1. Thin Films & InterferencePg

2. Thin Films
Everyone, at some point, has witnessed thin film interference
It occurs when you see the colour spectrum in gasoline or oil that has been spilled or in a soap bubble
The effect occurs due to optical interference

3. How does it work?
Consider a horizontal film like a soap bubble that is extremely thin, compared to the wavelength of light direction at it from above.
Some is reflected Some is refracted

4. How does it work?
The light rats that were refracted, and then reflected have travelled a longer distance
∆L causes light waves to go out of phase
This results in destructive interference that is hitting our eye

5. Real-world examples: Soap bubbles Oil ….also “Newtons Rings”
Bowl of water on top of reflective glass
When you look into the bowl you see a series of rings from constructive and destructive interference

6. Recall from Gr. 11 Properties of reflected waves
Fixed end (less dense medium to a move dense medium)
Free end (more dense medium to a less dense medium)
Ring so that the spring can move up and down…

7. 3-Cases for Thin Film Interference-Reflected Light
Remember: transmitted light wave are always in phase from the source
Comparing film thickness (t) to the wavelength of light ( )
For t << For t = / for t = /2
*fill so thin that *to go across the film *to go across film it has
*Minimal time lag and back is ½ lamda travelled an extra whole
Wave reflects back like *wave travels ½ *back to desctructive a fixed end (crest=trough) so the wave shifts interference
Destructive interference *constructive interference
Hit this dashed line at the same time

8. Summary for Reflected Light
Constructive interference occurs when:
λ / λ /2 (n – 1)
Started at λ/4 and occurs every 1/2λ
n is the maximum (if n = 1, then…..)
= λ/4 + 2λ(n-1)/ *common denominators
= ¼ (λ + 2λn – 2λ)
= λ(2n – 1)/4
Destructive interference occurs when:
0λ + λ/2 (n – 1)
0λ , λ/2, λ, ……started at 0 and occurs every 1/2λ
= λ(n – 1) /2

9. 3-Cases for Thin Film Interference – Transmitted Light
Now interested in light on the other side
Remember: transmitted waves are always in phase from the source
Comparing film thickness to the wavelength of light
For t << For t = λ/ For t = λ/2
*constructive *blue line is no longer a *shift by a wavelength
Interference crest b/c it has been shifted
½ wavelength
Crest
Reflects as atrough
Reflects as a crest
Reflects as a trough
Transmits as a crest

10. Summary for Transmitted Light
Opposite to the reflected light formulas
Constructive interference occurs when:
= λ(n-1)/2
Destructive interference occurs when:
λ(2n – 1) /4

11. Equations for Thin Film Interference
From v = fλ, we know v is the speed of light
So, c = fλ. If we assume the initial equation is the velocity of light in a different medium then we can take a ratio of c/v:
c = f1λ f1 = f2 because it is coming from v f2λ the same source
c = λ ….and we know n = c/v
v λ2
n = λ1 λ2

12. Practice
In the summer, the amount of solar energy entering a house needs to be minimized. We do this by applying a thin film coating to maximize reflection of light. If light (assume λ = 578 nm) travels into an energy-efficient window, what thickness of the added coating (n = 1.4) is needed to maximize reflected light?
“minimized” tells us that it is destructive interference
need to find the wavelength in the new film
t= n = λ1/λ t = λ(2n – 1)/4 (n =1 *max)
λ1 =
n =
**realistically that is too thin to apply, can “ramp it up” by increasing n to 10, etc. to get a thickness you can actually apply