A uniform film of TiO2, 1036 nm thick and having index of refraction 2.62, is sp

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 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.

Part A) What is the minimum thickness of TiO2 that you must add so the reflected light cancels as desired? T = 80.4 nm <- correct answer

Part 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. P = ?

Part C) Express your answer in wavelengths of the light in the TiO2 film. P = ?

Explanation / Answer

part A:

for destructive interference,

the required condition is 2*n_film*d*cos(theta)=m*lambda

where n_film=refractive index of the film =2.62

d=thick ness of the film

theta=angle of incidence=0 degrees in this case

m=an integer

lambda=wavelength=450 nm

let we have to add a thcikness of t nm.
so using the values:

2*2.62*(1036+t)*cos(0)=m*450

==>1036+t=85.8778*m

nearest integral multiple of 85.8778 is achived when m=13

so t=85.8778*13-1036=80.4114 nm

part b:

path difference=twice the thcikness=2*(1036+80.4114)=2232.8228 nm

part C:

wavelelngth of the light in TiO2 flim=wavelength in air/refractive index

=450 nm/2.62=171.755 nm
so 2232.8228/171.755=13

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