Transmission through thin layers. In the figure, light is incident perpendicular

Question

Transmission through 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.) Part of the light ends up in material 3 as ray r3 (the light does not reflect inside material 2) and r4 (the light reflects twice inside material 2). The waves of r3 and r4 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 n1, n2, and n3, the type of interference, and the thinlayer thickness L in nanometers. Give the wavelength that is in the visible range.

Explanation / Answer

For n2 > n1 and n2 > n3

For minimim transmission and destructive interference

2L = ( m + 1/2) lambda/n2 ------- where m = 0,1,2,3

as we need wavelength in visible range (390 nm to 700 nm)

lambda = 4Ln2 / 2m+1

for m = 0

lambda = 2282 nm ( not in visible range)

for m = 1

lambda = 2282 / 3 = 760.8 nm

for m = 2

lambda = 456.4 nm <-------- This is the wavelength in visible range

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