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?
Answer
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).
No comments:
Post a Comment