not for a specific lens

I think you can answer the question yourself. Can you indicate where the image will fall after the diaphragm, and which part of it will be blocked? Also, where is the focal point after the diaphragm, relative to the image.

That is what I mean by practical information, and that is why my preference goes to simple designs, or even  an optical setup if someone happens to have the optical gear in his lab.

Still, I really appreciate your efforts.

11 minutes ago, Strange said:

You are right. And that is also the position of the diaphragm.

Then I do not see why you find my question strange. The farther from the focal point, the more spread out the rays will be, which means that the position of the diaphragm is fundamental. As you said yourself, it is usually placed near the lens, but how near? Too near and many more rays will be blocked than when the diaphragm is placed a little further, when the rays have converged more, taking less space. Very close to the focal point would be more efficient. It would let more rays through even at small aperture. And so on.

But once again, I want to avoid a theoretical discussion, however interesting, and ask for practical information regarding the position of the diaphragm in an existing design.

edit: It would be in fact quite simple to emulate a simple setup. There are variable diaphragms in the market, and optical sets which would make it possible to do such experiments. The problem is their price which I cannot afford.

1 minute ago, Strange said:

Except the question doesn't make sense. What do you mean by "(last) focal point". For an object that is in focus, the focal point (more realistically, focal plane) is on the film.

The focal point is defined as the point where all rays coming out of a lens cross each other and invert the image. It is usually denoted by f. The image, when it is real, is formed usually after the focal point. But then you already knew that.

3 minutes ago, Strange said:

Are you looking for the details of how a camera lens with multiple elements works: what each element is there for, etc?

Not really. In fact, the simpler the better. Something like the diagram in your first reply, the one with a diaphragm.

The EF 200-400mm is also very interesting. I have seen many diagrams like this in my search, and the more complex they are the less information they give me.

What I really want to know is the location of the (last) focal point relative to the diaphragm. That will answer my question in a practical and definitive manner of how aperture and field of view relate. I already know the theory, what I would like is to see how it is put in practice.

Your explanation certainly makes sense. Still, as I said, I consider my question purely as a technical matter, and I would like more than a theoretical answer. I would be very interested for instance in the design of an existing objective, however old. I am not interested in classified or proprietary information and I wouldn't expect anyone of providing such anyway.

This is a purely technical matter. I have consulted a few textbooks on Optics and Photography, but found no clear answer. I hope someone in this forum will be able to help.

As anybody who has used a camera will know, you can regulate the amount of light that falls through a lens in three different ways:

- the sensitivity of the film (ISO-norm) or the sensors (digital cameras);

- by a longer or shorter exposure (shutter speed);

- a wider or smaller aperture.

My problem concerns the last method.

Obviously, the larger the aperture the more light will fall on the film or sensors.

What I do not understand is how the field of view can remain unaffected. Thinking of the way light rays are manipulated by the lens, the only explanation that makes sense to me is that the diaphragm must be placed at, or very close to, the focal point. This way, however small the opening or aperture, the view remains unchanged. 

Does it make sense?

Once again, I am looking for a technical answer, that is, the way(s) it has been solved by manufacturers since cameras exist. It must be common knowledge not protected by patents because it has always existed. But no clear answers can be found in textbooks, only vague allusions about the general location of the diaphragm.

2 minutes ago, studiot said:

 

Complete rubbish.

I just looked at Strange's link (+1 for finding this) and I note that is aimed at 11 year olds.

They are asked to measure (with a ruler) the half wavelength at about 6cm.

You can't get much more direct than that.

 

The (wave)length is the only part of the experiment that is an actual measurement and not taken on trust.

they are definitely measuring something, and calling it "wavelength". In this sense, it is a direct measurement.

But is it the same "wavelength" that is supposed to represent the distance between two peaks/troughs in the direction of the beam?

My answer is no, feel free to disagree.

But then, you have already done so. 

Phi? Would you please close this thread? It is going nowhere.

3 minutes ago, Strange said:

it is a direct measurement

No, it is not. It presupposes wavelength and frequency, and a speed of light. Once you accept that at least two of the terms are correct, you get the third one automatically.

But that is the whole point of the discussion here. Is the way the wave length is calculated legitimate? I did no read any compelling argument besides strong a strong belief in the method.

I am afraid that I have nothing new more to add.

27 minutes ago, Strange said:

Here is one: http://www.planet-science.com/categories/over-11s/physics-is-fun!/2012/01/measure-the-speed-of-light-using-chocolate.aspx (I have never tried this; I don't have a microwave oven.)

Another (much more accurate) method is to use an interferometer: http://physical-optics.blogspot.co.uk/2011/06/michelsons-interferometer.html

I had seen the chocolate story but never really watched it. I will now.

As far as the interferometer is concerned, it is the same kind of set up as discussed in this thread, only the way to produce the image on the screen is different.