Understanding Wavelengths and Fringes

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In a Young's double-slit experiment, the center of a bright fringe occurs where the waves from the slits differ in phase by a multiple of π. The equation d * Δy/L = m*λ is used to relate fringe positions to wavelength. One participant argues that each bright fringe corresponds to a full wavelength, equating one wavelength to 2π. While some suggest answers A or G, the consensus is that the correct answer is D, although there is disagreement with one participant believing it should be E. The discussion highlights the importance of understanding phase differences in wave interference patterns.
hidemi
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Homework Statement
In a Young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in phase by a multiple of:

the answer is 兀.
Relevant Equations
d * Δy/L = m*λ
I think the answer is E because each bright fringe is differed by a wavelength, in other words, one wavelength is equal to 2π.
(For example, the first bright fringe is d * Δy/L = 1*λ.)
 
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hidemi said:
Homework Statement:: In a Young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in phase by a multiple of:

the answer is 兀.
Relevant Equations:: d * Δy/L = m*λ

I think the answer is E because each bright fringe is differed by a wavelength, in other words, one wavelength is equal to 2π.
(For example, the first bright fringe is d * Δy/L = 1*λ.)
I'm thinking the correct answer is "A". Or maybe "G", but I can't be sure. (Because the multiple choice answers are not visible in your post...) :wink:
 
berkeman said:
I'm thinking the correct answer is "A". Or maybe "G", but I can't be sure. (Because the multiple choice answers are not visible in your post...) :wink:
sorry ,The multiple choices are :
A) π/4, B) π/2 C) 3π/4 D) π E) 2π
The correct answer is D, however I think it should be E.
 
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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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