Question

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.

Answer 1

This situation here is different from previous problem 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 (see Fig. 2.26). 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 wavelenght. For this, the path length difference 2L within the film must be equal to an odd nunber of half wavelength:
`2n_(2) L = (n + 1//2) lambda`
We want the least thickness for the coating, that is, the smallest L. Thus, we choose, `n = 0`, the smallest value of n. Solving for L and inserting the given data, we obtain
` L = (lambda)/(4n_(2)) = (550nm)/((4)(1.38)) = 96.6 nm`