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I've attached an image that describes the light-rays refracted through a thin lens.

enter image description here

P' represents the first real image that is created when light is refracted through the first convex mirror. But how does this translate to the second image point P'' through refraction? I understand the convex mirror formula

enter image description here

and how if you plug in the numbers it just works out that Q'' is where it is, but intuitively, this does not make sense. If the image P' is what P'' "sees", as it is the image, how can the refracted image (P'') be on the same side as the object image (in this case P')?

As far as I understand, P' acts as a real object for the second surface (the right side of the lens), and the rays that diverge from P' are what get refracted onto P''. But how is this possible, if they are both on the same side?

enter image description here

As far as I understand, the model above depicts virtual object through refraction. How is this shown in our case?

Emil Sriram
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2 Answers2

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Over time a set of constructions/formulae have been produced which enable you to predict the passage of light rays from "object" to "image".

The light rays from an object diverge and are then made to converge to form a real image which is one that can be seen on a screen although in fact no screen is necessary for the formation of a real image as described as an answer to the post Real images and their formation.

You then "graduate" to having a diverging beam coming out of an optical system.
When those diverging rays are back produced they meet at region which is called the virtual image.
You only "see" that virtual image when the rays which appear to come from it are made to converge to form a real image, for example using a lens as shown below.

enter image description here

The virtual mirror image acts as a real object as far as the lens is concerned and the lens produces a real image.

Now in you diagram incident on the second interface are converging rays which are never allowed to cross because of the second interface being present.
The second interface reduces the convergence of the rays but they still converge to form a real image.

So far we have a real object from which light rays diverge, a real image where convergent rays cross, and a virtual image where divergent rays appear to come from.
The fourth and final piece of the jigsaw is a virtual object, the $P'Q'$ in your diagram where convergent rays would have crossed and formed a real image but were prevented from doing so by being interrupted by an optical system which in your example was the second interface.

If the second interface was not there $P'Q'$ would be a real image but because of the presence of the second interface the rays never meet and to describe such a situation the concept of a virtual object is introduced.

Farcher
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  • But intuitively, how can the P'Q' image act as a virtual image for P''Q'' through refraction, intuitively? How does that make sense? I understand it has to be that way, but how? – Emil Sriram Oct 14 '23 at 08:00
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    As I explained P’Q’ acts as a virtual object which means that the incoming rays are converging whereas the rays from a real object are diverging. – Farcher Oct 14 '23 at 12:05
  • If it is a virtual object, it would be as if the rays are coming from it - as far as I understand. How is this true in this case? – Emil Sriram Oct 14 '23 at 13:55
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The answer is: the rays of the first surface are virtual rays of the second surface. The rules of geometric optics work the same for real and virtual rays. It is the same case as the reflection in a flat mirror: the image is formed in the brain because it perceives the real rays as if they came from the point where they are behind the mirror. Virtual rays are extensions of real rays.

Pedro Quistian
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  • Yeah but how does the image P' translate to P''? How does the image coming from P' turn into P''? – Emil Sriram Oct 13 '23 at 17:38
  • The image of the first surface becomes the object of the second surface. In terms of rays, there are two refractions: the first builds the image in P', although, before completing it, it is interrupted by the second surface and the rays then become virtual. By refracting these rays coming from the first transmission, the second interface constructs the final image in P´´. – Pedro Quistian Oct 13 '23 at 19:10
  • But in the equation provided (second one), plugging the numbers in, where s is the image distance of P' and s' is the image distance of P'', gives that s and s' are both positive and hence on the same side. How is this possible? It makes sense logically through the superposition of the two rays...but in the equation it makes no sense. – Emil Sriram Oct 13 '23 at 19:48
  • https://docs.google.com/presentation/d/1YIy2xtXaxSAqoMQuy-7KsT-CWYEImDi8/edit?usp=sharing&ouid=106173122441358100501&rtpof=true&sd=true – Pedro Quistian Oct 14 '23 at 01:01
  • when the object is on the same side of the reflecting or refracting surface as the incoming light (left in this case), the object distance So is positive; otherwise, negative. In summary: So>0 on the left side, the side where the light enters. – Pedro Quistian Oct 14 '23 at 01:02