Question

A thin film of ?g?2 has a refractive index n = 1.38 and it is deposited on glass as an anti-reflective layer/stratum for light with wavelength 580nm that falls in the normal direction towards the film. What wavelength is minimally reflected when the light falls at 45 ° towards the film?

Answer 1

To find the wavelength of minimally reflected light, we first find the thickness of the thin film,

The condition for destructive interference in case of thin films for normal incidence is given as follows:

n is refractive index of the thin film

t is thickness of the thin film

is wavelength of light incident at normal

rearranging above equation,

considering first order interference, m= 1,

,

equation 1 becomes,

In the given problem, the angle of incidence is

the term wavelength minimally reflected implies destructive interference for the wavelength incident at angle of

the condition for destructive interference for angle of incidence other than , which is the angle for normal incidence is

again considering first order interference, m=1,

t is the thickness of the thin film, .......(from the previous calculation)

n is the refractive index of the thin film

or the wavelength minimally reflected, is to be calculated

light is incident from air on the thin film

In equation 2, the solution requires angle of refraction , but we have angle of incidence and refractive index of medium from which light is incident (air) for which and refractive index of medium ( thin film of ?g?2) from which the light gets refracted ,

using snell's law we can find the angle of refraction,

Snell's law is given as follows:

rearranging the terms,

Substituting the values,

is the angle of refraction,

again consider the eq,

substituting values of m, t, n, and in above equation, we get,

that is,

rounding off,

is the wavelength that is minimally reflected when the light falls at 45 ° towards the film.