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 situation here differs from shown figure in that `n_(3)gt n_(2)gt n_(1)`. The reflection at point a still introduces a phase difference of `pi` but now the reflection at point b also does the same 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_(n_(2'))`
or , with `lambda n_(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`