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Physics Problem help?
Ok, here's question 1:
Light of wavelength 625 nm (in vacuum) is incident perpendicularly on a soap film (assume n = 1.33) suspended in the air. What are the two smallest nonzero film thicknesses (in nm) for which the reflected light undergoes constructive interference?
I know that the wavelength of the film would be 625/1.33. But after that, shouldn't this be the first smallest? I know it's not, but why? And how can you get the second smallest wavelength?
I tried that, but that's not correct. 625/1.33 is the film wavelength. So the shortest constructive interference would 1/2 of that wavelength or 625/2.66? Am I understanding that correctly?
1 Answer
- Anonymous1 decade agoFavorite Answer
First non-zero thickness= 1/2 wavelength
Second non-zero thickness= 1 wavelength
etc. for the third and so forth
To get reflection at a boundary, the wave must have a node. That happens every half wavelength.