How Can You Calculate the Wavelength in a Double Slit Experiment?

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SUMMARY

The wavelength in a double slit experiment can be calculated using the equation λ = d (x_M - x_m) / L (M - m). Given the angle to the eighth maximum at 1.12 degrees, a distance from slits to screen of 3.02m, and a distance between slits of 0.00025m, the wavelength is determined to be 610nm. This value is confirmed by using both the distance between the fifth and first minima and the angle to the eighth maximum, demonstrating consistency in the calculations.

PREREQUISITES
  • Understanding of the double slit experiment
  • Familiarity with wave interference patterns
  • Knowledge of trigonometric functions, specifically sine
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of the double slit interference formula
  • Learn about the significance of maxima and minima in wave patterns
  • Explore the impact of slit separation on interference patterns
  • Investigate the applications of wavelength calculations in optics
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Students studying physics, particularly those focusing on wave mechanics and optics, as well as educators looking for practical examples of wave interference phenomena.

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Homework Statement



The following data was obtained via the double slit experiment:

angle to the eighth maximum = 1.12 degrees
distance from slits to screen = 302cm = 3.02m
distance from first to fifth mimum = 2.95cm
distance between slits = 0.00025m

find the wavelength.

Homework Equations



supposed to use this equation to find wavelength:

xn/L=(n-0.5)(\lambda/d)
or
xm/L=(m)(\lambda/d)

The Attempt at a Solution



problem is, there's no given value for xm or xn

how can i get this value?

(oh, and i specifically have to use the equation listed above)
 
Last edited:
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##x_m## is the distance off axis of the mth maximum, numbering the central bright spot as ##m=0##. You can immediately write the distance between the Mth and mth maxima as ##x_M-x_m=(M-m)L\lambda/d##. A little manipulation gets you ##\lambda=d (x_M-x_m)/L (M-m)##. You provided the distance between the M=5 and m=1 maxima. Plugging that and your other data in gets you ##\lambda=610##nm.

You also gave the angle to the 8th maximum, which tells us that the distance from the axis to the 8th maximum is 3.02 sin(1.12)=0.059m. Plugging that in with M=8, m=0 also yields 610nm.
 

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