Diffraction: Study Guide (Physics 2 14.7)

Waves bend around edges

Stand outside an open door and you can hear people talking in the room, even if you cannot see them. The sound does not just travel in straight lines through the doorway. It spreads out into the space beyond. Water waves do the same thing when they pass the end of a harbour wall: behind the wall, where you might expect calm water, curved waves spread in.

This spreading is called diffraction. A wave diffracts when it passes through an opening (a gap or a slit) or past the edge of an obstacle. Every type of wave diffracts: waves on water, sound, light and other electromagnetic waves.

Diffraction does not change the wave’s speed, frequency or wavelength. The medium has not changed, so the waves beyond the gap have the same λ as the waves arriving. Only the direction in which the energy travels spreads out.

Gap size compared with wavelength

How much a wave spreads depends on one comparison: the width of the opening, a, compared with the wavelength, λ.

• a much larger than λ: most of the wave passes straight through. It spreads only a little at the edges, and there is a fairly sharp shadow on each side.
• a about the same as λ: the wave spreads strongly. Beyond the gap the wavefronts are nearly semicircles, as if the gap were a new point source.

So diffraction is most noticeable when the opening is comparable to the wavelength. Figure 1 shows the two cases in a ripple tank.

Two everyday comparisons make the rule concrete. A 500 Hz note in air (v = 340 m/s) has λ = 340 ÷ 500 = 0.68 m. A doorway about 0.9 m wide is only about 1.3 wavelengths across, so the sound spreads widely. Light of wavelength 500 nm meets the same doorway, which is now 0.9 ÷ (500 × 10⁻⁹) = 1.8 million wavelengths wide. The light hardly spreads at all, and the edge of the shadow looks sharp. To see light diffract clearly you need a slit a fraction of a millimetre wide.

Why a single slit gives bands

If the gap were a single point, the wave beyond it would just spread evenly. A real slit has a width, and every part of the opening acts as a source of its own small wavelets. These wavelets all start in step, because the same wavefront reaches the whole slit at once. Beyond the slit they overlap, so they interfere, using the superposition rule from Topic 14.6.

This is the standard light experiment. Monochromatic light (a single wavelength λ, for example from a laser) falls on a narrow slit of width a. A screen is a distance L away, with L much larger than a. On the screen you see:

• a wide, bright central band straight ahead,
• dark fringes where the light cancels, and
• dimmer bright bands further out, getting fainter with distance from the centre.

Whether wavelets add or cancel at a point on the screen depends on the path length difference ΔL: how much further one wavelet travels than another to reach that point.

Where the dark fringes are

Look at light leaving the slit at an angle θ to the straight-through direction (the normal to the slit). Because the screen is far away, the wavelets heading to one point on the screen travel along nearly parallel lines. Figure 2 shows the geometry.

The path length difference between the wavelets from the two edges of the slit is

ΔL = a sin θ

Now use a pairing argument. Split the slit into a top half and a bottom half. Pair each point in the top half with the point exactly a/2 below it in the bottom half. Each pair has a path difference of (a/2) sin θ.

• If (a/2) sin θ = λ/2, every pair arrives half a wavelength out of step. Each pair cancels, so the whole slit gives darkness. This happens when a sin θ = λ.
• Split the slit into four, six, eight … strips instead, and the same argument gives darkness whenever the edge-to-edge path difference is a whole number of wavelengths.

So the dark fringes (minima) are at

a sin θ = mλ, with m = ±1, ±2, ±3, …

m = 0 is not a dark fringe. At θ = 0 all the wavelets travel the same distance and arrive in step, so the centre of the pattern is the brightest point. Notice the surprise: when the edge-to-edge difference is a whole λ, the result is dark, not bright.

The small-angle form

On the screen, a point at distance y from the centre of the central band is at angle θ with tan θ = y/L. For angles under about 10°, sin θ ≈ tan θ ≈ θ (in radians), with an error of under 2%. Then a sin θ = mλ becomes a(y/L) = mλ, so the distance from the middle of the central bright band to the m-th dark fringe is

y_min = mλL/a

Consequences you can read straight from this equation:

• The dark fringes are evenly spaced, λL/a apart (for small angles).
• The central bright band runs from the first dark fringe on one side to the first on the other side, so its width is 2λL/a: twice the width of each side band.
• The pattern gets wider when λ or L increases, and when the slit gets narrower (a smaller). A narrower slit spreads the light more. This matches the rule from Figure 1.
• If a < λ, then sin θ = λ/a would be greater than 1, so there are no dark fringes at all. The light spreads over every forward direction.

Shape of the opening

The pattern depends on the shape of the opening, not only its size.

• A long, narrow slit spreads light mainly in the direction across its narrow width. The bands are stripes parallel to the slit. The long direction is many wavelengths across, so there is almost no spreading along it.
• A small circular hole spreads light equally in all directions around the beam. The pattern is a bright central disc surrounded by faint rings.
• A square hole spreads light along both of its widths, giving bright spots in a cross shape.

The rule is always the same: the narrower the opening in a given direction, the wider the spreading in that direction.

Reading a pattern

A picture or set of measurements of a diffraction pattern is evidence you can work backwards from. If you know λ and L and can measure the spacing of the dark fringes, you can find a. If you know a and L, the pattern gives λ. Measure across several fringes and divide, rather than measuring one small gap: it makes the result more precise.

Worked example 1: predicting the pattern

Question. Laser light of wavelength 520 nm passes through a slit 0.130 mm wide. A screen is 2.40 m away. Find (a) the angle to the first dark fringe, (b) the distances from the centre to the first and second dark fringes, and (c) the widths of the central bright band and of the first side band.

1. Convert to metres: λ = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m; a = 0.130 × 10⁻³ m = 1.30 × 10⁻⁴ m.
2. (a) First dark fringe, m = 1: sin θ₁ = λ/a = (5.20 × 10⁻⁷ m) ÷ (1.30 × 10⁻⁴ m) = 4.00 × 10⁻³. So θ₁ = 0.229°. This is far below 10°, so the small-angle form is valid.
3. (b) y₁ = λL/a = (5.20 × 10⁻⁷ m)(2.40 m) ÷ (1.30 × 10⁻⁴ m) = 9.60 × 10⁻³ m = 9.60 mm. y₂ = 2λL/a = 19.2 mm.
4. (c) Central band: from −y₁ to +y₁, width = 2 × 9.60 mm = 19.2 mm. First side band: from y₁ to y₂, width = 19.2 − 9.60 = 9.60 mm.

Check. Using the exact form y = L tan θ₂ with sin θ₂ = 8.00 × 10⁻³ gives 19.2006 mm, the same to 3 significant figures. The central band is twice the width of the side band, as expected. Units: (m × m) ÷ m = m.

Worked example 2: working backwards, then predicting a change

Question. A student shines monochromatic light through a slit 0.300 mm wide onto a screen 1.50 m away. She measures the distance between the second dark fringe on the left and the second dark fringe on the right as 12.0 mm. (a) Find the wavelength. (b) She then halves the slit width and changes to light with 1.5 times the frequency. Predict the new width of the central bright band.

1. (a) The two second-order dark fringes are each at y₂ = 2λL/a from the centre. The distance between them is 2y₂ = 4λL/a.
2. So λ = (12.0 × 10⁻³ m)(3.00 × 10⁻⁴ m) ÷ (4 × 1.50 m) = 6.00 × 10⁻⁷ m (600 nm). Then y₁ = λL/a = 3.00 mm, and the central band is 6.00 mm wide.
3. (b) The light still travels at the same speed, so from v = fλ, multiplying f by 1.5 multiplies λ by 1/1.5 = 2/3.
4. Width of the central band = 2λL/a. The new width is the old width × (2/3) ÷ (1/2) = old width × 4/3.
5. New width = 6.00 mm × 4/3 = 8.00 mm.

Check. Directly: λ = 4.00 × 10⁻⁷ m, a = 1.50 × 10⁻⁴ m, so 2λL/a = 2(4.00 × 10⁻⁷)(1.50) ÷ (1.50 × 10⁻⁴) = 8.00 × 10⁻³ m. The two changes pull in opposite directions: the shorter wavelength narrows the pattern, the narrower slit widens it. The slit change is larger, so the pattern gets wider overall.

Common misconceptions

• “a sin θ = mλ gives bright fringes.” For a single slit it gives the dark fringes. The centre (θ = 0) is bright, and m = 0 is not a dark fringe.
• “A narrower slit gives a narrower pattern.” The opposite. Pattern width ∝ λ/a, so narrowing the slit spreads the light more.
• “All the bright bands are the same width and brightness.” The central band is twice as wide as the others and much brighter.
• “Diffraction changes the wavelength.” The wave beyond the gap has the same λ and f. Only its direction spreads.
• “Only light diffracts” or “light does not diffract.” All waves diffract. Light diffracts too; it is just hard to notice unless the opening is close to its tiny wavelength.
• Using y = mλL/a at large angles. The small-angle form needs θ under about 10°. Otherwise use a sin θ = mλ and tan θ = y/L.
• Mixing units. Put λ, a, L and y all in metres. Writing λ in nm and a in mm gives an answer that is wrong by a factor of 10⁶.

Where this leads

In Topic 14.8 you will combine this single-slit spreading with the interference of light from two or more slits, and see the double-slit fringes sit inside the single-slit pattern you met here. Continue to Topic 14.8: Double-Slit Interference and Diffraction Gratings. First, test yourself with the practice questions, then use the revision notes and the topic checklist. For superposition itself, look back at Topic 14.6: Wave Interference and Standing Waves.