Reflection by thin layers. In the figure, light is incident perpendicularly on a

Question

Reflection by thin layers. In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays r_1 and r_2 Interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). The table below provides the Indexes of refraction n_1, n_2, and n_3, the type of interference, and the thickness L in nanometers. Give the wavelength that is in the visible range.

Explanation / Answer

since n2<n1 ,n2>n3 and the condition for destructive interference

2L=(m+1/2)(lambda/n2)

lambda=2Ln2/(m+1/2)

where m=0,1,2,3,------

m=0=>lambda=2*417*1.62/0+1/2)=2702 nm

m=1=>lambda=2*417*1.62/(1+1/2)=900.7 nm

m=2=>lambda=2*417*1.62/(2+1/2)=540.4 nm

For the wavelength to be in the visible range, we choose m = 2 with lambda=540.4 nm

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