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E. The Wave Nature of Light · The Wave Nature of Light

Air Wedges and Measuring with Light

E2.1 E2.3 E2.4 E3.2 A1.8 A1.12 A1.13About 65 minutes
Learning goals
  • I can explain the fringes formed by an air wedge between two glass plates.
  • I can derive and use Δx = Lλ/(2t) for the spacing of air-wedge fringes.
  • I can use an air wedge to measure a very small thickness, such as a hair or a sheet of foil.
  • I can explain how interference fringes are used to test the flatness of surfaces.

Words to know

air wedge
A thin, wedge-shaped layer of air between two flat glass plates that touch at one end and are separated by a thin spacer at the other.
fringe spacing (Δx)
The distance between neighbouring dark (or bright) fringes; for an air wedge, Δx = Lλ/(2t).
optical flat
A very precisely flat piece of glass used with interference fringes to test the flatness of other surfaces.
Newton's rings
Circular interference fringes formed in the thin air gap between a curved lens and a flat glass plate.
interferometry
Measuring lengths or changes in length by counting or analysing interference fringes.

A ruler made of light

How would you measure the thickness of a single hair? A ruler is useless: a hair is less than a tenth of a millimetre thick. A micrometer can do it, but light can do it even better. Lay two microscope slides on top of each other, slip the hair between them at one end, and look at the reflection of a lamp: you will see a row of evenly spaced fringes. Counting them measures the hair, with light's own wavelength as the ruler.

The air wedge

Two flat glass plates of length L touch along one edge. At the other edge a thin spacer of thickness t holds them apart. Between them is a thin wedge of air whose thickness grows steadily from 0 to t.

Light shining down reflects from two surfaces:

  • the bottom of the top plate (glass → air, lower index): no inversion;
  • the top of the bottom plate (air → glass, higher index): inverted.

So there is one inversion. Where the air layer has thickness y, the extra path is 2y. (The air layer's index is 1.00, so λ is the same as in air.)

  • Dark fringes: 2y = mλ (m = 0, 1, 2, …). At the contact line, y = 0: the wedge always starts with a dark fringe.
  • Bright fringes: 2y = (m + ½)λ.

Moving from one dark fringe to the next, the air thickness increases by λ/2. Because the wedge is a straight slope, y = (t/L)x at distance x from the contact line. Increasing y by λ/2 means moving a distance

Δx = Lλ / (2t)

The fringes are parallel to the contact line and evenly spaced. A thicker spacer makes a steeper wedge and closer fringes.

tplates touch here (dark fringe)spacer (hair, foil)Llight intwo reflections(glass→air: no flip;air→glass: flip)side view: air gap greatly exaggeratedview from above: equally spaced dark fringes, Δx = Lλ/(2t)
An air wedge. Light reflecting from the bottom of the top plate and the top of the bottom plate interferes. The air thickness grows by λ/2 from one dark fringe to the next, giving evenly spaced fringes Δx = Lλ/(2t) apart. The angle is hugely exaggerated: a real wedge is only a few hundredths of a millimetre thick at the open end.
Worked example

Example 1: Fringe spacing

Two glass plates 10.0 cm long are separated at one end by a hair 0.080 mm thick. They are lit from above by sodium light (λ = 589 nm). Find the spacing of the dark fringes and the number of dark fringes.

  1. Δx = Lλ/(2t) = (0.100)(589 × 10⁻⁹) ÷ (2 × 8.0 × 10⁻⁵).
  2. Δx = 3.68 × 10⁻⁴ m ≈ 0.37 mm.
  3. Number of dark fringes: the air thickness reaches t at the far end, so the largest m with 2t ≥ mλ is m ≤ 2t/λ = 1.6 × 10⁻⁴ ÷ 5.89 × 10⁻⁷ = 271.6, so m = 0 to 271.

Δx ≈ 0.37 mm, with about 272 dark fringes across the plates (you would need a magnifier to count them).

Worked example

Example 2: Measuring a thickness

Two plates 12.0 cm long are separated at one end by a strip of aluminium foil. With light of wavelength 600 nm, the dark fringes are 1.5 mm apart. How thick is the foil?

  1. Rearrange Δx = Lλ/(2t): t = Lλ/(2Δx).
  2. t = (0.120)(6.00 × 10⁻⁷) ÷ (2 × 1.5 × 10⁻³).
  3. t = 2.4 × 10⁻⁵ m.

The foil is about 0.024 mm (24 µm) thick.

Worked example

Example 3: Filling the wedge with water

The air wedge in Example 2 is filled with water (n = 1.33). What is the new fringe spacing?

  1. The light now travels through water in the gap, so the wavelength there is λ/n.
  2. Δx = Lλ/(2nt): the spacing shrinks by a factor of 1.33.
  3. Δx = 1.5 mm ÷ 1.33 = 1.13 mm. (The phase changes are still one inversion, since water's index is lower than glass's.)

Δx ≈ 1.1 mm: the fringes move closer together.

Testing flatness and other uses

If both plates are perfectly flat, the air wedge has a perfectly even slope and the fringes are straight and evenly spaced. If one surface has a bump or a hollow, the air thickness changes there and the fringes bend. Each fringe is a line of equal air thickness, like a contour line on a topographic map, with a "height" step of only λ/2 (about 0.3 µm) between neighbouring fringes.

  • Lens makers and precision machinists use an optical flat placed on a part to check flatness to a fraction of a micrometre.
  • A curved lens resting on a flat plate makes circular fringes called Newton's rings. Rings that are perfect circles show the lens has the correct shape.
  • Instruments called interferometers count fringes to measure tiny changes in distance. Very large interferometers detect ripples in space itself from colliding black holes, changes far smaller than an atom.
flat surfaces: straight, even fringes(top view, not to scale)a bump: fringes bend around it(top view, not to scale)
Fringes as contour lines. Two flat plates give straight, evenly spaced fringes (left). A bump changes the air thickness, so the fringes bend around it (right).
Just for fun

Measuring a hair with an air wedge is one of the few times in life when splitting hairs counts as good science.

Hands-on activity

Interference between two sheets of glass

You need: two clean, flat pieces of clear glass or rigid clear plastic (two glass slides, two small picture-frame glass panes, or two clear CD case lids), a single hair, a fluorescent light, LED desk lamp or a phone screen showing a plain bright green image, a magnifying glass, and a dark cloth.

Grown-up note: picture-frame glass has sharp edges. An adult should handle it, or tape the edges first. Clean the pieces with soapy water and dry them well.

  1. Lay one piece on the dark cloth. Put the hair across one end and lay the second piece on top, so the pieces touch at the other end.
  2. Look at the reflection of the light source in the glass, close to the direction the light is reflected. Use the magnifying glass.
  3. Look for thin, evenly spaced fringes running parallel to the hair. Press gently on the top piece and watch them move.
  4. If you can see fringes clearly, count how many lie across 1 cm and estimate Δx. With λ ≈ 530 nm (green), estimate t = Lλ/(2Δx).

What to look for: even with no hair, two pressed pieces usually show irregular coloured fringes where the gap varies, like a map of the gap. With the hair, the fringes become more regular and are very close together, often a few tenths of a millimetre apart (the magnifier helps). Pressing changes the gap and makes the fringes shift, showing how sensitive they are to tiny changes in thickness.

Your turn

1

In an air wedge, how many reflections are inverted?

2

Why is the fringe at the contact line dark?

3

By how much does the air thickness change from one dark fringe to the next?

4

Plates 8.0 cm long are separated at one end by a 0.020 mm spacer. Find Δx for 550 nm light.

5

How many dark fringes appear in the last question?

6

Plates 10.0 cm long, λ = 633 nm, Δx = 0.50 mm. Find t.

7

What do curved fringes tell you about the plates?

More practice: warm-up

1

Write the air-wedge fringe spacing equation.

2

If the spacer is made thicker, do the fringes get closer or farther apart?

3

Are air-wedge fringes evenly spaced?

4

What are Newton's rings?

5

Light reflecting from glass into air (inside the top plate): is it inverted?

More practice: on your own

1

Plates 15 cm long show dark fringes 1.2 mm apart in 600 nm light. Find the spacer thickness.

2

The wedge in the last question is filled with water (n = 1.33). Find the new spacing.

3

Thirty dark fringes (counting the contact line as m = 0) fit exactly across an air wedge, with a dark fringe at the spacer end, in 589 nm light. Find the spacer thickness.

4

Explain why each fringe in an air wedge is a 'line of equal thickness'.

5

Two plates are 5.0 cm long with a 1.0 × 10⁻⁵ m spacer, and the gap is filled with water (n = 1.33). Find Δx for 500 nm light.

Challenge

1

A student slowly lifts the spacer end of an air wedge while watching one spot. 12 dark fringes move past the spot in 600 nm light. By how much did the air thickness at that spot change?

2

In an air wedge lit with white light, why do you see only a few coloured fringes near the contact line before the pattern washes out?

3

For an air wedge, you put the hair at a distance of 6.0 cm from the contact line instead of at the end of 10.0 cm plates. Does Δx depend on the plate length or on the distance to the spacer? Explain.

Check your understanding

Quiz

1.

At the line where two glass plates of an air wedge touch, the fringe is…

2.

Between neighbouring dark fringes, the air thickness changes by…

3.

Plates 10 cm long, spacer 0.050 mm, λ = 500 nm. The fringe spacing is…

4.

Which change makes the air-wedge fringes farther apart?

5.

Curved fringes in an air wedge show that…

6.

Filling an air wedge with a liquid of index n makes the fringe spacing…

For parents and teachers

Big idea: air wedges (E2.4, E3.2: the curriculum names air wedges directly) show how interference can measure tiny thicknesses. Students should be able to explain the dark contact line (one inversion) and derive Δx = Lλ/(2t) from the λ/2 step.

How to help: handle any glass yourself. Even without a perfect result, seeing irregular fringes between two pressed glass pieces is a convincing demonstration.

Common difficulties:

  • Using λ instead of λ/2 for the thickness step between fringes.
  • Forgetting the dark fringe at the contact line when counting (m starts at 0).
  • Not converting the spacer thickness and wavelength to metres.