Thin Film Interference and reflection

AI Thread Summary
The discussion centers on determining the thinnest film of a coating with a refractive index of 1.43 on glass (n=1.52) that allows for destructive interference of red light (660 nm) by reflection. The key concept is that the film thickness must be 1/4 of the wavelength of light within the coating material. The wavelength of light changes when entering a medium with a different refractive index, requiring the calculation of the effective wavelength in the coating. The formula provided indicates that the thickness should be d = λ₀/(4n), where λ₀ is the wavelength in air. This understanding significantly aids in solving the problem.
pconn5
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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 :confused:
 
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pconn5 said:

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

Thanks again.
 

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