Question

# A uniform film of TiO2, 1036 nm thick and having index of refraction 2.62, is spread...

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 545 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. (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 . (C)Express your answer in wavelengths of the light in the TiO2 film.

The light reflected from the top of the TiO2 film interferes with the light reflected from the top of the glass surface. These waves are out of phase due to the path difference in the film and the phase differences caused by reflection.

There is a π phase change at the TiO2 surface but none at the glass surface, so for destructive interference the path difference must be mλ in the film.

(A) Let T be the thickness of the film so that the reflected light cancels

2T = mλ0/n,

=>T = mλ0/(2n)

=>T = m (520.0 nm)/[2(2.62)] = 99.237m

T must be greater than 1036 nm, so m = 11, which gives T = 1091.6 nm.

since we want to know the minimum thickness to add,

ΔT = 1091.6 nm - 1036 nm = 55.6 nm

(B) Path difference = 2T = 2(1092 nm) = 2184 nm = 2180 nm.

(C) The wavelength in the film is λ = λ0/n = (520.0 nm)/2.62 = 198.5 nm.

Path difference = (2180 nm)/[(198.5 nm)/wavelength] = 11.0 wavelengths

#### Earn Coins

Coins can be redeemed for fabulous gifts.