Young's double-slit experiment - Bright fringes occurring

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In 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 2π radians. The discussion highlights that each bright fringe corresponds to a difference of one wavelength, reinforcing that a phase difference of 2π is essential for the formation of bright fringes. Participants agree that the correct answer to the homework question is E, contrary to the official answer of D. The clarification emphasizes the relationship between wavelength and phase difference in this context. Understanding this relationship is crucial for accurately interpreting the results of the experiment.
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:

A) π/4, B) π/2 C) 3π/4 D) π E) 2π
The correct answer is D.
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:

A) π/4, B) π/2 C) 3π/4 D) π E) 2π
The correct answer is D.
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.)
Yes - you are correct (answer E). A difference of 1 wavelength corresponds to a phase difference of ##2\pi## radians. The 'official' answer is wrong.
 
Steve4Physics said:
Yes - you are correct (answer E). A difference of 1 wavelength corresponds to a phase difference of ##2\pi## radians. The 'official' answer is wrong.
Thanks for clarifying!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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