Question

A material having an index of refraction of 1.35 is used as an antireflective coating on a piece of glass (n = 1.50). What should be the minimum thickness of this film in order to minimize reflection of 460 nm light?

nm

If instead a material with an index of refraction of 1.75 is used for the coating, what are the two smallest thicknesses to minimize reflection (in order).

nm and nm

Answer 1

(a) Since the coating refractive index is less than the glass therefore there will be no phase shif.
hence for the minimum reflection the thickness is given by

Where t is thickness
is wavelength = 460 nm
n is refractive index of coating = 1.35
t = (460*10^-9) /(4*1.35) =85.185*10^-9 m
t = 85.185 nm
(b) Now in this case the refractive index of coating is more than the glass therefore phase shift will takes place.
Therefore for the minimum reflection the thickness is given by

t = (460)/(2*1.75) = 131.43 nm