Lens/Mirror question: as an object moves away from...

[Stephen2009]

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So as an object first starts out touching a concave mirror, then moves away from it, it's image:

1. increases in size then, once it passes the focal point, decreases in size.

My question: how do we prove this using the thin lens eq.?

Similar question for a convex mirror:

2. the image decreases in size.

Again, how do we prove this using the thin lens equation?

THANKS!🙂

 

[Rabolisk]

1/f = 1/o + 1/i. Size of image is the absolute value of i/o. Let's say that we have a concave mirror with the focal length of 10. If the object is close to the mirror (1), then 1/10 = 1/1 + 1/i. 1/i = -9/10, and i/o = -10/9. When the object approaches the focal length (5), then 1/10 = 1/5 + 1/i. 1/i = -1/10, and i/o = -2, which is greater (the absolute value, that is) than -10/9. When the object is placed at the focal length 1/i is 0, which means that i is infinite, and thus you get no image. Now at o = 20, you get i/o = 1. At o = 100, you get i/o of 1/11.

It's rather easy to do with the math, if not tiring. The better way is to remember this. Whenever you move away from the focal length, your image size decreases. In a concave mirror, the focal length is in front of the mirror, so you approach infinity and then go back down. In a convex mirror, the focal length is behind the mirror, so you never reach infinity.

 

[Stephen2009]

Thanks! I just printed your explanation out... I really appreciate the time you took explaining it!