Destructive interference is exploited in making non-reflective coatings for lenses. The coating material generally has an index of refraction less than that of glass, so both reflected waves have a half-wavelength shift. A film thickness of 1/4 the wavelength in the film results in destructive interference.

Combining steps 1-3:

D

=

Db - Dt

=

(

2t +

ln 2

)

-

ln 2

=

2t

Bringing in the destructive interference condition:

2t = (m + 1/2) l_n

The minimum film thickness is given by m = 0.

Therefore

tmin

=

ln 4

=

l 4n

The thickness is generally chosen to be 1/4 of the wavelength of light in the middle of the visible spectrum, often green light. Thus the film will not give completely destructive interference for other colors, particularly for colors at the ends of the spectrum, red and violet. This is why non-reflective coatings often look purple, because they do reflect some red and violet light.

One final philosophical note, to really make your head spin if it isn't already. In our analysis we consider the two reflected waves and see how they cancel. This means none of the wave energy is reflected back into the first medium. Where does it go? It must all be transmitted into the third medium (that's the whole point of a non-reflective coating, to transmit as much light as possible through a lens). So, even though we did the analysis by drawing the waves reflecting back, in some sense they really don't reflect back at all, because all the light ends up in medium 3.