Looking at the color vs line spectrum

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The discussion highlights the distinction between perceiving color and understanding line spectra, emphasizing that they are not directly related to emission lines from electron transitions. Color perception involves receptor cells in the eye responding to varying frequencies and intensities of light, which the brain interprets as color. The thread suggests exploring older discussions for more in-depth information on color perception. It concludes by stating that any further discussions should occur in non-homework forums after reviewing previous threads. The topic underscores the complexity of visual perception and its underlying mechanisms.
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Homework Statement
Are both related to emission lines following electron transition?
Relevant Equations
Rydberg formula and fresnel equation
The difference between the principle of looking at the color of an object and the principle of the line spectrum

Are both related to emission lines following electron transition?
 
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Neither are related to emission lines.

The receptor cells in the eye trigger neurons in different ways according to the frequency and intensity of the incoming light, and the brain interprets the signals from them as the perception of color. Search here and you’ll find some older threads on color perception with more details.
 
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Thread closed. Any further discussion in be in one of the non-homework forums, after searching for older threads on the subject.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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