Thin Films & Interference
Pg. 502 - 507
Thin Films
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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
How does it work?
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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
How does it work?
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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
Real-world examples:
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Soap bubbles
Oil
….also “Newtons Rings”
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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
Recall from Gr. 11
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Properties of reflected waves
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Fixed end (less dense medium to a move dense medium)
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Free end (more dense medium to a less dense medium)
Ring so that the spring can
move up and down…
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 <<
*fill so thin that..
*Minimal time lag
For t =
/4
*to go across the film
and back is ½ lamda
for t =
/2
*to go across film it has
travelled an extra whole
Hit this dashed
line at the same
time
•
•
Wave reflects back like *wave travels ½
*back to desctructive
a fixed end (crest=trough) so the wave shifts
interference
Destructive interference *constructive interference
Summary for Reflected Light
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Constructive interference occurs when:
λ /4 +
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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)/4
= ¼ (λ + 2λn – 2λ)
= λ(2n – 1)/4
*common denominators
Destructive interference occurs when:

0λ + λ/2 (n – 1)
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0λ , λ/2, λ, ……started at 0 and occurs every 1/2λ
= λ(n – 1) /2
3-Cases for Thin Film Interference –
Transmitted Light
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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 = λ/4
For t = λ/2
*blue line is no longer a
crest b/c it has been shifted
½ wavelength
*shift by a wavelength
For t <<1
Crest
Reflects as atrough
Reflects as a crest
Reflects as a trough
Transmits as a crest
*constructive
Interference
Summary for Transmitted Light
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Opposite to the reflected light formulas
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Constructive interference occurs when:
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= λ(n-1)/2
Destructive interference occurs when:
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λ(2n – 1) /4
Equations for Thin Film Interference
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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:
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c = f1λ1
v
f2λ2
c = λ1
v
λ2
n = λ1
λ2
f1 = f2 because it is coming from
the same source
….and we know n = c/v
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?
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t=
“minimized” tells us that it is destructive interference
need to find the wavelength in the new film
n = λ1/λ2
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