Thin Film Interference and reflection

In summary, the goal is to find the thinnest coating with a refractive index of 1.43 on glass (with a refractive index of 1.52) that will cause destructive interference of the red component (660 nm) of an incident white light beam in air. The equation used is 2t = m*lambda, where t is the thickness, m is an integer, and lambda is the wavelength. The solution is to create a layer of thickness equal to 1/4 of the wavelength of the incident light, which can be calculated using the equation d = lambda_0/(4n), where lambda_0 is the incident wavelength and n is the refractive index of the coating material.
  • #1
pconn5
9
0

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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  • #2
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
 
  • #3
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 [itex] \lambda_0 [/itex], that in the material is [itex] \lambda=\lambda_0/n [/itex]. So the thickness of the antireflecting coating should be [itex] d=\lambda_0/(4n)[/itex].

ehild
 
  • #4
Thank you very much. That helped me out a lot. I couldn't not figure that out at all.

Thanks again.
 

FAQ: Thin Film Interference and reflection

1. What is thin film interference and how does it occur?

Thin film interference is a phenomenon that occurs when light waves reflect off of two surfaces of a thin film material, causing constructive and destructive interference. This results in the formation of bright and dark fringes, and is caused by the difference in the path length of the reflected waves.

2. How does the thickness of a thin film affect interference?

The thickness of a thin film plays a crucial role in thin film interference. The path length difference between the two reflected waves increases as the thickness of the film increases, resulting in a larger phase difference and more pronounced interference patterns.

3. What factors influence the color of a thin film?

The color of a thin film is influenced by several factors, including the thickness of the film, the refractive indices of the materials involved, the angle of incidence of the light, and the wavelength of the incident light. The combination of these factors determines the color of the reflected light.

4. How is thin film interference used in practical applications?

Thin film interference has many practical applications, such as in the production of anti-reflective coatings for glasses and camera lenses, in the creation of colorful patterns for soap bubbles and oil slicks, and in the production of thin film solar cells. It is also used in interferometry, a technique used for measuring small changes in distance or refractive index.

5. Can thin film interference occur with other types of waves besides light?

Yes, thin film interference can occur with other types of waves besides light, such as sound waves, water waves, and electromagnetic waves. The principles of thin film interference remain the same, but the specific equations and calculations may differ depending on the type of wave.

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