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csirvi
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[Moderator's note: Moved from a technical forum and thus no template. Own effort in next post.]
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Which is the correct solution for Optical Image Overlap?
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SUMMARY
The discussion centers on the correct solution for Optical Image Overlap, specifically regarding sign conventions in optics. Participants agree that the sign of the focal length (f) is positive when the object is real and on the same side as the incoming light. The conversation highlights the distinction between real and virtual images, noting that a resonator can be formed when light retraces its path through mirrors, potentially leading to laser formation. Ultimately, the consensus is that there is only one valid solution to the problem presented.
PREREQUISITES- Understanding of optical sign conventions
- Knowledge of real and virtual images in optics
- Familiarity with concave mirrors and their properties
- Basic principles of laser resonators
- Research optical sign conventions in detail
- Study the behavior of light in concave mirrors
- Explore the principles of laser resonators and gain media
- Learn about the mathematical modeling of optical systems
Students and professionals in optics, physicists, and engineers working with optical systems, particularly those interested in image formation and laser technology.
[Moderator's note: Moved from a technical forum and thus no template. Own effort in next post.]
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Cutter Ketch
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Looks fine to me.
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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.
- #5
csirvi
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How many posible answer of this problem
- #6
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.
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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.
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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.
- #9
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.
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