Lesson 4 of 5The Wave Nature of Light

Unit 4 · Lesson 4 of 5

Thin Films and Polarization

Why soap bubbles and oil slicks shimmer with colour, how a half-wavelength flip on reflection decides which colours survive, how coatings cancel glare, and how polarizing filters prove that light is a transverse wave.

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Key things to remember

1Wavelength inside a film: $λ_{f i l m} = λ / n_{f i l m}$ ; the frequency does not change.
2Reflection off a higher-index medium flips the wave by half a wavelength; reflection off a lower-index medium does not.
3One phase change: bright when $2 t = (m + \frac{1}{2}) λ_{f i l m}$ . Zero or two: bright when $2 t = m λ_{f i l m}$ . The dark conditions swap.
4The thinnest non-reflecting coating (two phase changes) has $t = \frac{λ}{4 n}$ .
5Unpolarized light through one filter keeps half its intensity; crossed filters block it; Malus's law $I = I_{1} cos^{2} θ$ gives the in-between cases.
6Only transverse waves can be polarized, so polarization shows light is transverse.

A soap bubble is made of clear water and soap, yet it swirls with colour. A puddle with a trace of oil on it shows rainbow bands that shift as you move your head. Nothing in either film is dyed. The colour comes from interference: light reflecting off the front of a very thin layer overlaps with light reflecting off the back, and at each point some wavelengths reinforce while others cancel. The second half of this topic looks at a different wave property, polarization, which only transverse waves can show — and light shows it.

Two reflections from one film

Fig 1Part of the light reflects from the top surface and part from the bottom. The second ray travels an extra distance of about twice the thickness.

When light hits a film of thickness $t$ , a little reflects from the top surface and most continues in. At the bottom surface a little more reflects and comes back out, running alongside the first reflected ray. Near normal incidence, the second ray has travelled an extra distance of $2 t$ inside the film. Inside the film, light is slower but its frequency is unchanged, so its wavelength is shorter. Path differences inside the film must be compared with this wavelength in the film:

λ_{f i l m} = \frac{λ}{n _{f i l m}}

Lambda is the wavelength in air or vacuum; n is the refractive index of the film.

The phase change on reflection

Light reflecting off a medium with a higher refractive index (slower light) is flipped by half a wavelength, just as a pulse on a string is inverted at a fixed end. Light reflecting off a medium with a lower index (faster light) is not flipped, like a free end. Light that is transmitted is never flipped. Before using any thin-film equation, count how many of the two reflections are flipped.

Phase changes	Examples	Bright reflection when	Dark reflection when
Exactly one	Soap film in air; oil ( $n = 1.45$ ) on water ( $n = 1.33$ ); air wedge between glass plates	$2 t = (m + \frac{1}{2}) λ_{f i l m}$	$2 t = m λ_{f i l m}$
Zero or two	Coating ( $n = 1.38$ ) on glass ( $n = 1.50$ ); any film whose index lies between the indices on either side of it	$2 t = m λ_{f i l m}$	$2 t = (m + \frac{1}{2}) λ_{f i l m}$

Here m = 0, 1, 2, ... The half-wavelength flip swaps the bright and dark conditions.

In white light, a film of one particular thickness reinforces some wavelengths and cancels others, so the reflection is coloured. A soap film drains downward and is thinner at the top, so the colour changes in horizontal bands. Just before it bursts, the top becomes much thinner than any visible wavelength. Then $2 t \approx 0$ and only the single phase change matters: every colour cancels and the film looks black. A quarter-wavelength-thick coating on a lens does the reverse job deliberately. With a flip at both surfaces, the extra path of $2 t = \frac{1}{2} λ_{f i l m}$ makes the two reflections cancel, so less light is reflected and more passes through — the principle behind anti-reflection coatings on eyeglasses, camera lenses and solar panels.

Air wedges measure tiny thicknesses

Rest one glass plate on another with a thin spacer at one end and a wedge of air forms between them. Seen in single-colour light there is a set of straight dark and bright fringes. The contact edge is dark (one phase change, zero thickness), and each further dark fringe marks an increase of

\frac{1}{2} λ

in the air gap. Counting fringes turns a wavelength into a ruler for gaps of a few micrometres.

Polarization

Fig 2A vertical filter passes vertically oscillating light. A second filter at 90 degrees blocks it; one at an in-between angle passes some.

Light is an electromagnetic wave, and its electric field oscillates at right angles to the direction of travel. Ordinary light from the Sun or a bulb is unpolarized: its oscillations point in every direction perpendicular to the ray, changing randomly. A polarizing filter transmits only the part of the oscillation along its transmission axis, so the light that emerges is plane-polarized and has half the original intensity. Put a second filter after the first with its axis at $9 0^{\circ}$ and nothing gets through: the filters are crossed. Slide a third filter between them at $4 5^{\circ}$ and light reappears, because the middle filter turns part of the oscillation toward the last filter's axis. Sound, a longitudinal wave, cannot be polarized at all, so polarization is direct evidence that light is transverse.

I = \frac{1}{2} I_{0} (unpolarized through one filter) I = I_{1} cos^{2} θ (polarized light through a filter at angle θ)

The second relation is Malus's law, where theta is the angle between the light's polarization and the filter's axis.

Polarization by reflection. Light reflecting off water, snow or a road is partly polarized parallel to the surface, which for a horizontal surface means horizontally. At one particular angle, where $tan θ_{B} = n_{2} / n_{1}$ (about $5 3^{\circ}$ from the normal for water), the reflected light is completely polarized.
Polarized sunglasses have vertical transmission axes, so they block the horizontally polarized glare while still passing about half of the unpolarized light from everything else.
Stressed plastic placed between two polarizing filters shows bands of colour, because stress changes how the plastic affects each wavelength's polarization. Engineers use this to see where a transparent model of a part is most strained.
Liquid crystal displays (LCDs) use polarizing filters, which is why a phone or monitor can look dark through polarized sunglasses when tilted.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A soap film (

n = 1.33

) in air is lit from the front with white light. What is the smallest thickness that strongly reflects green light of wavelength

540 nm

? What is the next thickness that does?

Example 2

A camera lens made of glass (

n = 1.50

) is coated with magnesium fluoride (

n = 1.38

). What is the minimum coating thickness that cancels reflected light of wavelength

550 nm

Example 3

An air wedge between two glass plates is lit from above with light of wavelength

589 nm

. The contact edge is dark. Counting the contact edge as dark fringe zero, how thick is the air gap at dark fringe number 12?

Example 4

Unpolarized light of intensity

800 W/m^{2}

passes through a vertical polarizer and then a horizontal polarizer. (a) What emerges? (b) A third filter at

4 5^{\circ}

is placed between the two. What emerges now?

Where marks get lost

Skipping the phase-change count. The same film thickness can be bright or dark depending on whether one or two reflections are flipped, so always count first.
Using the wavelength in air for the film. The path difference is inside the film, so divide by the film's refractive index.
Forgetting that the path difference is $2 t$ , not $t$ . The second ray goes down through the film and back up.
Thinking every reflection flips the wave. Only reflection off a higher-index (slower) medium does.
Treating the first polarizer as a $cos^{2} θ$ problem. Unpolarized light has no single angle; one filter always passes half.

Flashcards 8

Phase change on reflection

Wavelength inside a film

Anti-reflection coating thickness

Why a soap film turns black before it pops

+ 4 more

Practice

Light travelling in air reflects off the surface of a pane of glass. What happens to the reflected wave?

Light of wavelength

600 nm

in air enters a film with refractive index

1.50

. What is its wavelength inside the film, in nanometres?

A thin layer of oil (

n = 1.45

) floats on water (

n = 1.33

). What is the minimum oil thickness that strongly reflects light of wavelength

600 nm

? Answer to the nearest nanometre.

A coating with

n = 1.25

is applied to glass with

n = 1.52

. What is the minimum coating thickness, in nanometres, that cancels reflected light of wavelength

500 nm

Just before a vertical soap film bursts, its top edge looks black in reflected light. Why?

Unpolarized light of intensity

120 W/m^{2}

passes through two polarizing filters whose transmission axes are

6 0^{\circ}

apart. What intensity emerges, in

W/m^{2}

W/m²

Two polarizing filters are crossed so that no light gets through. A third filter is slid between them with its axis at

4 5^{\circ}

. What happens?

Polarization is evidence that light is what kind of wave: transverse or longitudinal?

An air wedge lit with

632.8 nm

light has a dark fringe at the contact edge. How thick, in nanometres, is the air gap at the 8th dark fringe beyond the contact edge?

Glare from a wet road is mostly horizontally polarized. Which way should the transmission axes of polarized sunglasses point?

Match each film to the number of reflections that are flipped by half a wavelength.

Soap film surrounded by air

Magnesium fluoride coating on glass

Air wedge between two glass plates

Order the steps for solving a thin-film interference problem.

1Solve 2t = (condition) for the thickness or wavelength
2Find the wavelength inside the film using lambda over n
3Choose the bright or dark condition that matches that count
4Identify the refractive index on each side of both reflecting surfaces
5Count how many of the two reflections are flipped

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