Question

emergency physics question!

what is the minimum film of coating with index of refraction of 1.52 on a glass with index of refraction of 1.40 for which destructive interference of a red component of white light in air can take place by reflection? assume the wavelength of the red component in vacuum is 650nm. find the minimum visible wavelength in vacuum for constructive interference due reflection for the thickness of the coating film you found.

what should be the minimum additional coating to get constructive interference for the red component of 650nm?

not just an answer, I need help understand the answer and the question!

STEP BY STEP explanation please!

Answer 1

For this question, there are two rays that interfere.

The first ray which gets reflected from the air(n=1) to film(n=1.52) interface. Due to this a phase change of /2 will occur.

And then comes the refracted ray which enters the film and gets reflected from the film(n=1.52) to glass(n=1.4).

For this no phase change is there as the ray comes from a medium of higher refractive index(n=1.52) to a lower refrctive index(n=1.4).

So, a total of  /2 phase change occurs between the two rays.

Now, the extra distance traveled between the two rays,x = 2*t

where t = thickness of the film

So, for destructive interference, extra distance traveled must be equal to odd multiple of ( /2 ) = (2m+1) /2

For, m = 1,

x = 3* /2

But the two rays are already  /2 out of phase <----- due to phase changes on the interfaces

So, 2t = 3* /2 -  /2 =  

So, t =  /2

where   = wavelength in the film = 650 nm / 1.52

So, t = (650/1.52)/2 = 213.8 nm <-------answer