# How do parallel reflected rays meet to form image at infinity? If they never meet then how is image formed?

In my textbooks it is written that when an object is kept at focus, its image is formed at infinity and is real. But how is this possible because parallel lines never meet and it is necessary for rays to meet to form image.

Furthermore on youtube there are videos where the object is kept at focus of concave/convex mirror/lens respectively its image is formed very far. How is this posible as parallel rays never meet??

You are correct that parallel rays never meet. Saying "the image is formed at infinity" is a loose way to say that the image distance $$d_i$$ approaches infinity as the object distance $$d_o$$ approaches the focal length $$f$$:

$$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \implies d_i = \frac{1}{\frac{1}{f}- \frac{1}{d_o}} = \frac{d_o f}{d_o - f}$$ $$\implies \lim_{d_o \rightarrow f} d_i = \infty$$

It is indeed not possible to form an image when the object is in the focus plane of a lens. Normally, a lens maps point in a plane to other points in according the lens equation: $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ where $$d_o$$ is the distance of the object plane to the lens and $$d_i$$ is the distance from the image plane (Thanks for J. Murray for kindly providing the lens equation). The image that is formed is also magnified by a factor $$M$$, where $$M=d_i/d_o$$.

In the case that the object is in the focus, the rays don't intersect anymore. Points in the object plane are mapped to angles and the magnification becomes infinite:

The reason that those videos are able to create a focus by placing an object in the focus point is that the object is not truly in the focal point. Otherwise the image would be slightly out of focus. But, the farther away you get from the lens, the less this tiny bit matters. If you are far enough away, you won't see a difference between an object in the focal or an object slightly away from the focal point.

Finally, I would like to remind you that we always have a lens on us: our eyes. Our eyes are able to focus on objects "at infinity", which is why we can focus on mountains or stars even though their rays are basically parallel after travelling for kilometers or even light years. Well, at least my eyes can't anymore because I have glasses but you get the point.

Site used to create the image: https://phydemo.app/ray-optics/simulator/ You should check it out. It's really nice.

Not just in optics, but more generally in physics, the phrase "at infinity" has to be understood in a special way. This is because infinity is itself a special mathematical term. In the case of integers, for example, we should not think of "infinity" as a very large integer. It is a more subtle concept than that.

Coming now to optics, the idea of "forming an image at infinity" is a way of talking about a limiting case. For an ideal lens if the object is at a distance larger than the focal length then there will be a real image at the location given by the formula $$1/u + 1/v = 1/f$$, where $$u$$ is the object distance and $$v$$ is the image distance. Solving for $$v$$ we have $$v = \left( \frac{1}{f} - \frac{1}{u} \right)^{-1}$$ Mathematically, as $$u \rightarrow f$$ we have $$v \rightarrow \infty$$. Physically, the location of the image is at a distance further and further away as the object approaches the focal plane. The phrase "at infinity" is a useful shorthand for these facts. You are correct to say that, in Euclidean geometry (which is the appropriate assumption here), parallel lines do not meet. But if the distance $$u$$ is just a very small amount larger than $$f$$ then there will be a real image at a very large distance and this is a useful thing to know.

There is a point which can complete the previous answers: geometric optics is an approximate theory and the rays never really intersect at a point: the image of a point is a spot and what matters is the size of the image spot.

In the case of an object very close to the focus, the theoretical image is very distant but above all, the depth of field is enormous: if you move the screen further or closer, the image will remain sharp.

Let's take an example: a converging lens of 10 cm focal length and an object point $$A$$ on the axis 1 mm before the object focus $$F$$. In theory, the image is easily found using the Newton relation $$\overline{FA} \overline{F’A’} = -f^2$$. So $$\overline{F'A'}=\frac{10^{-2}}{10^{-3}}=10$$ m and the magnification is very close to $$\frac{10}{10^{- 1}}=100$$. In principle, if the object $$AB$$ perpendicular to the optical axis measures 1 cm, its image $$A'B'$$ is at 10 m from the lens and $$A'B'=$$ 1 m .

But what happens if I put the screen at 20 m from the lens. Suppose the lens's diameter is 5 cm. The beam coming from $$A$$ and converging towards $$A'$$ is a cone of angle at the vertex $$\frac{5^{-2}}{10}=10^{-3}$$ rad and therefore, if l If the screen is 10 m back, the size of the spot will be 5 cm (evidently !). We must compare the size of this spot to the centers of the beams : 2 m. 5 cm is not much compared to 2 m and the image will remain sharp.

Ultimately, by bringing the object closer to the focus, each point $$A$$ or $$B$$ will give a parallel beam of 5 cm in diameter and as soon as the distance between the centers of the two beams is large compared to 5 cm , the image will be sharp. For an object $$AB$$ of 1 cm placed 10 cm from the lens, the distance will be 1 m at 10 m from the lens: beyond that, the distance no longer matters: we are "at infinity".

In geometric optics, the "position of infinity" depends on the size of the object, the size of the lens and the resolution of the recording system.

This phenomenon is well known in the other direction, when we make an image on a screen of an object “at infinity”. Just look at the adjustment ring of an old camera: infinity is a few meters in front of the lens!

As a joke, we can note the difficulties our philosopher or mathematician colleagues have in mastering the infinite. In optics, it's much calmer: we can say, infinity is there, at 5 m !

To conclude, do not forget that all optical instruments have aberrations and that the notion of geometric image is an approximate concept.

Hope it can help and sorry for my poor english !