Asked by Neenah
Using light of wavelength 600.0nm in a double slit apparatus we find delta x = 1.0cm. If we use the same apparatus for light of wavelength 450.0 nm, what is the new delta x on the screen?
Delta x = L*lambda/d
But L and d are not given, so how do I approach this problem?
Delta x = L*lambda/d
But L and d are not given, so how do I approach this problem?
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