A thin transparent film coated on a surface of the lenses in order to suppress the reflections from it is called anti reflecting film.

A film can be an anti-reflecting film if it satisfies amplitude and phase condition. i.e Both the reflected rays have the same amplitude and they should be out of phase by П

Amplitude condition: amplitude condition says that amplitude of both the reflected rays should be same. According to Fresnel’s equation,

($μ_f$ -$μ_a$ ) / ($μ_f$ + $μ_a$ ) = ($μ_g$ - $μ_f$ ) / ($μ_g$ + $μ_f$ )

Let $μ_f$ =x , $μ_g$ = y and $μ_a$ = 1

⸫(x-1)/(x+1) = (y-x)/(y+x)

⸫(x-1)(y+x) = (y-x)(x+1)

⸫ xy + $x^2$ -y-x = xy + y -$x^2$ -x

⸫ x=$\sqrt y$

i.e $μ_f$ = $\sqrt{(μ_g )}$

The refractive index of the material suitable for anti reflecting coating should be approximately square root of refractive index of glass. Usually MgF2 and cryolite are widely used.

Phase condition: Phase condition says that both the reflected rays should be out of phase byП.that means there should be complete destructive interference between the reflected rays.

Δ = (2n±1)λ/2

2$μ_f$ t cosr ± λ/2± λ/2 = (2n±1)λ/2

Since reflection at B and C occurs at the surface of denser medium,

2$μ_f$ t cosr ± λ=(2n±1)λ/2

For normal incidence cos r =1 and let t= $t_{min}$ for n=0

⸫ 2$μ_f$ $t_{min}$ = λ/2

$t_{min}$ = λ /4$μ_f$

that means the thickness of the film coated on the lens should be quarter of wavelength. Such a thick coating eliminates reflections and allows more light to pass through the optical component.

Non-reflecting coating reduces reflections from 4 to 5 % to less than 1 %. This 1 % reflection takes places on longer as well as shorter wavelength side. Hence thin film appears purple.