- #1
mayer
- 38
- 0
Greetings,
I don't have a magnifying glass/convex lens otherwise I would try to confirm this by experiment. Maybe I'll get one soon but to get things started. Let's say you have an object, with its own light source, and a convex lens with focal point f. The object is beyond 2f away from the lens. You, the observer, are on the other side of the lens. If you put a screen at a position determined by the thin lens equation, which in this case will be somewhere between f and 2f on your side of the lens, you will get the inverted real image.
Alright that's my setup not the real question but can someone confirm that the above is correct anyways? Thanks, still knew to this topic. Now for the actual question.
How will your view of the image change at three different locations; behind where the real image is focused, at the position of where the real image is focused, and in front of where the real image is focused.
1) When you are behind the real image focus point, the rays are diverging and you will view the object upside down, correct?
2) I am not sure how to approach the other two positions, however. At the point where the real image is focused, its been, well, focused, and our eyes are made to converge diverging rays. Similarly, in front of the real image, the rays are converging, which I am not sure how our eyes would respond(by attempting to converge less?).
3) Also, how do we perceive the magnification of the image when the observer is moving towards/away from converging vs diverging rays.
Very long question, I don't expect you to respond to all of it but any/all response is appreciated. Would clear things up on this topic quite a bit. Thanks!
I don't have a magnifying glass/convex lens otherwise I would try to confirm this by experiment. Maybe I'll get one soon but to get things started. Let's say you have an object, with its own light source, and a convex lens with focal point f. The object is beyond 2f away from the lens. You, the observer, are on the other side of the lens. If you put a screen at a position determined by the thin lens equation, which in this case will be somewhere between f and 2f on your side of the lens, you will get the inverted real image.
Alright that's my setup not the real question but can someone confirm that the above is correct anyways? Thanks, still knew to this topic. Now for the actual question.
How will your view of the image change at three different locations; behind where the real image is focused, at the position of where the real image is focused, and in front of where the real image is focused.
1) When you are behind the real image focus point, the rays are diverging and you will view the object upside down, correct?
2) I am not sure how to approach the other two positions, however. At the point where the real image is focused, its been, well, focused, and our eyes are made to converge diverging rays. Similarly, in front of the real image, the rays are converging, which I am not sure how our eyes would respond(by attempting to converge less?).
3) Also, how do we perceive the magnification of the image when the observer is moving towards/away from converging vs diverging rays.
Very long question, I don't expect you to respond to all of it but any/all response is appreciated. Would clear things up on this topic quite a bit. Thanks!
