Thin Films & Interference

Pg. 502 - 507

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

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

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

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

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…

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 = /4 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 desctructivea fixed end (crest=trough) so the wave shifts interference

• Destructive interference *constructive interference

Hit this dashed line at the same time

Summary for Reflected Light Constructive interference occurs when:

λ /4 + 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 *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

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 <<1 For t = λ/4 For t = λ/2

*constructive *blue line is no longer a *shift by a wavelengthInterference crest b/c it has been shifted ½ wavelength

CrestReflects as atroughReflects as a crestReflects as a trough Transmits as a crest

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

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 = f1λ1 f1 = f2 because it is

coming from v f2λ2 the same source

c = λ1 ….and we know n = c/v v λ2

n = λ1 λ2

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/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