Jump to content

Aperture and Field of View


Dalo

Recommended Posts

9 minutes ago, Dalo said:

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.

Beams as in collimated light?

Yes, there will be a difference, because there is little light that's not near the center of the aperture, if you have a fundamental Gaussian mode.

Two lenses with a pinhole at the common focal point is a spatial filter. The spatial profile of the beam in the plane of the pinhole becomes a Fourier transform of the modes of the light that hit the lens. If it's just the fundamental (0,0) mode, the light goes through. But higher-order modes don't go through the center, and are filtered out. It's a way of "cleaning up" laser beams.

250px-Hermite-gaussian.png

Link to comment
Share on other sites

5 minutes ago, Dalo said:

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

You don't want to discuss what the outcome of your actual experiment is expected to be? Or why you expect the results of your actual experiment to be different from that predicted by optics?

And why isn't taking pictures a good enough experiment? If you want to see if there is any difference between a general scene and light from point sources/lasers then why not take pictures of those two things?

Link to comment
Share on other sites

Or how about just use your eyes. If your pupils dilate or constrict, does your field of view change suddenly? Neither theoretical nor empirical evidence point to a change in the field of view.

Edited by CharonY
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

2 hours ago, Dalo said:

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

!

Moderator Note

This attitude is inconsistent with our science discussion forum. You have an understanding problem with mainstream science, and in order to discuss it with you, you can't exclude information that might bring clarity to your understanding. That isn't how real life works, and it's certainly not how science works. 

It's hard for most of the members who engage in discussion with you to understand why current science seems like so much magic to you. You have over 200 posts questioning basic physics and it still seems like you're fairly willful about "not getting it". Still, others are trying to help.

So you will answer questions put to you, or your threads will get closed. That's pretty easy, it's in the rules. The rest is becoming less and less up to you. Your posts are being reported more and more for their abuse of reason. A change of strategy is recommended.

 
Link to comment
Share on other sites

10 hours ago, swansont said:

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.

If this wasn't the case the diameter of the lens used (both in bench optics and in photography) would matter to the field of view. It doesn't. 

I am struggling to see how this is such a long thread! The general problem was solved for both a simple case and with a little extrapolation to some camera optics in the first few posts. 

 

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

1 minute ago, Dalo said:

 

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

No, we can apply the well understood and incredibly well tested theory's of electromagnetic propagation that we have. This isn't just guesswork or assumption. 

Scientist don't just run experiments because they seem like a good idea or some guy on the internet has a pile of misconceptions. 

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

3 hours ago, Dalo said:

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

Anyone with a basic knowledge of science knows exactly what the outcome will be. 

9 minutes ago, Dalo said:

But then how come we are still able to get an image of the blocked source?

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

9 minutes ago, Dalo said:

Others apparently do not need such an assurance. I feel I do.

You can do the experiment for a few dollars, so let us know how you get on.

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

5 minutes ago, Dalo said:

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.

Because, from every point on the subject, rays can go though every part of the lens to form the image. So if you block some of those rays, you reduce the total number of rays forming the image but you don't change where they come from.

To put that in terms of your lasers, if you were to position one at the extreme edge of the field of view and aim it through the edge of the lens then it would be blocked when the diaphragm is shut down. If you leave the diaphragm in that position and, without changing the position of the laser, rotate the laser to go through the centre of the lens then it will again appear in your image. 

The same thing is true for a laser position in the middle of the field of view. If you aim it at the edge of the lens than it can be blocked by the diaphragm. If you aim it near the middle of the lens, then it won't be blocked by the diaphragm.

What that shows is that some rays from every part of the scene are blocked by the diaphragm (so it becomes universally darker as you stop down) but some rays from every part of the image will always reach the film/sensor.

This is what my diagram was supposed to illustrate. Apparently you missed the point but I'm afraid I can't think how to make it any simpler. (Which seems to be a recurring theme on your threads.)

Again, maybe if you draw the paths of the rays yourself, it might make more sense to you.

Or perhaps you can get a 5 year old to explain it to you? :)

13 minutes ago, Dalo said:

(with the proper filters)

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

Link to comment
Share on other sites

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.

Edited by Dalo
Link to comment
Share on other sites

Just now, Dalo said:

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.

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

You need to draw a diagram showing where these light sources are and where the rays from them are going. How many light sources do you have in your setup, for example? And are do you mean 3 beams from each light source? Or 1 from each? Or ...? Are they directional (lasers) or just light bulbs?

Quote

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

Five what "passing through the lens"?

Quote

Use a gray filter, or if the light sources can be dimmed, reduce the intensity of the beams to show the lamps themselves. 

Why do you need a filter? Why do you think that the lamps can't be seen? 

Quote

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

There isn't enough information in your vague description to say. 

Quote

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.

Draw a diagram and it might be possible to answer and explain why.

Link to comment
Share on other sites

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.

 

Link to comment
Share on other sites

2 minutes ago, Dalo said:

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.

I think you have avoided answering any questions which would make your intention clearer. 

I can only analyse my example, because it is the only one on offer.

3 minutes ago, Dalo said:

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

So you have 5 parallel beams going through the lens? Is that right?

So, with no diaphragm we get 5 spots from the 5 lasers:

Untitled.png.e5416df75c853944e045cb65f523a559.png

With the diaphragm closed down a bit, you get 3 spots:

Untitled2.png.65c72c17e611886a17f67d3679a06c89.png

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

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

2 minutes ago, Dalo said:

Yes, and my description was very clear on that. As were my explanations afterwards. 

As many people failed to understand, that may be debatable. But let's not.

Any other comments on my diagrams? Or shall I just cut straight to reporting this for another trolling attempt.

Link to comment
Share on other sites

1 hour ago, Strange said:

To put that in terms of your lasers, if you were to position one at the extreme edge of the field of view and aim it through the edge of the lens then it would be blocked when the diaphragm is shut down. If you leave the diaphragm in that position and, without changing the position of the laser, rotate the laser to go through the centre of the lens then it will again appear in your image. 

The same thing is true for a laser position in the middle of the field of view. If you aim it at the edge of the lens than it can be blocked by the diaphragm. If you aim it near the middle of the lens, then it won't be blocked by the diaphragm.

What that shows is that some rays from every part of the scene are blocked by the diaphragm (so it becomes universally darker as you stop down) but some rays from every part of the image will always reach the film/sensor.

This seems key to the missing understanding. 

Light from every point you are imaging is radiating in every direction, so in ray optics you can drawn a ray hitting any part of the lens. 

Link to comment
Share on other sites

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?

Link to comment
Share on other sites

22 minutes ago, Dalo said:

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.

Because you have only considered a single ray (beam) from each source. You need to consider all the possible paths from that rays can take each source through the lens to the film.

See, for example, the second post in this thread.

To relate this more specifically to your example, lets look at a few of the other rays (highlighted in red) from the position of the top laser (assuming it were not actually a laser but a normal light source just an object being photographed):

Untitled.png.7313b0481cedfd5e8bf238113b6186de.png

This is important because otherwise you would get light sources disappearing but you wouldn't see the rest getting dimmer. However what we actually see is all the sources getting dimmer because they all have rays that go through all parts of the lens and so all of them have some of their rays blocked. Let's look at a couple of the rays from the position of the central laser:

Untitled.png.ae1c0513b49ef08bc712eb044f5b8b84.png

They get blocked so the light or object at that point will also be dimmer than it would be without the diaphragm.

Edited by Strange
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

Just now, Dalo said:

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

No. You are comparing two different things. (That is the trouble when you make vague and confusing/confused descriptions instead of being explicit and, you know, drawing a diagram or something.)

With lasers aimed parallel to the lens you will see them disappear as you close the diaphragm. Because there is only a single path.

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

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.