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.

Question

Anneliese

Physics

24 June, 20:01

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.

+1

Answers (1)

Know the Answer?

Powell · 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)