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?

Answer 1

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