Wavelength used in double slit experiment

  • Thread starter songoku
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  • #1
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Homework Statement:
In two separate double slit experiments, two different wavelengths are used. First wavelength is 708 nm and it is observed the second order bright fringe occurs at same position as third order dark fringe of second wavelength. Determine the second wavelength
Relevant Equations:
Bright fringe: d sin theta = m.lambda

Dark fringe: d sin theta = (m - 1/2).lambda
m × lambda for bright = (m - 1/2) × lambda for dark so:
2 × 708 = 2.5 × second lambda
Second lambda = 566.4 nm

But the answer is 495 nm. Where is my mistake? Thanks
 

Answers and Replies

  • #2
Charles Link
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Looks like they did it for 3rd order bright fringe. Did you copy the problem correctly? Looks like they may have made an error.
 
  • #3
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I agree with your answer. You could round it to 566 nm, however, the original wavelength was not given with more digits either.

708/495 = 1.43 and 495/708 = 0.70 both don't look like plausible fractions and I don't find an easy typo that would lead to this answer.

@Charles Link: That would be 472 nm.
 
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  • #4
Charles Link
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@mfb has the 472 correctly computed. I should have multiplied it out, instead of estimating it. (495 is not correct for any choice).
 
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Thank you very much for the help Charles Link and mfb
 
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