Calculate Distance Between Maxima of 500 nm Wavelength Light

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SUMMARY

The calculation of the distance between maxima for light with a wavelength of 500 nm, using two narrow slits spaced 0.10 mm apart and a viewing screen 1.20 m away, results in a distance of 6.0 x 10-3 m between each maximum. The fourth maximum is located 2.4 x 10-2 m from the central maximum. The formula used for this calculation is delta x = (lambda x L) / d, where lambda is the wavelength, L is the distance to the screen, and d is the slit separation.

PREREQUISITES
  • Understanding of wave interference principles
  • Familiarity with the formula for calculating maxima in double-slit experiments
  • Knowledge of units of measurement in physics (meters, nanometers)
  • Basic algebra for manipulating equations
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  • Study the principles of wave interference in more depth
  • Learn about the effects of varying slit separation on interference patterns
  • Explore the application of the double-slit experiment in quantum mechanics
  • Investigate the impact of wavelength changes on the distance between maxima
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a) A light with a wavelength of 500 nm illuminates two narrow slits that are 0.10 mm apart. If the viewing screen is 1.20 m from the slits calculate the distance between each maximum.

b)how far would the fourth maximum be from the central maximum?



a) Given: lambda = 500 x 10e-9 m
d= 0.1 x 10e-3 m
L= 1.2 m

delta x = lambda x L/d
=(500 x 10e-9 m) (1.2 m)/(0.1 x 10 e-3 m)
=0.0006 m
=6.0 x 10 e-3

Therefore, the distance between each maximum is 6.0 x 10 e-3 m

b) Since the distance between each maximum is 6.0 x 10 e-3 m the fourth maximum would be 4 x (6.0 x 10 e-3 m)

Therefore, 4 x (6.0 x 10 e-3 m) = 2.4 x 10 e-2

Therefore, the fourth maximum would be 2.4 x 10 e-2 m away from the central maximum.


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