# Why do we see a blurred image when the screen is not placed at the position where the image is formed by a convex lens?

Regarding the proposed duplicate:
Why does the image lose its clarity when it is formed in front of the retina?

Thank you for sharing that question. The only answer there was informative. However, I find my question to be an extension of that answer. I understood that we observe blurred images as point sources do not focus at a single point on the screen. But I don't understand how real image is formed when light rays diverge from a point in front of the screen as explained in this question using the second diagram. Here, I'm looking for an answer with rigorous explanation of physics behind blurred images.

In our practical session, in order to determine the approximate focal length of a converging lens we were asked to focus a distant object on a screen and measure the distance between the lens and the screen. We had to adjust the distance of the screen from the lens until we obtained a sharp image. When the screen was not at the focal plane we obtained a blurred image. This is how we had to find the image location for all other positions of the object.

I don't understand "Why do we see a blurred image when the screen is not placed at the position where the image is formed by a convex lens?". I tried to reason my observations using the following information.

I've learnt that a real image forms when the refracted light rays actually converge or appear to converge behind the screen, and a virtual image is formed when refracted light rays actually diverge or appear to diverge from a point in front of the screen. I've also verified this fact using ray diagrams and from experiments.

Let us consider the following diagram:

The red dotted lines represent different positions of the screen at different points of time. When the screen is placed at A, real image is formed on its surface as the rays appear to converge behind the screen. As the rays from one point on the object fall on the screen at different points, we get a blurred image. When the screen is at B, we obtain a sharp image as all rays from a particular point on the object converge into a single point at the screen's surface.

But, I was unable to explain why I get a blurred image on a screen when it's placed at C. The real image is formed in front of the screen. The real image formed can be considered to be a real object which emits light in only certain direction away from the lens. Having said that, it's obvious we couldn't obtain a real image on a screen placed at C. This is further supported by the ray diagram itself.

Let us consider the following diagram which consists of only the image and screen C from the previous one:

It can be seen that the rays emitted from a particular point on the object do not meet at all. They diverge from that point on the object and for the screen, it would finally result in a virtual image. And since, virtual image couldn't be obtained on a screen, I don't think we could see even a blurred image. But this contradicts the observations. And so, my reasoning is incorrect. I don't find any inconsistencies with my reasoning. No real image would be formed on a screen if an object if placed in front of it without any optical devices between.

In short, it would be helpful if you could explain why we observe blurred real images on the screen when the screen is not placed where image is formed.

Image Courtesy: My own work :)

• @Ruslan: Thank you for sharing that question and answer. Your answer there was really informative. However, I find my question to be an extension of your answer. I understood we observe blurred images as point sources do not focus at a single point on the screen. But I don't understand how real image is formed when light rays diverge from a point in front of the screen as explained in the question using the second diagram. Thank you. – Guru Vishnu Feb 4 at 14:37
• @JonCuster: No. It was suggested previously by another user and my above comment explains why I'm unable to understand blurring. Also please see the beginning of the question in block quotes. Thank you. – Guru Vishnu Feb 5 at 4:00

If the screen were a mirror, you would see a reflection of the point source and there would be no intrinsic blur; but since you're looking at a screen which scatters light, you see the distribution of intensity at the screen. The image you see on a screen - whether the image is blurred or sharply focused - is the distribution of light intensity on the screen.

You can think of an object - or a real image of the object, as a collection of point light sources. When light from a point light source is brought to a focus on a screen, light intensity from that point source is all concentrated at one point on the screen.

But when light from a point source is not brought to a focus on the screen, it is spread out over some region on the screen. When you look at the screen, you see light (from that point source) over that region instead of at a single point; and you see it as a blurred point.

A real image can only be seen if its rays enter your eye, which means that you need to be on the right-hand side of your diagram, looking back toward the lenses, to see the real image. You can't see the real image if there is a screen intercepting the rays. If there is a screen, you can only see the light intensity at the screen -- which is blurred unless the image is focused precisely on the screen.

If the screen were, for example, lightly scratched plastic, and you looked through the screen back at the real image, you would see both the real image and a blurred image because some of the rays would go straight through the plastic without being scattered (allowing the real image to be seen), and others would be scattered at the plastic.

• Thank you for your answer. I understood why we observe a blurred real image when the screen is placed at A. Here light rays from a single point on the object focus at a point behind the screen and hence a real image or screen compatible image is produced. But when the screen is at C, the light rays appear to diverge from a point in front of the screen. And I don't understand how it could form a real image in the first place and hence a blurred real image? I hope you understood my question. – Guru Vishnu Feb 4 at 5:34
• Please see my edited answer. – S. McGrew Feb 4 at 14:02
• Sure, but I'm unable to see the change. – Guru Vishnu Feb 4 at 14:03
• Sorry, I clicked the wrong button and lost the initial edit. It's fixed now. – S. McGrew Feb 4 at 14:18
• Thank you for your edits. I have positively voted your answer for your kind help and quick response. But, I'm sorry. I still do not understand why a real image must form when rays from a particular point on the object do not converge at all when the screen is not placed at the position where the image is supposed to form in accordance with the lens equation. – Guru Vishnu Feb 4 at 14:50

Actually, your first image is misleading. It makes you think that the only light that is emitted by the effective object is a thin cone that gives you a rather small blurry spot at the plane $$C$$. But that's not quite correct: you've forgotten that the rays from the actual object are emitted not only parallel to the optical axis and through the lens center, but also quite away from the lens center. Let's redraw the ray diagram:

Here I've made the lens much larger (more ideal in the sense of preserving more of the ray directions), so as to catch the rays you forgot. Now this does look quite like your second image, doesn't it?

So, the first answer is that the smaller the lens, the larger the depth of focus.

But that's not all there is to it. Consider a flat picture that we are to take (sort of) a photo of. If we put it on a scanner, which would play the role of the photographic sensor at the plane $$B$$, we'll capture the image perfectly in focus. Let's now put some air gap between the scanner working surface and the picture. This would mean that we've displaced the scanner from plane $$B$$ to plane $$C$$. Do you think we'll get nothing at all? Actually no, the closer the picture to the scanner's working surface, the less blurry the scan will be. It won't abruptly lose all the detail just because we've moved the scanner a tiny bit from $$B$$.

Why is it so? It's because of Lambert's cosine law: as the angle of scattering increases, the amount of scattered light decreases, reaching zero at 90°. So if our picture at the plane $$B$$ were the Malevich's Black Circle*, with a white hole made somewhere in the middle (don't try this at the museum), the light scattered from this hole would reflect mostly in the specular direction, with less light in other directions, so that at plane $$C$$ we'd have more light at the point that the specularly-reflected ray crosses, and some less light around it.

But even if our picture is not a Lambertian scatterer (as isn't e.g. the Moon's surface), there's another factor. Our plane $$C$$ is planar, not circular, so the light emitted by the effective object at $$B$$ would spread out more at the periphery than at the point where the shortest line segment between the effective object and $$C$$ crosses $$C$$. This will also reduce the circle of confusion.

*Why not the more famous Black Square? Because it's now not in a good shape: it's covered in noticeable craquelure, so the it'd be a bad example of uniformly black surface.

• +1 Thank you for your answer. Now I see that the criterion for the formation of real and virtual images based on the convergence or divergence of the light beam is not totally correct. Since the condition for a real image (rays either converge or appear to converge behind the screen) is not met when the screen is at C, do we call the illumination pattern obtained a "real" image? – Guru Vishnu Mar 25 at 13:17
• @GuruVishnu I think we consider there being no image at $C$. In geometric optics, an image is a pattern that is virtually indistinguishable from the object (aside from the size; not sure of the distorted images due to curved mirrors and the like). – Ruslan Mar 25 at 14:19
• Fine. Then, I feel it's just a matter of terminology and the definitions have some loop holes. Finally, with reference to the first section of your answer, may I know whether there exists any discrepancy in replacing the real image at plane B as a real object of same size and position for further analysis towards the right? – Guru Vishnu Mar 25 at 14:23
• @GuruVishnu in geometrical optics, I don't think there's any problem (barring finite lens size). I don't know of any theorem stating this, but I suppose one should be able to prove this from Huygens–Fresnel principle in the appropriate limit. – Ruslan Mar 25 at 14:33
• Thanks. It was certainly useful to think, the wave fronts in the advancing side of the focal plane just looks like that of a point source and so even from wave optics point of view, this must hold true. I understood this fact. I was just puzzled because the ray diagram in your post didn't make any difference in my understanding of the image-object replacement described in my question. I just wanted to be sure not to miss any minor points from your answer. – Guru Vishnu Mar 25 at 14:38

So, your question is why do we call something an image when the light rays do not meet at all, isn't it?

There is a term called resolution, it is defined as number of pixels per inch (pixels you can understand as the light source). See these two images (image means picture, not the physical definition)

The first picture is blurry but the second one is seemingly nice, the number of pixels per inch (resolution) in the first picture is less than the second one, but you see the image is still forming. The spacing between the pixels is more (in the blurry one) and hence we are getting it not exactly, the analogy is same as the actual drawing and those dot drawing

You see, the dots still represent the picture but the spacing between the dots is causing us to differentiate it between the actual drawing, because when draw smooth curves or lines there are points everywhere on that curve sketch or line and hence no spacing, therefore a clear image.

When we put the screen just after the image (here I'm talking about your case) we get the images of every point of the object but they are a little further than they were in actual object and hence we get the image as the blurry one.

Hope this helps !

• Thank you for your answer. "So, your question is why do we call something an image when the light rays do not meet at all, isn't it?" : Short answer: no; Long answer: I understood we obtain a blurred image when different light rays from a single point on the object are incident at different points on the screen when it's at "A" but not when the screen is at "C". This is because in order to obtain a real image the rays must either converge on the surface or behind the screen. In "C" the rays appear to diverge from a point "in front" of the screen and this is the source of my confusion. – Guru Vishnu Feb 4 at 5:49
• In "C" the rays appear to diverge from a point "in front" of the screen and this is the source of my confusion. what's the confusion in this? – user240696 Feb 4 at 5:53
• When rays appear to diverge from a single point in front of the screen we could very well replace the real image by a hypothetical real object, which emits light rays only in a specific region towards right. To put simply, if we place a well lit candle in front of a screen how could we see it's image? I'm not saying the illuminance created by the flame but the image from which we could interpret different features of the flame. – Guru Vishnu Feb 4 at 5:57
• which emits light rays only in a specific region towards right why? What made you to conclude that? – user240696 Feb 4 at 6:02
• Obviously, the image as an entity in space cannot emit light rays in all directions. Incident rays travel from left to right in the question. And so without any reflection a ray cannot travel from right to left. Why "specific region"? This region or solid angle through which the rays are emitted depends on the size of object, aperture of lens, positions of object and image, etc. In short, you can visualise the boundary lines of this region as the rays passing through the upper and lower extremes of the aperture of the lens. – Guru Vishnu Feb 4 at 6:07