csirvi
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Which is the correct solution for Optical Image Overlap?
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csirvi
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Cutter Ketch
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Looks fine to me.
Cutter Ketch
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I notice the sign isn’t consistent with all the other signs. You didn’t do the signs the way I would have, but I think the way you did them you would have to still write f as positive.
Normally the sign conventions are relative to the actual direction of the light. If the object is on the same side of the optic as the light is coming from the object is real and the sign is positive. If the image is on the side of the optic the light is going toward, the image is real and the sign is positive. If the optic is focusing the focal length is positive. In that convention x is positive, x+10 is positive, and f is positive.
Normally the sign conventions are relative to the actual direction of the light. If the object is on the same side of the optic as the light is coming from the object is real and the sign is positive. If the image is on the side of the optic the light is going toward, the image is real and the sign is positive. If the optic is focusing the focal length is positive. In that convention x is positive, x+10 is positive, and f is positive.
csirvi
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How many posible answer of this problem
Cutter Ketch
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Ha! That feels like a trick question. All right, I’ll bite. I believe there is only one. In fact, I believe it even only works on axis as the virtual source and the real image move in opposite directions when you move the real source.
csirvi
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Like after reflecting concave mirror if the ray retrace its paththen also we can get a value of x that is also a probability.
Cutter Ketch
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csirvi said:
Well, that does put the image on top of the object, and that is all the question explicitly asks. So I guess that is a valid answer.
However you should note that those are fundamentally different. In the solution using both mirrors the light not only returns to the object, but it is also traveling in the original direction. If it can pass through the object it will happily go around again and again. This constitutes a resonator. You can stick gain in where the source is and make a laser.
csirvi
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Still my doubt is that out of these two which is correct. If anyone solution is wrong then why ? Out of this.