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

The situaltion here differs from figure `(9.1)` in that `n_(3)gtn_(2)gtn_(1).` The reflection at point a still introduces a phase difference of `pi` but now the reflection at point `b` also does the same ( see figure `9.2`). Unwanted reflections from glass can be, suppressed (at a chosen wavelength ) by coating the glass with a thin transparent film of magnesium fluoride of a properly chosen thickness which introduces a phase change of half a wavelength. For this, the path length difference `2L` within the film must be equal to an odd number of half wavelengths:

`2L=(m+1//2)lambda_(n2),`
or, with `lambdan_(2)=lambda//n_(2),`
`2n_(2) L=(m+1//2)lambda,`
We want the least thickness for the coating, that is, the smallest `L.` Thus we choose `m=0,` the smallest value of `m.` Solving for `L` and inserting the given data, we obtain
`L=(lambda)/(4n_(2))=(550nm)/((4)(1.38))=96.6 nm`