Say I have a single convex lens. I set it up like a camera obscura, shine a light on my object and project the image onto a projection plane. Given the focal length of my lens, the projection plane, object and lens need to be certain specific distances from each other for the projected image to be in focus. This is given by the equation:

1/f = 1/s1 + 1/s2

Where f is the focal length, s1 is the distance from lens centre to the projection plane, and s2 is the distance from lens centre to an object.

Now say I remove the projection plane and replace it with my own eyeball. I can see (part of?) the real image through the lens. I know its not the virtual image because the image is upside down (whereas the virtual image created by a convex lens is the right way around). Now I take a step backwards from the projection plane, and the real image as seen through the lens is still in focus to my eyeball. I take a few more steps backwards and the image is still in focus.

Why can my eye see a focused image at many different s1 values, but a projection can only be focused at a specific s1?


2 Answers 2


You don't really see a real image. You always see a virtual image, just an upside-down one. When you say you see a real image, this just means that there is a focusing plane where you could place a translucent screen and still get an image.

Now your premise is that you can place your "eyeball" in the focusing plane and see an image. That's actually exactly when you can't see an image since you cannot focus on your eyeball (and a good thing you can't). The lens in your eyeball allows you to focus on virtual images before your eye by creating a real image on your retina from them. But you cannot look at real images that would appear on your eyeball itself.

In fact, when the real image passes your eyeball and consequently the virtual image flips upside down you only see an indiscriminate blur (the image of a single point that cannot decide whether it's upside down or not).

  • \$\begingroup\$ huh, interesting. So this image on the wikipedia page for lenses is wrong then. It calls that image the real image. You're saying if you're looking through a convex lens with your eyeball, its always going to be a virtual image. This image flips depending on where the object is to the focal point. And only images with objects beyond the focal point can be projected onto a plane. \$\endgroup\$ Jan 1, 2019 at 17:45
  • \$\begingroup\$ Think of it this way... the difference between a real image and a virtual image is whether or not all of the collected light is in focus at that point... I.e. a projected image that is in focus on the wall is a real image. But your eye also collects light and focuses it on the retina. Which means, in order for that to happen the real image that was projected onto the wall has to scatter (reflect off of the wall) as many virtual images in order for the eye to collect/focus (some of) it again. \$\endgroup\$ Jan 2, 2019 at 17:47
  • \$\begingroup\$ Thanks @StevenKersting! I actually have a followup question here that is related to my confusion about real and virtual images! I would appreciate it if you had some time to take a look: photo.stackexchange.com/questions/103958/… \$\endgroup\$ Jan 3, 2019 at 13:01

Any image is where all the light appears to be coming from. When you capture a real image on a screen or paper, you collect all the rays of light arriving there and redistribute them in all directions again. That means that you are dependent on all rays coming from the same light source or you'll be mixing things up. If you remove the screen, light rays will leave in the same direction they have been arriving and will not mix. Wherever your eye views the scene, the light from some object's point will pass through the corresponding point of the real image, allowing you to focus on it.

The difference between a real and a virtual image is basically that the point where your "viewing lines" to a point of a visualised object converge are in front of the lens rather than behind it, meaning you can place a physical screen there and capture and redistribute light at that plane's point without mixing light from different points of the object.


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