A glass lens is coated on one side with a thin film of magnesium fluoride `(MgF_(2))`to reduce reflection from the lens surface (Fig. 2.26). The Index of refraction of `MgF_(2)` is 1.38, that of the glass is 1.50. What is the least coating thickness that eliminates (via interference) the reflections at the middle of the visible specturm `(lambda = 550nm)`? Assume that the light is approxmately perpendicular to the lens surface.

Explanation

KEY IDEA
Reflection is eliminated if the film thickness L is such that light waves reflected from the two film interfaces are exactly out of phase. The equation relating L to the given wavelength `lambda` and the index of refraction `n_(2)` of the thin film is either Eq. 35-50 or Eq. 35-51, depending on the reflection phase shifts at the interfaces.
Calculations: To determine which equation is needed, we fill out an organizing table. At the first interface, the incident light is in air, which has a lesser index of refraction than the `MgF_(2)`, (the thin film). Thus, we fill in 0.5 wavelength under `r_(1)` in our organizing table (meaning that the waves of ray `r_(1)` are shifted by `0.5 lambda` at the first interface). At the second interface, the incident light is in the `MgF_(2)`, which has a lesser index of refraction than the glass on the other side of the interface. Thus, we fill in 0.5 wavelength under `r_(2)` in our table.
Because both reflections cause the same phase shift, they tend to put the waves of `r_(1)` and `r_(2)` in phase. Since we want those waves to be out of phase, their path length difference 2L must be an odd number of half-wavelengths:
`2L=("odd number ")/(2) xx (lambda)/(n_(2))`

This leads to Eq. 35-50 (for a bright film sandwiched in air but for a dark film in the arrangement here). Solving that equation for L then gives us the film thicknesses that will eliminate reflection from the lens and coating:
`L=(m+(1)/(2))(lambda)/(2n_(2))`, for m = 0, 1, 2,.... `" "(35-52)`
We want the least thickness for the coating-that is, the smallest value of L. Thus, we choose m = 0, the smallest possible value of m. Substituting it and the given data in Eq. 35-52, we obtain
`L=(lambda)/(4n_(2))=(550 nm)/((4)(1.38)) =99.6 nm.` (Answer)