Thin Film Interference formula

In summary, the correct formula for the optical path difference in thin film interference is \delta=2nt + \frac{\lambda}{2}, taking into account the phase shift upon reflection from a medium with a higher refractive index. The mistake in the first formula was not accounting for the phase shift in the second wave.
  • #1
Dr. Jekyll
35
0
Hi everyone!

Here's the problem related to thin film interference. I mean, it's not quite a problem. I've been looking for the formulae related to this type of interference and fount two different things in two different textbooks.

It's the formula for the optical difference of the wave reflected from the top (air-film) surface and the wave reflected from the bottom (film-air) surface. Now, I ran into these two formulae (for the same thing):

[tex]\delta=2nt - \frac{\lambda}{2}[/tex] (1)

and

[tex]\delta=2nt + \frac{\lambda}{2}[/tex] (2).

Now, I'm somehow sure that formula (2) is the right one but I can't find any mathematical proofs for it.

Now, I can get the (1) formula but I'm probably wrong somwhere.
This is how I would do it:

We have a thin film (t is the thickness) and the light beams vertically on the film. I also took a spot somewhere above the film (x will be distance from the film). It's actually not relevant but makes things a bit more clearer (at least for me).

The wave 1 reflects from the upper surface and has a phase shift [itex]\frac{\lambda}{2}[/itex]. The second, wave 2, reflects from the lower surface (without any phase shift). Now, this would be the difference when they both reach the spot:

[tex]\delta=x+2nt+x - (x+\frac{\lambda}{2}+x) \Rightarrow \delta = 2nt - \frac{\lambda}{2}[/tex].

Please tell me where did I go wrong.

Sorry for maybe the bad english. It's not my native language.

Thanks in advance!
 
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  • #2


Hello!

Thank you for bringing up this issue with thin film interference formulas. It's actually a common confusion for many students and researchers alike. The correct formula for the optical path difference in thin film interference is indeed the second one you mentioned, \delta=2nt + \frac{\lambda}{2}.

The reason for this is that when light reflects off a medium with a higher refractive index (in this case, the film), there is a phase shift of \frac{\lambda}{2}. This is known as the phase change upon reflection. This phase shift needs to be taken into account when calculating the optical path difference between the two waves reflected off the top and bottom surfaces of the film.

In your calculation, you correctly accounted for the phase shift in the first wave, but you forgot to take into account the phase shift in the second wave. This is why your calculation resulted in the first formula with a negative sign for the phase shift. However, when you include the phase shift in both waves, you get the second formula with a positive sign for the phase shift.

I hope this clarifies things for you. Let me know if you have any other questions or concerns. Good luck with your studies!
 
  • #3


Hi there!

Thank you for sharing your question about the thin film interference formula. I can understand your confusion as there are different formulae for the same concept in different sources. However, I can assure you that both formulae are correct and can be used to calculate the optical difference in thin film interference.

Formula (1) is known as the "phase difference formula" and it calculates the phase difference between the two reflected waves. This is why it includes the phase shift of \frac{\lambda}{2}. On the other hand, formula (2) is known as the "path difference formula" and it calculates the difference in the optical path length between the two reflected waves. This is why it does not include the phase shift.

In your explanation, you have correctly used the path difference formula to calculate the difference in the optical path length. However, if you want to use the phase difference formula, you would need to take into account the phase shift of \frac{\lambda}{2} for the first reflected wave as well. So the correct calculation using the phase difference formula would be:

\delta=x+2nt+\frac{\lambda}{2}+x-(x+\frac{\lambda}{2}+x) \Rightarrow \delta=2nt+\frac{\lambda}{2}.

I hope this helps to clarify the difference between the two formulae. Both can be used to calculate the optical difference in thin film interference, but they are based on different principles. It's always good to check with multiple sources and understand the underlying principles to avoid confusion. Keep up the good work in your studies!
 

1. What is the basic formula for thin film interference?

The formula for thin film interference is given by:
I = (4 * t * n) / λ
Where I is the intensity of the reflected light, t is the thickness of the film, n is the refractive index of the film, and λ is the wavelength of incident light.

2. How does the thickness of the film affect the interference pattern?

The thickness of the film directly affects the path difference of the light waves, which in turn affects the constructive and destructive interference of the waves. A thicker film will result in a larger path difference and a more pronounced interference pattern.

3. Can the refractive index of the film be greater than that of the incident medium?

Yes, the refractive index of the film can be greater than that of the incident medium. This is known as a high-index film and can result in a phase shift of the reflected light waves, leading to a different interference pattern compared to a low-index film.

4. How does the wavelength of incident light affect thin film interference?

The wavelength of incident light determines the spacing of the interference fringes. A shorter wavelength will result in narrower fringes, while a longer wavelength will result in wider fringes. This can be seen in the equation, where a higher value of λ will lead to a smaller value of I and vice versa.

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

Thin film interference is used in a variety of applications, such as anti-reflective coatings on eyeglasses and camera lenses, thin film solar cells, and in the production of colorful coatings on CDs, DVDs, and other reflective surfaces. It is also an important concept in the study of optical properties of materials in physics and engineering.

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