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 450 nmfalls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels.

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

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.

Express your answer in wavelengths of the light in the TiO2 film.

Answer 1

a) For destructive interference of reflected light

where n is the refractive index of the TiO2 film and d is the thickness.

m is the integer and lambda is the wavelength

Therefore the thickness is

Fpr m = 12, d = 1030. 53 nm which is less than the given thickness.

For m = 13

Therefore the minimum thickness that you must add is

b) The path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film is

In terms of wavelengths of the light in the TiO2 film, the path difference is

whre is the wavelengths of the light in the TiO2 film