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Waves- Interference/Double-slit experiment

  • Thread starter Jodi
  • Start date
23
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Hi; Could someone please help me with the following question: In a double-slit experiment it is found that blue light of wavelength 460 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? Can i use the equation dsin(theta) = m(lamda)? If I use this, i plug in the answer I get for dsin(theta) into this equation: dsin(theta) = (M+0.5)(lamda)? Is this right, because it doesn't work. Thanks for your help.
 

Doc Al

Mentor
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Looks good to me. Realize that there are many wavelengths of light that will have a minimum at that location, but not all will be visible. Plug in a few values of M and solve for [itex]\lambda_2[/itex] until you find one in the visible range.
 

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