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.

Answer 1

path difference between r3 and r4=L=331 nm

for minimum interference,

path difference=(m-0.5)*wavelength/n2

(division by n2 is because of the fact that

in a medium with refractive index n,

wavelength reduces by n)

hence 331 nm=(m-0.5)*wavelength/1.82

==>(m-0.5)*wavelength=1.82*331 nm

for m=1, wavelength=1204.8 nm

for m=2, wavelength=401.6 nm

for m=3, wavelength=240.97 nm

hence wavelength in visible range is 401.6 nm