Posted by Anonymous on Thursday, March 26, 2009 at 11:00pm.
For constructive interference
d sinQ=ml=2x460nm=920nm
For destructive interference of the other light, we have
d sinQ=(m’+1/2)l
When the two angle are equal, then
920nm=(m’+1/2)l
l=1.84x103 nm for m’=0
l=613 nm for m’=1
l=368 nm for m’=2
The only wavelength here that is visible is 613 nm
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