To produce a minimum reflection of wavelengths near the middle of visible spectrum (550 nm) how thick should a coating of ${\mathrm{MgF}}_2 (\mu =1.38)$ be vacuum-coated on a glass surface?

Consider light to be incident at near normal incidence. We wish to cause destructive interference between rays r₁ and r₂ so that maximum energy passes into the glass. A phase change of $\lambda /2$ occurs in each ray for at both the upper and lower surfaces of the MgF₂ film the light is reflected by a medium of greater index of refraction. When striking a medium of lower index of refraction, the light is reflected with no phase change. Since in this problem both rays 1 and 2 experience the same phase shift, no net change of phase is introduced by these two reflections. Hence, the only way a phase change can occur, is if the two rays travel through different optical path lengths. The optical path length is the product of the geometric path difference a ray travels through different media and the refractive index of the medium in which it is travelling. 
For destructive interference the two rays must be out of phase by an odd number of half wavelengths. Hence, the optical path difference needed for destructive interference is,

$2\mathrm{\mu d}=(2\mathrm{n}+1)\frac{\mathrm{\lambda }}{2},\mathrm{n}=0,1,2\dots \dots \dots$

Note that $2\mu d$ is the total optical path length that the rays traverse when n = 0;

$\Rightarrow \mathrm{d}=\frac{\mathrm{\lambda }/2}{2\mathrm{\mu }}=\frac{\mathrm{\lambda }}{4\mathrm{\mu }}=\frac{550}{4\times 1.38} \mathrm{nm}\approx 100\mathrm{nm}=1\times {10}^{-7}\mathrm{m}$