Question #109355
White light containing the wavelengths between 420 and 720 nm traveling in the air falls on a thin layer n1 = 1.5 and a thickness of 0.000001 m. If this layer is placed on another thin film n2 = 1.4, determine the wavelengths of the light that is not reflected in the air.
1
Expert's answer
2020-04-14T09:40:02-0400

In this case n0<n1;n2<n1(1<1.5;1.4<1.5)n_0<n_1; n_2<n_1(1<1.5;1.4<1.5).


So, the phase changes are not common to both surfaces. For destructive interference


λ=2dn2m,where\lambda=\frac{2dn_2}{m}, where m=1,2,3,4,5,....m=1,2,3,4,5,....


We have


λ5=600nm\lambda_5=600nm


λ6=500nm\lambda_6=500nm


λ7=428nm\lambda_7=428nm.






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Comments

Maica
13.05.20, 23:40

Thank you, but how do we determine d in that formula?

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