Having trouble with this question. Correct answer is D, but after reviewing optics I thought that for a converging lens if you decrease do that causes di and m to increase?
TPR FL optics
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thanks frodo but that didn't really help much
Actually this does a good job of demonstrating what is going on. The question states that the original object is halfway between the focal point and mirror, and that the new placement of the object is closer to the mirror than that. Reflections in this area decrease in size, and move closer to the mirror, as you can see not only from the gif, but also if you measured out the dimensions yourself.
Thank you. For some reason I was misreading the entire question and thinking about this in the context of a lens and not a mirror. On that note, is there a difference between how a lens would affect the path of light versus a lens (assuming they are both converging or both converging)?
do image location (di) and magnification change with distance for mirrors in the exact same way that they do would do for a lens? (when comparing a converging lens and a converging mirror)
No. For a diverging lens (concave), the image is always virtual, so while the size of the image will change, it will not change in the same manner as a diverging mirror. I found the page that is the same about lenses as the one about mirrors. Hopefully it helps to visualize it since drawing it out can be a bit confusing sometimes.
http://www.acs.psu.edu/drussell/Demos/RayTrace/lenses.html
Honestly, and it took me a while to see this myself, but EVERY question about lens/mirrors can be answered with the equation: 1/f = 1/do + 1/di; m = -di/do, and P = 1/f seriously EVERY single question.
Try plugging in an object distance for the concave (converging mirror) and then shortening the object distance. Make up some arbitrary focal length, and see what happens when the object is at 2f, at 1f, at 1/2f, at 1/5f, etc.
This will truly explain, in mathematical terms, why when you are at the focal length there is no image, because then 0 = 1/di and di = infinity. Play with this equation and the magnification equation and you will literally be able to understand perfectly why and when an image will be virtual, real, enlarged, shrunk, no image, etc.
Try plugging in an object distance for the concave (converging mirror) and then shortening the object distance. Make up some arbitrary focal length, and see what happens when the object is at 2f, at 1f, at 1/2f, at 1/5f, etc.
This will truly explain, in mathematical terms, why when you are at the focal length there is no image, because then 0 = 1/di and di = infinity. Play with this equation and the magnification equation and you will literally be able to understand perfectly why and when an image will be virtual, real, enlarged, shrunk, no image, etc.
