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Thin Films, Newton's Rings and Other Interference · Lesson 1 of 5

Phase Change on Reflection

13 min read2 ideas6 checks3 figures to playBy Asheshanand Acharya
What you will learn 0 / 2
Stokes treatment of reflection
Look around
The black top of a soap film

At a mela, a child dips a wire ring into soapy water and holds it up. A flat film fills the ring, and bands of colour drift slowly down it. The water drains downward, so the film grows thinner at the top. Just before it bursts, the top of the film turns black. There is still a film there, because the colours below it stay put. It simply reflects no light.

you will find out why the thinnest part of a soap film looks black, though light still reflects from both of its faces.

Two reflections from one film

A thin film, such as a soap film or a layer of oil on water, has two surfaces. Light reflects a little from each of them. So the eye receives two reflected waves from every point of the film. Both come from the same incident wave, so they are coherent, and they interfere.

Whether the two waves add or cancel depends on two things. The first is the extra path that one wave travels inside the film. The second is any change of phase that a wave suffers when it reflects. This lesson settles the second question; the next lesson finds the first.

What the string already told us

The chapter Superposition and Reflection showed a pulse reflecting from the end of a string. At a fixed end the pulse came back upside down, a phase change of \(\pi\). At a junction of a light string with a heavy one, the reflected pulse also came back inverted. Going from heavy to light, it came back upright. On the heavy string the wave is slower.

Light behaves the same way. In an optically denser medium, one of larger \(\mu\), light is slower. So we expect a light wave reflected from the surface of a denser medium to come back turned over. A wave reflected from a rarer medium should come back upright. George Stokes proved this for light with nothing more than the principle of reversibility.

Stokes' argument

Let a wave of amplitude \(a\) meet the surface of glass from air. Let \(r\) be the fraction of the amplitude that is reflected and \(t\) the fraction transmitted. So the reflected wave has amplitude \(ra\) and the transmitted wave \(ta\). For light going from glass into air, call the same fractions \(r'\) and \(t'\).

Stokes' argument from reversibilityAir is above a flat surface and glass below, with a dashed normal at the point O. In the first chip a wave of amplitude a comes down from the upper left to O. A part r a is reflected to the upper right and a part t a is transmitted into the glass towards the lower right. In the second chip both of these rays are reversed and run back into O. The reversed reflected wave splits into r squared a going to the upper left and r t a going into the glass to the lower left; the reversed transmitted wave splits into t t prime a going to the upper left and r prime t a going to the lower left. Readouts state what reversibility demands: r squared plus t t prime equals 1 along the upper left, and r prime equals minus r along the lower left. O a ra ta air glass air to glass: r = −0.2, t = 0.8 4 % of the light is reflected
Fig 1.1Choose the chips. In the real rays, a wave of amplitude \(a\) splits at O into \(ra\) reflected and \(ta\) transmitted. Reverse both rays. Each splits again, and reversibility demands that the pieces rebuild the original wave and nothing else.

Now reverse both rays, as in the second chip of Fig 1.1. By reversibility they must recombine into the original wave, travelling back along the upper left, and send nothing along the lower left. The reversed reflected wave \(ra\) splits into \(r^2a\) up and \(rta\) down. The reversed transmitted wave \(ta\) splits into \(tt'a\) up and \(r'ta\) down.

Deriving \(r' = -r, \qquad tt' = 1 - r^2\) step 1 of 4
1

\[ r^2a + tt'a = a \]

Why. the two waves leaving along the upper left must add up to the original wave \(a\).

RememberStokes' relations
\[ r' = -r, \qquad tt' = 1 - r^2 \]
Holds whena wave meeting the boundary of two transparent media that absorb nothing; unprimed from the first medium, primed from the second.
Symbols\(r, t\) the fractions of the amplitude reflected and transmitted for light from the first medium; \(r', t'\) the same for light from the second medium
Unitsall are pure numbers

The minus sign is the whole result. One of \(r\) and \(r'\) is negative, so one of the two reflections turns the wave over. Which one? At normal incidence the reflected fraction is \(r = (\mu_1 - \mu_2)/(\mu_1 + \mu_2)\). From air into glass, \(r = (1 - 1.5)/(1 + 1.5) = -0.2\). From glass into air, \(r' = +0.2\). So the reflection from the denser side turns the wave over; the one from the rarer side does not.

Deeper

The formula \(r = (\mu_1 - \mu_2)/(\mu_1 + \mu_2)\) comes from the electromagnetic theory of light, the Fresnel equations, and is beyond the syllabus. It also gives \(t = 2\mu_1/(\mu_1 + \mu_2)\): \(0.8\) from air into glass and \(1.2\) from glass into air. Check Stokes: \(tt' = 0.8 \times 1.2 = 0.96\) and \(1 - r^2 = 1 - 0.04 = 0.96\). The intensity goes as the square of the amplitude, so bare glass reflects \(0.2^2 = 4\,\%\) of the light at each surface.

principle 6.1
Phase change on reflection (Stokes' rule)

Light reflected at the surface of an optically denser medium, one of higher refractive index, suffers a phase change of \(\pi\). Light reflected at the surface of a rarer medium suffers no change of phase. A transmitted wave never changes its phase.

A reflected wave, in phase or turned overA light wave travels from the left towards a boundary in the middle of the figure. The top row shows the incident wave. The bottom row shows the reflected wave travelling back to the left, drawn three times larger than its true size. A dot on each row marks the wave at the boundary. With the chip air to glass, the second medium is denser; when the incident dot rises the reflected dot falls, so the reflected wave is turned over, a phase change of pi. With the chip glass to air, the second medium is rarer, and the two dots rise and fall together, so there is no change of phase. A readout gives the amplitude ratio r, minus 0.2 or plus 0.2. incident wave reflected wave air glass, μ = 1.5 air to glass: r = −0.2, phase change π
Fig 1.2Press play and watch the two dots at the boundary. From air to glass, the reflected dot falls when the incident dot rises: the reflected wave is turned over. Choose glass to air, and the dots rise and fall together. The reflected wave is drawn three times its true size.

In Fig 1.2 the boundary itself shows the rule. From air into glass, a crest arrives just as the reflected wave leaves with a trough. From glass into air, a crest leaves as a crest. Both reflected waves are weak, a fifth of the incident amplitude, but only one is inverted.

A phase of \(\pi\) is half a wavelength of path

A wave that falls behind by one wavelength of path is back in step, a phase of \(2\pi\). So an extra path \(\Delta x\) gives a phase \(2\pi\,\Delta x/\lambda\), as the chapter Interference and Reflection of Sound found. A phase of \(\pi\) is therefore the same as an extra path of \(\lambda/2\). We shall count the turn at a denser surface as an extra half wave.

RememberThe half wave on reflection
\[ \text{reflection from a denser medium:}\quad \Delta\varphi = \pi \;\Longleftrightarrow\; \Delta x = \dfrac{\lambda}{2} \]
Holds whenreflection at a boundary into a medium of higher refractive index; no change for reflection from a rarer medium or for transmission.
Symbols\(\Delta\varphi\) the phase gained on reflection; \(\Delta x\) the equal extra path; \(\lambda\) the wavelength in the medium the wave travels in
Units\(\Delta\varphi\) in rad; \(\Delta x, \lambda\) in nm or m
CheckpointUnanswered

A ray in water (\(\mu = \tfrac43\)) meets the flat surface of a glass block (\(\mu = 1.5\)) and part of it is reflected back into the water. What happens to its phase?

Checkpoint · numericUnanswered

Light of wavelength \(600\,\text{nm}\) reflects from a glass surface in air. The phase change of \(\pi\) is the same as what extra path in air, in nm?

nm± 2 %
Checkpoint · numericUnanswered

For light going from air into a certain glass at normal incidence, \(r = -0.2\). Use Stokes' relations to find the product \(tt'\).

± 2 %
Story · George Gabriel Stokes, 1849
The perfect blackness of a spot

Newton had seen that the centre of his rings, where a lens touches a glass plate, is black. In 1802 Thomas Young explained the dark centre by guessing that one of the two reflections loses half a wave. Nobody could prove the guess, or say why the spot was perfectly black and not merely dim. In 1849 George Stokes, a young professor at Cambridge, published a short paper on exactly this question. He reversed the rays at a surface and showed that the two reflections, from inside and outside, must differ in sign. His argument needed no theory of what light is, only that its paths can be retraced.

Which surface gives the half wave

Two surfaces, two answers

Now apply the rule to a film. A soap film hangs in air. At its top surface, light goes from air towards soapy water, a denser medium, so that reflection gains \(\pi\). At its bottom surface, light inside the film meets air, a rarer medium, so that reflection gains nothing. The two reflected waves therefore differ by half a wave on top of any path difference.

Which reflections of a film gain half a waveA thin film lies between air above and a second medium below; its thickness is exaggerated. A ray comes down from the upper left. Part of it reflects at the top surface as ray 1; the rest crosses the film, reflects at the bottom surface and leaves as ray 2, parallel to ray 1. Labels on the right say whether each reflection adds a phase of pi. Chips choose a soap film in air, oil on water, and a coating of refractive index 1.38 on glass. In the first two only the top reflection, from a denser film, gains pi, so the rays differ by an extra half wave. On the coated glass both reflections are from a denser medium, both gain pi, and the two half waves cancel. A readout says which case it is. 1 2 air soap, μ = 4/3 air top: phase π bottom: no change one π: the rays differ by 2μt + λ/2
Fig 1.3Use the chips. In a soap film and in oil on water, only the top reflection gains \(\pi\). On a lens coating of \(\mu = 1.38\) over glass of \(1.5\), both reflections are from a denser medium, and the two half waves cancel. The thickness of the film is exaggerated.

Oil on water behaves like the soap film. Oil has \(\mu\) about \(1.4\), more than water's \(\tfrac43\), so the top reflection gains \(\pi\) and the bottom one does not. A coating of \(\mu = 1.38\) on glass of \(1.5\) is different. Both of its reflections are from a denser medium, both gain \(\pi\), and the two extra half waves cancel.

principle 6.2
Counting the half waves of a film

Look at the two surfaces of the film. If exactly one reflection is from a denser medium, add \(\lambda/2\) to the path difference of the two reflected waves. If both are, or neither is, add nothing.

Common mistake

"Denser" here means optically denser, a larger refractive index. It has nothing to do with the density of mass. Oil floats on water because it is lighter, yet it is the optically denser of the two.

ExampleA film of water on a window pane

After rain, a thin film of water (\(\mu = \tfrac43\)) lies on a glass window (\(\mu = 1.5\)). Which of the two reflections gain a phase of \(\pi\), and what extra path do the half waves add between the two reflected waves?

Try it on paper first. The solution is one tap away when you are ready.
Checkpoint · select all that applyUnanswered

In which of these films does exactly one of the two reflections gain a phase of \(\pi\)?

CheckpointUnanswered

A soap film hangs in air. Which of its two surfaces gives the reflected wave a phase change of \(\pi\)?

Checkpoint · in your own wordsUnanswered

The top of a draining soap film is far thinner than a wavelength, and light still reflects from both of its surfaces. Explain why it looks black.

Now return to the soap film at the mela. At its thin top, the two surfaces are so close that the extra path inside the film is almost zero. Only the half wave of the front reflection is left between the two reflected waves. They meet turned against each other and cancel, for every colour at once. So the thinnest part reflects nothing and looks black. The next lesson adds the path inside the film and finds the colours below.

Exam traps
  1. Giving a phase change to the reflection at the back surface of a soap film, because it is "inside" the film.
    Ask only which way the light is going at that surface. From the film into air is towards a rarer medium: no change.
  2. Thinking that transmitted light also changes its phase.
    Only a reflection from a denser medium turns the wave over. A transmitted wave never does.
  3. Deciding "denser" by the density of mass.
    Compare the refractive indices. Oil is lighter than water but optically denser.
  4. Adding \(\lambda/2\) twice when both surfaces are denser.
    Two half waves are one whole wave, which changes nothing. Add \(\lambda/2\) only when exactly one reflection is from a denser medium.