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Synchrotron question

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Matt89:
Could someone help me with Question 6 in Synchrotron on the 2006 Exam?

(The exam can be found here: http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/2006/2006physics2-w.pdf)

According to itute, the solution is found using this formula:

Wavelength Shift ∆λ = (1 - cosϴ)h/mc[/b]

I've NEVER seen this formula before. Could someone give me a brief explanation of it and where you would use it PLEASE

Daniel15:
Me neither, I don't think I've ever seen that formula before O.o

Ninox:

--- Quote from: "Daniel15" ---Me neither, I don't think I've ever seen that formula before O.o
--- End quote ---


I haven't seen it either. Neither me nor my teachers could get it using, so we let it be an open question to the class. The most logical explanation we could find was simply: more deflection, more energy lost and longer subsequent wavelength. We've already got 0 and 90deg, so 135deg is the answer.

Galelleo:
Daniel, lol, you probably arent doing detailed study - synchotron and applications.

Im doing sound, so i cant help you, sorry  matt

Daniel15:

--- Quote from: "Galelleo" ---Daniel, lol, you probably arent doing detailed study - synchotron and applications.

Im doing sound, so i cant help you, sorry  matt
--- End quote ---

Nope, that's the detailed study I am doing, which was why I was concerned with not seeing that formula.

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