Finding the Minimum Thickness for Destructive Interference in Thin Films

In summary, the conversation discusses the calculation of the minimum thickness of TiO2 that must be added to a uniform film on crown glass in order to cancel the reflected light from air at a wavelength of 515 nm. The formula for destructive reflection from a thin film is used, and it is determined that adding 45 nm to the existing film will achieve the desired result.
  • #1
Sofija Zdjelar
4
0

Homework Statement


A uniform film of TiO2, 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 515 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels.

What is the minimum thickness of TiO2 that you must add so the reflected light cancels as desired?

Homework Equations


Am i using the wrong formulas? The examples in my book seem to solve these kind of exercises easily... How should i solve this exercise?

The Attempt at a Solution


I have used the formula for destructive reflection from thin film, half-cycle phase shift (2t = mλ with λ = λair/nfilm). However, it is not correct. I have also tried to use the formula for destructive reflection from thin film, no relative phase shift ((2t = m + 1/2)λ), which also did not work. I have mainly used m = 1 because i read somewhere that it is 1 when calculating the minimum thickness.
 
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  • #2
Hello and welcome to PF.

If you put the original value of the thickness of the film into the formula 2t = mλ, what would you get for m? Do you get an integer?
 
  • #3
Hi! Thank you for answering. No, i do not get an integer...
 
  • #4
Sofija Zdjelar said:
Hi! Thank you for answering. No, i do not get an integer...
OK. So, that means the original thickness of the film gives neither constructive nor destructive interference. As you increase the thickness of the film beyond the initial thickness, what would be the first value of m for which you would get destructive interference?
 
  • #5
I get m = 11 (when i use the formula 2t = mλ and λ being λair/nfilm
Not sure if 11 is correct, or what I am supposed to do with it.
 
  • #6
I think m = 11 is right. What is the thickness of the film that corresponds to m = 11? How would you use this to answer the question?
 
  • #7
The thickness that corresponds to that m is 1081, which means that i have to add 45 nm to the existing film. I tried it and it was right! Thank you so much :-)
 
  • #8
OK. Good work.
 

1. What is interference in thin films?

Interference in thin films is a phenomenon that occurs when light waves reflect off of two surfaces of a thin film and interact with each other. This interference can result in the production of bright or dark fringes, depending on the thickness of the film and the wavelength of the light.

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

The thickness of a thin film plays a crucial role in interference. When the thickness of the film is an integer multiple of the wavelength of the light, constructive interference occurs and bright fringes are produced. When the thickness is a half-integer multiple, destructive interference occurs and dark fringes are produced.

3. What is the difference between thin film interference and bulk interference?

The main difference between thin film interference and bulk interference is the number of surfaces involved. Thin film interference occurs between two surfaces, while bulk interference involves multiple surfaces or layers of different materials.

4. What are some real-world applications of thin film interference?

Thin film interference has many practical applications, including anti-reflective coatings on eyeglasses and camera lenses, thin film solar cells, and the colorful patterns seen on soap bubbles and oil slicks.

5. How does the refractive index of a thin film affect interference?

The refractive index of a thin film, which is a measure of how much a material slows down the speed of light, can significantly impact interference. A higher refractive index can cause a phase shift in the light waves, which can alter the interference pattern and result in different fringe patterns.

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