Chapter 35, Problem 043 Reflection by thin layers. In the figure, light is incid

Question

Chapter 35, Problem 043 Reflection by thin layers. In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials for clarity.) The the indexes of refraction ny, n2, and n, the type of interference, and the thinlayer thickness L in nanometers. Give the wavelength that is in the visible range. 1 and 3. (The rays are tilted only waves of rays ri and r2 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). The table below provides n1 n2 n3 Type 1.60 1.43 1.90 min Number Units

Explanation / Answer

At both interfaces there is reflection off of a higher index material. Consequently, there are two additional
factors of n/2, and the condition for destructive interference is 2t = (m + 0.5)n.

The thicknesses for which there is destructive interference is are = n*n = 2nt/(m + 0.5).

Using n = 1.43 and m = 1 ,
= n*n = 2nt/(m + 0.5)
= (2*1.43*222e-9)/(1.5)
= 423.3 nm
which is in the visible region.

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