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 thin layer thickness L in nanometers. Give the wavelength that is in the visible range.

Explanation / Answer

Here given

n1 = 1.59

n2 = 1.35

n3 =1.51

L = 401

m= 0, 1, 3, 5,7,9..... (odd numbers)

2L =(m/ 2) (/n2)

We have to take visible light range is usually 390 to 790nm

2L= 2 x401 = 802nm

For m=3

802 = (3/2) (/1.35)

= 534.67x1.35

=721.8nm

for m = 5

802 =5/2 (/1.35)

=433.08nm

for m= 7

802 = 7/2 (/ 1.35)

=309.34nm

here odd numbers 3 and 7 are too low and too high of visible region

so m= 5 will be the correct odd number and the wavelength is 433.08nm

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