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As I understand, farsightedness glasses use convex lenses to create virtual image that is farther than the object (and thus past the near point of the user). However, this only happens if the object is within the focal length of the lens. If the object is beyond the focal length of the lens, no virtual image is formed (in fact real image would be formed on the other side of the lens). So what does someone see when wearing convergent glasses and looking beyond the glasses focal length? Can they see clearly there at all? If so, how?

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    $\begingroup$ I wear glasses that have a focal length of about one meter, and let me assure you that I can clearly see things that are more than one meter from my face when I wear them. Remember! It's not just the eyeglasses that form the image on my retinas. It's the system consisting of my eyeglasses and the corneas and the lenses and the fluids in my eyes. $\endgroup$ Commented Jan 22, 2023 at 15:28

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The trick with converging eyeglasses is that the eye is closer to the lens than the focal length. The wearer is never in a position to view the (inverted) real image.

In the thin lens equation, $$ \frac1f = \frac1{d_i} + \frac1{d_o}, $$

the convention is that a “virtual image” has a negative image position $d_i$. A person who is wearing converging lenses to look at an object that’s far away will be trying to focus converging light rays, which would form an image somewhere behind the wearer’s head. In the thin lens equation for the eye, these converging rays correspond to a “virtual object,” with negative object position $d_o$.

If you wear converging lenses and hold a book near the focal length, the rays from that book exit the lens nearly parallel: slightly converging if the book is a little farther, and slightly diverging in the book is a little nearer. The image position position $d_i$ is bouncing around between negative infinity and positive infinity, but that’s a mathematical artifact due to asking about a hypothetical meeting point instead of the behavior of the light at the lens. The actual quantity in the thin lens equation, $\frac1{d_i}$, is varying smoothly around zero.

Note however that if you move the converging lens so that the real image forms before the eye, there is definitely a weird region. This is “easier done than said,” to misquote the adage. Go to a drugstore and get some inexpensive reading glasses, or get the magnifying lens from your soldering iron or your sewing kit. Hold it far from your eye and look out the window at a distant object: you’ll see the inverted real image. Hold the lens right up to your eye and look out the window: you’ll see the upright virtual image. Now smoothly move the lens from near your eye to far from your eye, and watch what happens as the image inverts.

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  • $\begingroup$ I m still confused though, you said that if I hold the lens right up to my eye and look far, I would see a virtual image. I thought that in this case virtual image cannot be formed. Or can it? I understand the part where I move the lens far from my eye and see inverted real image though. $\endgroup$
    – Yevgeniy P
    Commented Jan 23, 2023 at 12:21
  • $\begingroup$ Oh, you’re right. When I teach single-lens optical systems to reluctant students, I get a little sloppy about equating “I see that and it’s upright” with virtual images. Let me think about how to edit that sentence so that it’s more correct without getting in the weeds of details. $\endgroup$
    – rob
    Commented Jan 23, 2023 at 13:43
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Glasses, both near and far, are more often actually convex on the outside of the lens and concave on the users side of the lens.

As a glasses wearer, you are correct that glasses are always a bit of a compromise, far sighted lenses tend to be 'blury' at closer distances, but at a certain distance and beyond always look sharp, whereas near sighted lenses are still blury if too close, and rapidly are blury as the distance increases, but at a typical reading distance will be sharp.

I have different glasses for IT work and screens than for reading a book for exmple as the typical distances are different. Likewise I have seperate far sighted glasses for driving.

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    $\begingroup$ The question uses “convex lens” to mean “converging lens.” A convex-concave lens, like most eyeglasses, is usually converging if its middle is thicker than its edges, and diverging if its edges are thicker than its middle. $\endgroup$
    – rob
    Commented Jan 22, 2023 at 13:32

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