Unit 14 · Topic 14.9

14.9 Thin-Film Interference

In a thin film, such as a soap bubble or an oil slick, light reflected from the top surface interferes with light reflected from the bottom surface. Whether that reflected light is strengthened or canceled depends on the film's thickness, the wavelength, and whether each reflection flips the wave by 180°. Antireflection coatings use this to cancel reflections, and changing thickness gives soap bubbles their colors.

Key terms

thin-film interference
phase change on reflection
index of refraction
antireflection coating
film thickness

Two reflections that interfere

When light hits a boundary, some reflects, some is transmitted and some may be absorbed. In a thin film, light reflects from the top surface (ray 1), and some light goes into the film, reflects from the bottom surface and comes back out (ray 2).

Ray 2 travels an extra distance: down through the film and back up. For light hitting the film straight on (normal incidence), that extra path is 2t, where t is the film's thickness.

\"Thin\" means the film is only about as thick as a wavelength of light, a few hundred nanometers. In a thick pane of glass the two reflections don't produce visible colors.

Phase changes on reflection

A reflection can flip the wave, a 180° phase change, which is the same as shifting it by half a wavelength. The rule is like the rope rule in 14.3:

Reflecting off a material with a higher index of refraction (slower light): 180° phase change.
Reflecting off a material with a lower index of refraction: no phase change.
Refraction (light passing into the next material) never changes the phase.

Wavelength inside the film

Inside the film, light is slower, so its wavelength is shorter: $\lambda_{\text{film}} = \frac{\lambda}{n_{\text{film}}}$ , where λ is the wavelength in air. The extra path 2t must be compared with this shorter wavelength.

Putting it together

Count the phase flips at the two surfaces, then compare 2t with $\lambda_{\text{film}}$ . These conditions aren't on the equation sheet, so build them from the reasoning each time.

Situation	Constructive (bright reflection)	Destructive (no reflection)
One reflection flips (soap film in air; oil on water)	$2t = (m + \frac{1}{2})\lambda_{\text{film}}$	$2t = m\lambda_{\text{film}}$
Both flip or neither flips (coating on glass)	$2t = m\lambda_{\text{film}}$	$2t = (m + \frac{1}{2})\lambda_{\text{film}}$

Real-world examples

Soap bubbles and oil films: the film's thickness varies from place to place, so different colors are reinforced at different spots, making swirling rainbow bands. Where a soap film gets much thinner than the wavelength, the extra path is nearly zero and the one flipped reflection cancels the other, so the film looks dark just before it pops.

Antireflection coatings: a thin layer on a lens or solar panel, with an index between air and glass, so both reflections flip and the flips cancel out. Making the coating a quarter of the wavelength thick, measured inside the coating, gives an extra path of half a wavelength, so the two reflections cancel. The minimum thickness is $t = \frac{\lambda}{4n_{\text{coating}}}$ . Less reflected light means more light gets through.

Calculations in this course use light hitting the film straight on. At other angles the colors shift, but you only need to explain that qualitatively.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Antireflection coating
A camera lens (n = 1.50) is coated with magnesium fluoride (n = 1.38) to cut reflection of 550 nm light, near the middle of the visible spectrum. Find the minimum coating thickness.
Show the solutionHide the solution
1. Step 1: Top surface: air (1.00) to coating (1.38), a higher index, so that reflection flips. Bottom surface: coating (1.38) to glass (1.50), also higher, so that reflection flips too.
2. Step 2: Both flip, so the flips cancel out. To get destructive interference, the extra path must be half a wavelength in the coating: $2t = \frac{1}{2}\lambda_{\text{film}}$ (m = 0).
3. Step 3: $\lambda_{\text{film}} = \dfrac{550\text{ nm}}{1.38} = 399$ nm, so $t = \dfrac{\lambda_{\text{film}}}{4} = \dfrac{550}{4(1.38)} = 99.6$ nm, about 100 nm.
Answer: About 100 nm (99.6 nm)
Example 2Calculator allowed
Color of a soap film
A soap film (n = 1.33) in air is 300 nm thick. Viewed straight on, which visible wavelength (400 to 700 nm in air) is most strongly reflected?
Show the solutionHide the solution
1. Step 1: Top surface: air to soap, higher index, so that reflection flips. Bottom surface: soap to air, lower index, no flip. One flip.
2. Step 2: With one flip, bright reflection needs $2t = (m + \frac{1}{2})\lambda_{\text{film}}$ . Write $\lambda_{\text{film}} = \frac{\lambda}{n}$ and solve: $\lambda = \dfrac{2nt}{m + \frac{1}{2}} = \dfrac{2(1.33)(300)}{m + \frac{1}{2}} = \dfrac{798\text{ nm}}{m + \frac{1}{2}}$ .
3. Step 3: m = 0 gives 1596 nm (infrared), m = 1 gives 532 nm (green), m = 2 gives 319 nm (ultraviolet).
4. Step 4: Only 532 nm is visible, so the film looks green where it's 300 nm thick.
Answer: 532 nm (green)
Example 3Calculator allowed
Oil on water (trap)
A thin layer of oil (n = 1.45) floats on water (n = 1.33). What is the thinnest oil layer that strongly reflects 580 nm yellow light, viewed straight on?
Show the solutionHide the solution
1. Step 1: Top: air to oil, higher index, flip. Bottom: oil to water, lower index (1.33 < 1.45), no flip. So there's one flip, just like a soap film.
2. Step 2: The trap is assuming that because water is below, it acts like glass under a coating. Compare the actual indices.
3. Step 3: One flip, so bright reflection needs $2t = (m + \frac{1}{2})\lambda_{\text{film}}$ . The thinnest is m = 0: $2t = \frac{1}{2}\lambda_{\text{film}}$ , so $t = \dfrac{\lambda}{4n} = \dfrac{580}{4(1.45)} = 100$ nm.
Answer: 100 nm

Common mistakes

Using the wavelength in air instead of in the film. Divide by the film's index first.
Forgetting to check both surfaces for a phase flip. Compare the indices at each surface separately.
Thinking refraction causes a phase shift. Only reflection off a higher-index material does.
Using 2t = mλ for every film. Which condition gives bright or dark depends on how many flips there are.

On the exam

Questions usually ask for a minimum thickness, or which color is reflected or canceled. Show the phase-flip reasoning at each surface; that reasoning earns points even before the arithmetic.
You may be asked to explain why a soap bubble shows changing colors, or why it looks dark at its thinnest. Tie the answer to thickness, wavelength and phase changes.

Connected topics

Videos

AP Physics 2 - Unit 15 - Lesson 9 - Thin Film Intereference
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Thin Film Interference part 1 | Light waves | Physics | Khan Academy
khanacademymedicineWatch on YouTube (opens in a new tab)
Spectra Interference: Crash Course Physics #40
CrashCourseWatch on YouTube (opens in a new tab)
Thin Film Interference | 24.2 General Physics
Chad's PrepWatch on YouTube (opens in a new tab)
Thin Film Interference part 2 | Light waves | Physics | Khan Academy
khanacademymedicineWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 14.9 Thin-Film Interference. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A glass lens (n = 1.50) is coated with a thin layer of magnesium fluoride (n = 1.38) to reduce reflection of 550 nm light arriving straight on. What is the thinnest coating that works?

Question 2 of 4Calculator allowed

A layer of oil (n = 1.45) 300 nm thick floats on water (n = 1.33). White light shines straight down on it. Which visible wavelength (400–700 nm) is most strongly reflected?

Question 3 of 4Calculator allowed

A vertical soap film (n = 1.33) in air drains and becomes very thin at the top, much thinner than a wavelength of light. In reflected light, the top looks black. Why?

Question 4 of 4Calculator allowed

Light in air hits a thin layer of water (n = 1.33) lying on glass (n = 1.50). Which reflections undergo a half-wavelength phase change?

0 of 4 answered