Thin Films Interference Pg 502 507 Thin Films

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 � Bowl “Newtons Rings” 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 � Fixed � Free of reflected waves end (less dense medium to a move dense medium) 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 *fill so thin that. . has *Minimal time lag whole *to go across the film and back is ½ lamda for t = /2 *to go across film it travelled an extra Hit this dashed line at the same time • Wave reflects back like desctructive *wave travels ½ *back to

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

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 = λ/4 For t <<1 Crest For t = λ/2 Reflects as atrough Reflects as a crest Reflects as a trough Transmits as a crest *constructive Interference *blue line is no longer a crest b/c it has been shifted ½ wavelength *shift by a wavelength

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

Equations for Thin Film Interference 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: � From �c = f 1λ 1 v f 2λ 2 � c = λ 1 v λ 2 n = λ 1 λ 2 f 1 = f 2 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? “minimized” tells us that it is destructive interference � need to find the wavelength in the new film t= n = λ 1/λ 2 t = λ(2 n – 1)/4 (n =1 *max) λ 1 = 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