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Consider the case of image formation in the convex lens when the object is placed between f and 2f its image will be formed beyond 2f on the other side. And the image position is said to be the point where all the rays intersect and if we observe the image from a position beyond image position we can say rays are diverging from the image position, but what if we observe from a position that is before the image position surely we will see the image but we are observing the rays before it gets chance to intersect how we will explain image formation in this case ?enter image description here

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  • $\begingroup$ Why do you say that Observer 2 will see a focused image? They won't! $\endgroup$ – Bill N Jan 3 at 5:00
  • $\begingroup$ So image position also depends upon observer? $\endgroup$ – user204283 Jan 3 at 5:01
  • $\begingroup$ Well what you are saying is that the the observer will observe an image even when he sees it from a location different than the point from where the rays converge. Is that correct? $\endgroup$ – harshit54 Jan 3 at 6:17
  • $\begingroup$ Yes he should see some image as light from the object is entering his eyes but I don't think he will observe same image as the observer 1 in the image $\endgroup$ – user204283 Jan 3 at 6:20
  • $\begingroup$ Is the point of this question to take into account that the eye is itself a lens? $\endgroup$ – Aaron Stevens Jan 3 at 11:48
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The ray diagram does not only provide the image position. It also provides the image size. The figure itself shows that. This is a real image. So observer 1 sees an enlarged real image as expected. Observer 2 is obviously preventing the formation of the real image. So you have to draw ray diagram including the eye lens to see what kind of an image will form.

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If an eye is placed in position 2 it will hardly be able to see anything. Might be exceptions, but to see that we must elaborate the matter.

Normal (emmetropic) human eye is able to accomodate (to see sharply) objects situated between about 25 cm and infinity. This happens because thanks to the adjustable curvature of the crystalline lens rays coming by a source within that range can be made to converge to a point on the retina.

Of course rays coming from an object in front of the eye arrive at the pupil as a divergent beam. These are the kind of beam a normal eye is able to focus. But the eye placed in position 2 does not receive a beam of that kind, it receives a convergent beam. Such a beam will be brought to a point well before reaching the retina and will again diverge, thus forming on the retina an extended spot.

There is a kind of refractive defect allowing to bring to a focus converging beams too: it's hypermetropia. In any case within a limit. Since the questioner gives no numbers it's impossible to say if a moderately hypermetropic eye would see that image.

But try an example. If f = 50 cm, real image is at 3f = 150 cm from the lens. Eye 2 is in between, say at 75 cm, i.e. 75 cm before image. So the image to be seen is 75 cm behing eye, which requires a far-sightedness of at least 1/0.75 = 1.33 diopters - a low hypermetropia.

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