White light passing through prism

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Shorter wavelengths of light, such as ultraviolet, refract more than longer wavelengths like infrared when passing through a prism, which is why they do not rejoin to form white light upon exiting. The discussion emphasizes the importance of understanding the principles of light refraction and dispersion. Participants are encouraged to conduct their own research to better grasp the concepts involved. The conversation also touches on the visual representation of light behavior using examples from popular culture. Ultimately, the phenomenon of light not rejoining is attributed to the differing angles of refraction for each color.
hendrix7
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Homework Statement
Why don't the individual colours of the spectrum rejoin to form white light on exit from a prism?
Relevant Equations
n/a
I don't really have any idea. I know that shorter wavelengths refract more on entry to prism; e.g. ultra-violet refracts more than infra-red, but I don't know why they don't join up again when speeding up on exit from the prism.
 
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BvU said:
'No idea' is not enough. Do some googling !
Or listen to more Pink Floyd... :wink:

1700620704125.jpeg

https://news.artnet.com/news/squaring-the-circle-hipgnosis-documentary-anton-corbijn-2317872
 
hendrix7 said:
Homework Statement: Why don't the individual colours of the spectrum rejoin to form white light on exit from a prism?
Relevant Equations: n/a

why they don't join up again when speeding up on exit

Which way does the refraction upon exit work ? Towards joining up or towards even further splitting ?

Consider three cases:
  • as in @berkeman Pink Floyd gramophone record sleeve picture,
  • when sides are parallel (flat glass plate)
  • as in @berkeman Pink Foyd gramophone record sleeve picture upside down
##\ ##
 
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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