A camera manufacturer wishes to deposit a non-reflective coating onto a glass lens for a wavelength of 550 nm

Solution:

There are two interfaces that we need to pay attention to.
They are, air-coating interface and the coating - glass interface.

At the first interface light travels from lower to higher refractive index medium.
At the second interface also, light travels from lower to higher refractive index medium.

Whenever light is reflected at such interface, the phase of the reflected beam is shifted by 180 degrees. In this case since two such phase shifts are involved, total phase shift due to reflection = 0

Phase difference therefore is only due to the thickness of the coating. For destructive interference,

2d = (m+1/2) (lamda)'

Where lamda' is the wave length of the light in the medium and d is the thickness of the coating.

Wave length in vacuum is lamda= n * lamda'

2d = (m+1/2) (lamda)/n where n is the refractive index of the coating material.

m = 0, 1, 2, ............

Three lowest thicknesses are: (using m =0, 1 and 2)

d = (lamda)/4n , 3 (lamda)/4n , 5 (lamda)/4n

d = 99.6 nm, 298.9 nm, 498.2 nm