*18. mmh A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light (wavelength = 640.0 nm in vacuum). Assuming that the visible spectrum extends from 380 to 750 nm, for which visible wavelength(s) in vacuum will the film appear bright due to constructive interference?

 

This is for destructive interference which is what we have for the red light.  So we are told we have the smallest thickness so with 1 phase shift we get

 

 

So the minimum thickness would be m= 1.  That means the thickness of the film is given by

 

 

Now we need to consider constructive interference, so our phase equation is now

 

 

 

This is the condition for constructive interference.  We do not use (m-1/2), because we do not want negative signs showing up.  Now substitute what we found for thickness

 

So the possible wavelengths are

 

So now we just put in m’s and see what we get for a wavelength.  We only count wavelengths between 380 nm and 750 nm

 

 

Only visible light is

 

 

 

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This page last updated on July 12, 2018