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I was using a convex lens and placed the object at principal axis at a distance from optical center lesser than focal length (between $F_1$ and optical center). Then I started observing the size of the image from other side of lens.

At first I had placed my eye close to the $F_2$ and between $F_2$ and $2F_2,$ then moved it away from that towards $2F_2.$ I found that as I moved away from lens, the image was getting bigger and bigger.

That's where my confusion comes in. What I understand is that the size of the image formed at any point is only dependent on its position from lens and lens. It should not be dependent on observer but the size of object seen by observer can get smaller and smaller as he moves away from lens just like a tree when seen from a distance appears small as compared to looking at it from closer distance.

Why is the size of image of object increasing?

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    $\begingroup$ Don't forget that your eye is an adaptive optical system. In particular, it can change its focal length by reshaping eye lens. $\endgroup$
    – Ruslan
    Commented Jan 18, 2022 at 14:41
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    $\begingroup$ How are you judging the size of the image? Is it relative to the lens? $\endgroup$
    – Farcher
    Commented Jan 18, 2022 at 14:53
  • $\begingroup$ The size of image was geting large as compared to lens and also from its intial size when i was at my intial location $\endgroup$
    – user316791
    Commented Jan 19, 2022 at 10:18
  • $\begingroup$ @Ruslan if you are right then real objects would also seem to get enlarged when you move away because our eye can't predict if the rays were coming out of a lens or directly from a object. It means that your statement was wrong. You can do this experiment because it is easy to do and maybe you can then help me $\endgroup$
    – user316791
    Commented Jan 19, 2022 at 10:25

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It's really not easy to judge the absolute angular size of an object (see Moon illusion). The image you see may get larger relative to the lens frame, but a bit smaller in angular size due to perspective.

In any case, with a perfect lens you're watching the virtual image at a fixed distance behind the lens, farther from the lens than the object, and magnified. This is equivalent to an experiment with a window (without optical power, just flat glass) and an object behind it. As you go away from the window, the object will seem larger—compared to the window frame. But it actually becomes smaller, as you can confirm if you try to measure its angular size e.g. by using a coin at an arm's length as a reference.

I've done the experiment you describe, and indeed the image grew relative to the lens but shrank relative to a SIM card I placed at a fixed distance to my eye.

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  • $\begingroup$ Thanks i will test this for sure $\endgroup$
    – user316791
    Commented Jan 19, 2022 at 14:13
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I suspect you are looking through the lens rather than looking at the physical image, aka "real image," that the lens would form. The Lensmaker's equation, $ \frac{1}{f} = \frac{1}{p} + \frac{1}{q}$ leads to a smaller image only for a real image (p and q both positive).

Since you are looking through the lens, you would have to analyze a two-lens system comprising the convex lens plus your eye's lens. If you do that, e.g., with a simple ray trace, you will see why the image size grows. See, e.g. Ray trace matrix method for a simple way to calculate image magnification.

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    $\begingroup$ I don't recall having a problem using the lens maker's equation with a virtual image. $\endgroup$
    – R.W. Bird
    Commented Jan 18, 2022 at 15:23
  • $\begingroup$ @R.W.Bird let me edit and rephrase that. $\endgroup$
    – Carl Witthoft
    Commented Jan 18, 2022 at 17:31
  • $\begingroup$ Well because the rays were diverging, it means that it is virtual image. And we can treat the rays as if it were coming from a real object. So by this thought method i can predict that image would look smaller as i move away. It's size ratio with lens should remain constant. But it was not the case. You can try yourself $\endgroup$
    – user316791
    Commented Jan 19, 2022 at 10:22
  • $\begingroup$ $ \frac{1}{f} = \frac{1}{p} + \frac{1}{p}$? Did you intend to write $ \frac{1}{f} = \frac{1}{p} + \frac{1}{q}$? $\endgroup$
    – Cross
    Commented Jan 19, 2022 at 11:00
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    $\begingroup$ @Cross ooops; edited to correct the equation $\endgroup$
    – Carl Witthoft
    Commented Jan 19, 2022 at 12:56

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