Phase difference between waves in two-slit model

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SUMMARY

The discussion centers on calculating the phase difference between waves in a double-slit experiment with a slit distance of 0.2 mm and a screen distance of 150 cm, using a wavelength of 500 nm. The phase difference at an angular distance of 10 degrees from the central peak is determined to be 165 degrees. The calculation involves the formula σ / λ = ΔΦ / 2π, leading to a phase shift of 436 radians, which is simplified to 165 degrees by accounting for complete oscillations. The final result confirms the correct conversion of the phase shift into the desired range.

PREREQUISITES
  • Understanding of double-slit experiment principles
  • Familiarity with wave mechanics and phase difference calculations
  • Knowledge of trigonometric functions and angular measurements
  • Proficiency in using formulas involving wavelengths and phase shifts
NEXT STEPS
  • Study the derivation of phase difference formulas in wave mechanics
  • Learn about the implications of angular separation in interference patterns
  • Explore the effects of varying slit distances on interference results
  • Investigate the relationship between phase shifts and oscillation counts in wave phenomena
USEFUL FOR

Students and educators in physics, particularly those focusing on wave mechanics, optics, and experimental physics. This discussion is beneficial for anyone looking to deepen their understanding of interference patterns and phase differences in wave phenomena.

vetgirl1990
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Homework Statement


In a double slit experiment, the distance between the slits is 0.2mm and the distance to the screen is 150cm. What is the phase difference in degrees, between the waves from the two slits arriving at point P, when the angular distance of P is 10 degrees relative to the central peak? The wavelength is 500nm. Convert your result so that the angle is between 0-360°

Answer: 165°

Homework Equations


σ / λ = ΔΦ / 2π

The Attempt at a Solution


σ = dsinθ
(dsinθ) / λ = ΔΦ / 2π

ΔΦ = (dsinθ)2π/λ = (0.00003473)2π / 5x10-7
ΔΦ = 436 rads * 180°/π = 24980°

Was it necessary to take into account the distance to the screen, since we were given angular separation? Also, I'm unsure of how to "convert the result so that the angle is between 0-360°).

EDIT: Figured this out!
If the phase shift is 436 rad, we know that the phase shift is also 69.39 oscillations, because 436/2pi = 69.46. Essentially, we can ignore the 69 complete oscillations and just focus on that .46. 436 - (2pi x 69) = 2.89. If you convert that to degrees, you get 165.
 
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