Question

A thin sheet of plastic (n = 1.60) is inserted between two panes of glass to reduce infrared (l = 700 nm) losses. What thickness (in nm) is necessary to produce constructive interference in the reflected infrared radiation?

a. 55

b. 318

c. 109

d. 218

e. 443

please show work

Answer 1

We have, for reflected radiation the ray which hits the leading edge has an inverted phase because the plastic has a higher refractive index from glass.

At the other side the reflected ray is right way up (passing from heavy to light)

So there is a half wavelength phase change by the material.

The wavelength of light is inversely proportional to the refractive index so if you mean that 700nm is the wavelength in air then it is 700/1.6 in the plastic.

Let the thickness be d

For constructive interference we need a further half wavelength path difference ( to add to the half already calculated by the reflection)

2d = 1/2 lambda

d = 1/4 lambda = 1/4 * 700/1.6 = 109 nm (option c)