Fringes and Phase in Optics: Solving Problem 35.61

AI Thread Summary
The discussion focuses on solving problem 35.61 related to optics, specifically calculating the number of fringes from the extra phase accumulated in a glass rod. The user successfully calculated the phase gained per minute but struggles to understand the conversion to fringe count. They arrived at the answer of 22.7 through a formula involving the phase difference, but they express uncertainty about the reasoning behind it. Participants are encouraged to clarify the relationship between phase and fringe count to enhance understanding. The need for a deeper grasp of this fundamental concept in optics is emphasized.
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Homework Statement


Hi all

Please take a look at problem 35.61 in this PDF (it is on page 5): http://physics.wustl.edu/classes/SP2009/212/homework/YF35.pdf

What I can do is to find the "extra" phase accumulated in the glass rod per minute. But I do not know how to transform this into a number of fringes?
 
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Ok, I got the answer 22.7, which is also correct. I got it by finding

(phase gained per minute) - (phase gained at T=20 degrees) / 2Pi.

But this was pure luck - I cannot see, why this gives me the correct answer.
 
Guys, I am sorry about doing this, but I really need help with this. This is such a fundamental thing that I must understand it. Can you give me a hand?
 

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