From https://www.physics-chemistry-interac...
Two words here on the notion of magnification, gamma, which is the ratio of the size of the image to that of the object, in algebraic value. Here is an example here, where object A B and its image A prime B prime have the same size.
So, I'm going to write that gamma is negative1, or minus one if you prefer.
In absolute value it is 1, since they have the same height, but as the image is reversed, I will have a negative magnification. Note that the magnification has no unit since it is the ratio of two magnitudes of the same kind.
I will submit to you a second example, where the image is twice the size of the object. So, I write that gamma is two in absolute value, otherwise it is negative 2, minus 2, again to mean that the image and the object are not oriented in the same way. I can show you a smaller image, because the image is not always bigger, it depends. And here, for example, the image is twice as small. So, I write that gamma, in this situation, is worth negative one-half. The magnification is therefore not characteristic of the lens, since for the same lens, we obtained 3 different magnifications. Here is even a fourth, positive, when the image is virtual. This is the case with a magnifying glass, a microscope, a refracting telescope or corrective lenses.
There is actually an infinity of magnifications. I would like to show you two triangles, here, that's why I left the particular ray which passes through the optical center which we know is not deviated, which are the triangles O A B and O A prime B prime.
So why am I showing you these two triangles?
This is because they allow us, being homothetic, to write this relation in another way.
Indeed, according to Thales' theorem, A prime B prime over A B equals O A prime over O A equals O A prime over O A.
That is to say that the magnification is the ratio of the sizes, but it is also the ratio of the distances, from the lens to the image, over the distance from the lens to the object.
An exercise, to illustrate all of this. You can stop the video if you want to do the exercise, otherwise I will continue. We have information on distances here. Not on the sizes. So the object is 225mm while the image of this object is 450mm from a lens, which we don't know anything about yet.
But we can answer the first question about magnification.
So, precisely, thanks to the second part here of my relation, by making the ratio of the distances O A prime by O A.
So, I know that A prime and A are not on the same side of the lens, we saw that in the constructions, and so I'm going to put the minus sign here and find that the magnification is negative 2.
If I am given the object size, now I can find the image size by doing gamma times AB and I find minus 150 mm. Hey, no?
If I am asked to find the position of the image focus F prime by construction and to deduce the focal length from it, at that moment I take a sheet with small squares, a scale, here, which is suitable, that is to say 1 cm for 50 mm and I can place the object A B, its image A prime B prime since I have everything I need.
I can even represent them since I have the size of the object, and in question 2, I found the size of the image.
So I can do this diagram. How to locate F prime? Well I know that the ray coming from B, parallel to the optical axis, will emerge passing through F prime. Of course I extend it to B prime. The point of intersection here is F prime, the secondary focal point. This is what I am asked. This allows me to find the focal length, here, which is therefore 150 mm, by construction.
Contrary to magnification, the focal length characterizes the lens. Even though the magnification varies, the focal length remains the same.
🎧 Thanks to Gilles FOURNAT for his contribution 😲🎸
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