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.

[tex]2T=\dfrac{m\lambda_0}{n}\\\Rightarrow T=\dfrac{m\lambda_0}{2n}\\\Rightarrow T=m\dfrac{545\times 10^{-9}}{2\times 2.62}\\\Rightarrow T=m104\ nm[/tex]

T must be greater than the film thickness so m = 10 where T=1040 nm

Minimum thickness to add must be

[tex]\Delta T=1040-1036=4\ nm[/tex]

Minimum thickness to add is 4 nm

Path difference is given by

[tex]2T=2\times 1040=2080\ nm[/tex]

The path difference is 2080 nm

The wavelength of the film is given by

[tex]\lambda=\dfrac{\lambda_0}{n}\\\Rightarrow \lambda=\dfrac{545}{2.62}\\\Rightarrow \lambda=208.01526\ nm[/tex]

The wavelengths of the light in the film is 208.01526 nm