Question

A uniform film of TiO2, 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 460 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels.

A) What is the minimum thickness of TiO2 that you must add so the reflected light cancels as desired?

ΔT= ____ nm

B) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in nanometers.

ΔP= _____ nm

Answer 1

The minimum occurs because there is destructive interference between the light reflected from the top of the film (A) and light from the glass boundary (B).

Note that there is a 180degree phase change (equivalent to an increase or decrease of half a wavelength path length) at the TiO2 boundary because the TiO2 has a higher refractive index than glass.

Let film thickness = d.

The extra distance travelled by a ray entering the film, travelling through the film and returning to the surface is twice the film thickness =2d (because it passes through the film and back). The 180degree phase change will result in a change equivalent to a half wavelength path length. Total path length can be considered to be = 2d + λ/2 or 2d-λ/2. This must equal λ/2 (or 3λ/2 or 5λ/2 etc) for destructive interference

2d - λ/2 = λ/2 or alternatively
2d + λ/2 = 3λ/2

These give d = λ/2. So the film thickness is 1/2 of wavelength = 230nm.

B) Path difference = (2/lambda)* phase difference

or is 2dsin theta= 2*230 sin 90 = 460nm