Question

Lenses are often coated with magnesium fluoride (refractive index n = 1.38) to reduce reflections. How thick should the layer of magnesium fluoride be if reflections from the coating surface interfere destructively with those from the coating-glass interface for a wavelength in the center of the visible spectrum? (Say, 550nm). Assume that the glass has refractive index n = 1.5.

Answer 1

Answer: 99.64 nm (≅ 100 nm)

Explanation: In order to explain this problem we have to obtain destructive inteference from two waves that reflect in the film and lens surface so both waves have a λ/2 shift then we have to get a difference path (2L) equal to an odd number of the half wavelegth.

Then we have the following expression:

L=(m+1/2)* λ/(2*n2);  where n2 is the refractive index of the coating

for m=0 we have the minimum thickness for the coating

L=λ/(4*n2)=99.64 nm (≅ 100 nm)