406 nm is the second thinnest coating of fluorite.
Given data:
Wavelength; λ = 775 nm
Refractive index of Calcium fluoride with wavelength of 775 nm is approximately 1.4308.
n = 1.4308
Formula for the thickness of the film that would destruct the light is;
t = (m + 0.5)(λ/2n)
Where m is the order of the thickness.
The first smallest thickness is at m = 0 while the second smallest thickness is at m = 1.
Thus;
t = (1 + 0.5)(775/(2 × 1.4308))
t ≈ 406 nm
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