Young's double-slit experiment - Bright fringes occurring

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SUMMARY

The discussion centers on the phase difference required for bright fringes in Young's double-slit experiment. Participants agree that the correct answer to the question regarding the phase difference for bright fringes is E) 2π, as each bright fringe corresponds to a phase difference of one wavelength, equating to 2π radians. The equation d * Δy/L = m*λ is referenced to illustrate this relationship. The consensus is that the official answer D) π is incorrect.

PREREQUISITES
  • Understanding of Young's double-slit experiment
  • Knowledge of wave interference principles
  • Familiarity with phase difference in wave mechanics
  • Ability to apply the equation d * Δy/L = m*λ
NEXT STEPS
  • Study the implications of phase differences in wave interference
  • Explore the mathematical derivation of the Young's double-slit experiment
  • Learn about the impact of slit separation on fringe patterns
  • Investigate real-world applications of wave interference in optics
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Physics students, educators, and anyone interested in wave mechanics and optical experiments will benefit from this discussion.

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!
 

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