Optics: Focusing Explained with an Owl

[cronnin]

I've made a drawing of what is confusing me here.
1. This should be a case of focusing into infinity or not?
2. Can anyone explain what happens with the owl?

 

[Ibix]

You haven't marked on the focal length of the lens, but as drawn it does seem to be bringing rays from infinity to a point, yes. That means it will not form an image of the penguin or the owl, contrary to the diagram.

Do you know how to construct a ray diagram?

What's the point of all the background stuff? It just makes the lens and the rays difficult to see.

 

[cronnin]

You haven't marked on the focal length of the lens, but as drawn it does seem to be bringing rays from infinity to a point, yes. That means it will not form an image of the penguin or the owl, contrary to the diagram.

Yes, the focal plane should be where parallel rays converge into point. The distance between the lens and that plane should be the focal distance?

Why won't it form an image? If it focuses to infinity it must form some kind of image somewhere on the right side?

Do you know how to construct a ray diagram?

That's the only thing we ever did. That's why I don't understand real examples.

What's the point of all the background stuff? It just makes the lens and the rays difficult to see.

That's the only image I could find with parallel rays of light that not parallel with optical axis of the lens at the same time.
I'll try to make a drawing next time.

 

[Andy Resnick]

Why won't it form an image? If it focuses to infinity it must form some kind of image somewhere on the right side?

The drawing is missing the location of an image plane; loosely speaking the image plane is where a screen is located for you to view the image. As presented, only the focal plane of the lens is shown (the common plane where the initially parallel rays converge). The focal plane is not generally an image plane.

 

[Ibix]

Yes, the focal plane should be where parallel rays converge into point. The distance between the lens and that plane should be the focal distance?

Yes.

Why won't it form an image? If it focuses to infinity it must form some kind of image somewhere on the right side?

It will form an image there, but only of something infinitely far to the left. Or, at least, a very long way to the left. It will also form an image of the owl, just not in the focal plane because that's not where initially diverging rays cross after they pass through the lens.

That's the only thing we ever did. That's why I don't understand real examples.

You are misunderstanding ray diagrams then, because they're all you need to understand this. Just pick a point on your penguin, draw one ray straight through the centre of the lens, and one ray parallel to the axis to the lens then down through the focus. Where they cross is the plane where the image will form. You should get the same plane for rays starting from any point in the same initial plane.

Alternatively you can use that, if the distance from the object to the lens is $u$ and the focal length of the lens is $f$ then the distance from the lens to the image is $v$, where $$\frac 1f=\frac 1u+\frac 1v$$Be careful with the sign convention - $u$ gets more positive to the left and $v$ more positive to the right in the way I have written this.

That's the only image I could find with parallel rays of light that not parallel with optical axis of the lens at the same time.
I'll try to make a drawing next time.

Fair enough - you have to work with what you've got. That said, I don't like the diagram. As a general rule when drawing diagrams only draw what you need.

 

[cronnin]

The drawing is missing the location of an image plane; loosely speaking the image plane is where a screen is located for you to view the image. As presented, only the focal plane of the lens is shown (the common plane where the initially parallel rays converge). The focal plane is not generally an image plane.

1. Let's put a lens beside the white wall and fix the distance to be same as the lens' focal distance . If we turn on a huge LCD screen at large distance, it should form a sharp image on the wall? The rays coming from infinity should converge at focal distance?

2. If we move LCD closer to the wall, the sharp image would form on the plane that's inside the wall?

3.So, now, if we want to shift it back to the wall plane, we should insert one lens between the wall and the lens, that would make rays converge faster?

4. Now we pull the LCD back to infinity and make fine adjustment to lens-wall distances make the system focus to infinity, and the minimum focusing distance will depend on the system properties (lens diameter, curvature, refraction index, etc)?

 

[cronnin]

You are misunderstanding ray diagrams then, because they're all you need to understand this. Just pick a point on your penguin, draw one ray straight through the centre of the lens, and one ray parallel to the axis to the lens then down through the focus. Where they cross is the plane where the image will form. You should get the same plane for rays starting from any point in the same initial plane.

At the moment I think I'm having a problem understanding what this actually means?


Are the images at 5:00 and 5:20 cases of focusing object at infinite distance or not?

 

[Ibix]

1. Let's put a lens beside the white wall and fix the distance to be same as the lens' focal distance . If we turn on a huge LCD screen at large distance, it should form a sharp image on the wall? The rays coming from infinity should converge at focal distance

Assuming it's a convex lens, yes. This is basically how you use a magnifying glass to start a fire on a sunny day.

2. If we move LCD closer to the wall, the sharp image would form on the plane that's inside the wall?

Yes.

3.So, now, if we want to shift it back to the wall plane, we should insert one lens between the wall and the lens, that would make rays converge faster?

Yes.

4. Now we pull the LCD back to infinity and make fine adjustment to lens-wall distances make the system focus to infinity, and the minimum focusing distance will depend on the system properties (lens diameter, curvature, refraction index, etc)?

This seems to me to be asking if the behaviour of a system depends on the properties of the system. That is trivially true. If you had a more specific question in mind I didn't see what it was.

Are the images at 5:00 and 5:20 cases of focusing object at infinite distance or not?

Both represent light coming from infinity. The first one is light from a point on the optic axis - hence the image is on the optic axis. The second is light coming from two points, one far above the optic axis and one far below - hence the image is two points, one below and one above the optic axis (the image is upside down).

 

[cronnin]

This seems to me to be asking if the behaviour of a system depends on the properties of the system. That is trivially true. If you had a more specific question in mind I didn't see what it was.

It was just me, thinking loud, and I agree that my comment was a bit redundant. I'm trying to understand magnification and how projector lens zooms the picture on the wall, and I first had to be sure about what seems trivial for the rest of you :) so that I could read further about it.

Both represent light coming from infinity. The first one is light from a point on the optic axis - hence the image is on the optic axis. The second is light coming from two points, one far above the optic axis and one far below - hence the image is two points, one below and one above the optic axis (the image is upside down).

Oh, so the drawing is a theoretical case? There is no compression of cubic reality (@5:00) into a single point, those parallel rays are coming from a single point.

 

[Andy Resnick]

1. Let's put a lens beside the white wall and fix the distance to be same as the lens' focal distance . If we turn on a huge LCD screen at large distance, it should form a sharp image on the wall? The rays coming from infinity should converge at focal distance?

If you set the image plane to be coincident with the back focal plane, then you are indeed imaging objects located at infinity. Now, the image-object relationship is better described in terms of angular rather than linear magnification. And by 'sharp', you mean 'in focus' (there's still diffraction, which is most pronounced in this particular imaging arrangement).

2. If we move LCD closer to the wall, the sharp image would form on the plane that's inside the wall?

As the object plane moves closer to the lens, the image plane moves *away* from the lens, until the LCD is at the front focal plane and the image is now projected at infinity.

3.So, now, if we want to shift it back to the wall plane, we should insert one lens between the wall and the lens, that would make rays converge faster?

This is poorly defined. I think you are asking to add a second lens element to make the compound lens have a shorter effective focal length, which is IIRC the definition of a telephoto lens. Telephoto lenses typically have a rear negative element to accomplish this.

4. Now we pull the LCD back to infinity and make fine adjustment to lens-wall distances make the system focus to infinity, and the minimum focusing distance will depend on the system properties (lens diameter, curvature, refraction index, etc)?

Not sure what you are getting at here. The minimum [object] distance is typically set by design metrics (maximum allowable aberration, for example).

 

[Ibix]

Oh, so the drawing is a theoretical case? There is no compression of cubic reality (@5:00) into a single point, those parallel rays are coming from a single point.

Depends what you mean. Yes, those rays are coming from a single point. You can think of them as diverging at 0.000000001° if that helps. Regarding the "cubic reality", though, you are taking a photo. The photo is a 2d representation of the 3d world, so what you see is a projection of the world, rather than a compression.

 

[cronnin]

Depends what you mean. Yes, those rays are coming from a single point. You can think of them as diverging at 0.000000001° if that helps. Regarding the "cubic reality", though, you are taking a photo. The photo is a 2d representation of the 3d world, so what you see is a projection of the world, rather than a compression.

Yes, that was also my mistake in thinking. It seemed for a moment that 2D owl (from the first post ) would get compressed into a single point, but that's not the case since a whole owl cannot fit in there. Thanks for all the input ppl. It's all much clearer now.