Double Slit Diffraction Problem

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SUMMARY

The double-slit diffraction problem discussed involves determining the wavelength of visible light that produces a minimum at the same location as a second-order maximum created by blue light with a wavelength of 478 nm. The correct approach requires using the formula for diffraction minima, which is d*sin(θ)=(m+1/2)λ. The user initially miscalculated by assuming the order of the second wavelength should be three but later corrected it to one, leading to the correct wavelength calculation.

PREREQUISITES
  • Understanding of the double-slit experiment
  • Familiarity with diffraction patterns and orders
  • Knowledge of the wavelength of light and its measurement in nanometers
  • Basic trigonometry for calculating angles in diffraction
NEXT STEPS
  • Study the principles of the double-slit experiment in detail
  • Learn about the mathematical derivation of diffraction minima and maxima
  • Explore the relationship between wavelength, frequency, and energy of light
  • Investigate the effects of different slit widths and separations on diffraction patterns
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Students of physics, educators teaching wave optics, and anyone interested in the principles of light diffraction and interference patterns.

Moxin
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In a double-slit experiment it is found that blue light of wavelength 478 nm gives a second-order maximum at a certain location on the screen. What wavelength of visible light would have a minimum at the same location?


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I have no clue how to approach this problem but looking in the book I assumed the wavelength with a minimum at that same location should be 3rd order (m=3) so I set up an equation as follows:


m1*Lambda(max) = m2*Lambda(min)
Lambda(min) = m1*Lambda(max)/m2 = (2*478*10^-9)/3 = 319 nm

Apparently it's wrong. I'm lost on where to go from here.. Any help ??
 
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The minimum from the second wavelength needn't be of the same order. Also, you need to note that diffraction minima are not described by the same equation as diffraction maxima.

The minima are located by:

d*sin(θ)=(m+1/2)λ

You have d and θ. You also need to note that m must be an integer.

Can you take it from there?
 
yeh i actually tried somethin like that before but i just now realized the other wavelength's m should be an order less (1, instead of 3) and i got it right, thanks tho !
 

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