I have a few fundamental questions in optics about the focus and blurry images.

Each text book says that according to this picture the object image is the same.

1) If we move the screen to point further then the image will be blurry. What causes it to be blurry?

2) Why do we need the rays to meet in the same point for focus?

3) What does blurry mean at lowest level possible?

4) As far as I see the objects top will now be in 3 different places. We should see the reflection of the top 3 times then?

5) Also, not so relevant to the to the main topic, but from that position, suppose we move the object right until point f. At point f, we stop seeing the object image on screen or without. What is happening at point f so we don't see the object anymore? Im asking this because I think it is really connected.

enter image description here



In the paraxial model of geometrical optics a lens can make all the infinitely many rays from an object point hitting the lens converge to a single infinitely small geometrical point. This is the definition of the image point in the world of paraxial optics. We can find the position of the image point by computing where the rays cross. Since all paraxial rays from the object point cross at this point, we can find the position of the image point by choosing only two arbitrary rays out of all the infinitely many rays and compute where these two rays cross. Two paraxial rays are enough to find the position of the image point, but they are not enough to make an image. To make the image itself, we need to have a crossing of all the infinitely many paraxial rays from the object point. In real life we may see images at other places, but this can only be studied using the model of wave-optics.

If a receiving plane is displaced a little from the sharp image point, all the infinitely many paraxial rays go on, spreading out from the image point, filling a cone of light, making a circular or elliptical patch of illumination in the displaced receiving plane.

An extended object consists of infintely many object points. If we consider two points close to each other, we now see two circular patches in the displaced receiving plane, each patch contributing its own illumination. At the places where these patches overlap, the resulting illumination is the sum of the illumination due to each patch. The overlap of the patches is given by the distance between the two points in the object. If the two points are very close to each other, the patches will overlap almost completely, and may seem to be one patch instead of two. We may therefore think that there is only one object point instead of two close points. In the displaced receiving plane we have therefore lost the ability to detect small detail, namely that there are really two close points instead of one, and we would feel that the image in the displaced receiving plane is blurry. If the intensity distributions of the patches are known, it is possible to use computer image processing techniques to retrieve the sharp image from the blurry picture.


Thought I'd pop it into an answer for clarity - and more words:

Your picture answers these well. Look at those blue lines. These are light rays from the point at the top of the Object.

At the real image point, they are all together, in another nice sharp point. This is in focus.

Point 4 needs some clarification - the diagram is a construct to demonstrate how light rays are affected by the lens. At the screen they are spread apart - with light rays from that point appearing anywhere between the blue lines, and light from all other parts of the object doing the same. So the image of the top of the object now becomes a smear across the screen.

Oh, and at point F - it will be a small blur on the screen. It doesn't vanish.

To try and help with an analogy, imagine drawing a picture using a sharp pen. You can draw a perfect copy of an image. Now try with a bundle of paintbrushes tied together - you'll have a very blurry picture.

  • $\begingroup$ Please look at M .Enns's response comment $\endgroup$ – The Capacitor Aug 23 '16 at 19:17
  • $\begingroup$ Ivan - what do you mean? Your comment on his post still makes no sense. It's the same thing you said in your point 4. It's not 3 or 5 places - it's smeared between them $\endgroup$ – Rory Alsop Aug 23 '16 at 20:00
  • $\begingroup$ Why not 3 or 5 place? $\endgroup$ – The Capacitor Aug 24 '16 at 4:02
  • $\begingroup$ There are not 3 or 5 lines. That is just a diagram. Really there are infinitely many "rays" of light $\endgroup$ – Rory Alsop Aug 24 '16 at 6:12
  • $\begingroup$ Of course, I get that. I'm trying to understand it at the lowest level possible. $\endgroup$ – The Capacitor Aug 24 '16 at 7:29

An image looks like an object because rays of light come from the some point in space (where the image is) in the same way that they come from the object.

Consider a single point on the object, say the top of the object as shown in the diagram, rays will spread out all originating from that point. To have something that looks like the top of the object you would have to have all these rays of light coming from another single point. This point is the image of the top of the object. There is only one place where this happens. Your diagram shows only the three easiest rays to trace but we could draw any number of rays all originating from the top of the object and passing through the image.

Near, but not at, the image these rays will all pass through a small area but not exactly the same point. If you put a screen there you get a blurry, out of focus, image. What is happening is that rays from each point in the object arrive at the screen spread out. Another way of looking at it is that at each point on the screen you are getting rays from a little area on the object rather than a single point.

The reason the light rays all converge to a point and you do get an image is a result of the special geometry of the lens. Other shapes wouldn't work.

If you put the object at the focal point then the rays refracted by the lens end up parallel and thus never converge and no image is formed.

  • $\begingroup$ " To have something that looks like the top of the object you would have to have all these rays of light coming from another single point." - I have a problem with this sentence. Why do we need all rays from the top? why not 1? or 3? Also, same about the blurry picture. Why exactly do we see blurry picture? Because not all rays from certain points meet at certain point? Why do we need all those rays at one point for that? why don't we see 5 tops? $\endgroup$ – The Capacitor Aug 23 '16 at 19:14
  • $\begingroup$ The diagram which you show, which is a standard type of diagram, only shows 3 rays, called the principal rays. Actually there are thousands of rays leaving that point and travelling through the lens. If those rays don't meet at a common point, you won't have a sharp image. The more rays that don't meet, the more distorted/blurry the image will be. When you change the position of the lens without changing the object or screen position, you DO see thousands of "tops," which means you see a blurry or smeared image that your eye and brain cannot resolve into distinct images. $\endgroup$ – Bill N Aug 23 '16 at 20:09
  • $\begingroup$ @BillN Ok, but why if I see thousands of "tops" it is smeared then? Is it really the brain resolving or more collissions between rays coming from different parts of the object? This is also what M. Enns sais - "Another way of looking at it is that at each point on the screen you are getting rays from a little area on the object rather than a single point." $\endgroup$ – The Capacitor Aug 24 '16 at 4:29
  • $\begingroup$ No. It's not "collisions" - it's lots of light from all the points of the object hitting the screen in different places. $\endgroup$ – Rory Alsop Aug 24 '16 at 6:14
  • $\begingroup$ @Rory Alslop so if look on a really small point, what we acctually may see is 1 ray from the top, 2 rays from the top - 0.1 mm, 5 rays from top - 0.2 mm and so on? And thats why the point is blurry? Of course the numbers are arbitrary. And the reflections of these rays we acctually see, is is propotional to the rays incoming that point? $\endgroup$ – The Capacitor Aug 24 '16 at 7:28

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