Thin Film Interference and reflection

[pconn5]

Homework Statement

What is the thinnest film of a coating with n = 1.43 on glass (n=1.52) for which destructive interference of the red component (660 nm) of an incident white light beam in air can take place by reflection?

Homework Equations

2t = m*lambda?

The Attempt at a Solution

There really is none. Honestly I have been reading through the book for like an hour and trying to figure out what everything means but I just don't understand what to do at all. Any help at all is really appreciated. This just doesn't make sense to me

 

[Andrew Mason]

Homework Statement

What is the thinnest film of a coating with n = 1.43 on glass (n=1.52) for which destructive interference of the red component (660 nm) of an incident white light beam in air can take place by reflection?

Homework Equations

2t = m*lambda?

The Attempt at a Solution

There really is none. Honestly I have been reading through the book for like an hour and trying to figure out what everything means but I just don't understand what to do at all. Any help at all is really appreciated. This just doesn't make sense to me

The purpose of the coating is to create two reflecting surfaces separated by a small distance (some light will always reflect when it passes from one medium into another with a different index of refraction). If this distance is 1/4 of a wavelength of the incident light, the light reflecting off the coating/lens surface destructively interferes with the light reflecting off the air/coating surface (ie the wave reflected from the coating/glass boundary is a 1/2 wavelength out of phase with the light reflecting from the air/coating boundary). This reduces light reflection and increases the amount of light passing through the lens.

So you have to create a layer of thickness equal to 1/4 of the wavelength of the light.

AM

 

[ehild]

The thickness of the coating should be 1/4 of the wavelength inside the layer. The wavelength changes when the light enters from air to an other material with refractive index n. If the wavelength of the incident light is $\lambda_0$, that in the material is $\lambda=\lambda_0/n$. So the thickness of the antireflecting coating should be $d=\lambda_0/(4n)$.

ehild

 

[pconn5]

Thank you very much. That helped me out a lot. I couldn't not figure that out at all.

Thanks again.