How Does Soap Bubble Thickness Affect Light Interference?

In summary: Also, why exactly does this only occur for odd multiples of the thickness?Constructive interference only occurs for odd multiples of the thickness because this is the minimum thickness at which destructive interference ceases and constructive interference begins.
  • #1
vetgirl1990
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Homework Statement


(a) Calculate the minimum thickness of a soap bubble film that results in constructive interference in the reflected light if the film is illuminated with light whose wavelength in free space is λ=600nm. The index of refraction of the soap film is 1.33.

(b) What is the film is twice as thick? Does this situation produce constructive interference?

Homework Equations


Constructive interference occurs here, so: 2nt = (m + ½)λn

The Attempt at a Solution


(a) Part A is pretty straightforward, as it just involves recognizing that this is a constructive interference case and then plugging in what we know.
2nt = (m + ½)λn
t = (0 +½)λ / 2n = λ / 4n = 600nm / 4(1.33) = 113nm

(b) The answer for Part B shows:
t = (m+½)λ/2n = (2m + 1) λ/4n
m = 0, 1, 2...
"The allowed values of m show that constructive interference occurs for odd multiples of the thickness corresponding to m=0, t=113nm. Therefore, constructive interference doesn't occur for a film that is twice as thickness

I'm confused as to what exactly they did to arrive at this conclusion. What is meant by "odd multiples of the thickness"? The answers doubled every value in the above equation except for thickness (t). Why wouldn't this be done, as the question is specifically asking "what if the film was twice as thick"?
 
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  • #2
vetgirl1990 said:
I'm confused as to what exactly they did to arrive at this conclusion. What is meant by "odd multiples of the thickness"? The answers doubled every value in the above equation except for thickness (t). Why wouldn't this be done, as the question is specifically asking "what if the film was twice as thick"?
They mean odd multiples, (2m+1), of λ/4n. I.e. odd multiples of the minimum thickness for destructive interference.
They did not double all values except t. They just multiplied the top and bottom of a fraction by 2.
 
  • #3
haruspex said:
They mean odd multiples, (2m+1), of λ/4n. I.e. odd multiples of the minimum thickness for destructive interference.
They did not double all values except t. They just multiplied the top and bottom of a fraction by 2.

I'm still confused as to why exactly this means that constructive interference doesn't occur for a film that is twice as thick.
 
  • #4
vetgirl1990 said:
I'm still confused as to why exactly this means that constructive interference doesn't occur for a film that is twice as thick.
Let the minimum thickness for complete destructive interference be t0=λ/4n. So in general it occurs for all odd multiples, (2m+1)t0. 2t0 is an even multiple of t0.
 
  • #5
haruspex said:
Let the minimum thickness for complete destructive interference be t0=λ/4n. So in general it occurs for all odd multiples, (2m+1)t0. 2t0 is an even multiple of t0.
Sorry just to clarify... isn't the minimum thickness for complete destructive interference 2nt = mλ (m=0), so t = λ/2n?
And instead, for complete constructive interference be t0=λ/4n. That's what my textbook says, at least.

Also, why exactly does this only occur for odd multiples of the thickness?
EDIT: I figured out why it needs to be ODD multiples of the thickness. For anyone else that may be having trouble understanding, I really dumbed it down for myself, but this is how I conceptualized what's going on...

For us to see constructive interference at the surface of the bubble, the TOTAL phase shift needs to be 2π (ie. 360°)
If the reflected wave already produces a 180° phase shift, then we know that the wave of light in the film needs to ALSO produce a 180° phase shift -- but from the extra distance traveled (t) in the film, rather than due to the refractive index.

180° (or π) corresponds to HALF a wavelength. But since the light in the film travels a total distance of 2t, the thickness of the film must correspond to a QUARTER of the wavelength.
Therefore, in general then, we can say that in order for a shift of additional distance of 180° to be produced, the thickness of the film needs to be t = ¼λfilm. This occurs at t=¼λfilm, t=3/4λfilm, t=5/4λfilm (odd integer multiples of the thickness!)

Now, if we add up all the phase shifts that have happened:
Reflected ray -- phase shift of 180°
Refracted ray in the film -- phase shift of 90°+90° = 180°
Total shift = 360° or 2π --> our condition for constructive interference
 
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  • #6
vetgirl1990 said:
Sorry just to clarify... isn't the minimum thickness for complete destructive interference 2nt = mλ (m=0), so t = λ/2n?
And instead, for complete constructive interference be t0=λ/4n. That's what my textbook says, at least.

Also, why exactly does this only occur for odd multiples of the thickness?
Sorry, I was thinking in terms of the reflected wave. When there's destructive interference in the reflected wave there's constructive interference in the propagated wave. Just swap con- with de- in what I wrote.
 
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1. What is the concept of interference in thin films?

Interference of thin films refers to the interaction of light waves as they pass through a thin film of material. This interaction results in the formation of bright and dark fringes, known as interference patterns, due to the superposition of the waves.

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

The thickness of a thin film plays a crucial role in determining the interference pattern. When the thickness is equal to a specific fraction of the wavelength of light, constructive interference occurs, resulting in bright fringes. Conversely, when the thickness is equal to half the wavelength, destructive interference occurs, resulting in dark fringes.

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

Thin film interference and diffraction are two related phenomena, but they differ in the source of the interference pattern. In thin film interference, the interference pattern is caused by the superposition of two light waves reflected from the top and bottom surfaces of the thin film. In thin film diffraction, the interference pattern is caused by the bending of light waves as they pass through a slit or opening in the thin film.

4. How can the colors observed in a thin film interference experiment be explained?

The colors seen in a thin film interference experiment are a result of the different wavelengths of light interfering with each other. As light passes through the thin film, some wavelengths are enhanced by constructive interference, while others are canceled out by destructive interference. The remaining wavelengths are the ones that are visible to our eyes, resulting in the observed colors.

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

Thin film interference is a crucial concept in various fields, such as optics, materials science, and engineering. It is used in the production of anti-reflective coatings for eyeglasses and camera lenses, as well as in the design of thin film solar cells. It is also essential in understanding the colorful patterns seen on soap bubbles and oil slicks.

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