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Thin film interference is given by: $2d\sin\theta=n\frac{\lambda}{n_i}$, where $d$ is the thickness of film, $\theta$ is the angle between incident light and tangent to the surface, $n$ is an integer, $\lambda$ is wavelength of light in the film and $n_i$ is the refractive index of the film.

Bragg Diffraction is given by: $2d\sin\theta=n\lambda$

The question is why $n_i$ is not necessary in the equation of Bragg diffraction.

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Brag diffraction is for crystal, the ni of which is very close to 1. For thin film, ni can be quite large since it is a dielectric material.

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Promoted from a comment:

Because the index of refraction at x-ray wavelengths for most materials is awfully close to 1.0, while the index of refraction in the visible can easily be 1.5 and above (1.46 for SiO2 and 2.0 for silicon nitride). So, for the visible, you need to account for the index of refraction, and in the x-ray you really don't.

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