The figure above shows three situations in which light reflects almost perpendic

Question

The figure above shows three situations in which light reflects almost perpendicularly from the top and bottom surfaces of a thin film, with the indices of refraction as shown. For a fulm to appear dark, the waves represented by rays 1 and 2 must be out of phase. This can arise from phase shits associated with the reflection at the top an bottom surfaces of the film and the extra distance traveled by ray 2 as it moves through the film. (a) For which situation(s) is there a net phase shift(due to reflection) between waves 1 and 2 that is equivalent to either zero wavelengths or one wavelength (lambda_firm), where lambda_firm is the wavelength of the light in the film? (b) For which situation(s) will the film appear dark when the thickness of the film is equal to 1/2 lambda_firm? (c) For all 3 situations, write the relation between thickness and wave length for constructive and destructive interference.

Explanation / Answer

part a )

2nt = m*lambda

a and c

part b )

2nt = (m+1/2)*lambda

b

part c )

for constructive

2nt = m*lambda

for destructive

2nt = (m+12/) lambda

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