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 desctructive a 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 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

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