Double-Slit Interference and Diffraction Gratings: Study Guide (Physics 2 14.8)

Two slits, one light source

Shine a laser at a card with two narrow slits close together. On a far screen you do not see two bright lines, one behind each slit. You see a row of evenly spaced bright and dark bands, called fringes.

This is the double-slit experiment, named after Thomas Young, who studied the interference of light in the early 1800s. Its result is strong evidence that light behaves as a wave:

• Particles going through two openings would just pile up behind each opening. Two streams of particles cannot add up to nothing.
• Waves can cancel. At a dark fringe, light from one slit plus light from the other gives darkness. That only makes sense if the light from each slit is a wave that can arrive crest-on-trough with the other.

Two things combine to make the pattern:

1. Diffraction. Each slit is narrow, so light spreads out from it (Topic 14.7). Without this spreading, the light from the two slits would never overlap on the screen.
2. Interference. Where the waves from the two slits overlap, they add (superposition, Topic 14.6). The same wavefront reaches both slits, so the two slits send out waves that start in step.

Path difference decides bright or dark

Take a point on the screen at angle θ from the straight-through direction. The slits are a distance d apart (centre to centre), and the screen is far away (L much larger than d), so the two paths are nearly parallel. Figure 1 shows that the wave from the lower slit travels further by

ΔL = d sin θ

• Bright fringe (constructive): the path difference is a whole number of wavelengths, so the waves arrive crest-on-crest. d sin θ = mλ, with m = 0, ±1, ±2, …
• Dark fringe (destructive): the path difference is an odd number of half-wavelengths (λ/2, 3λ/2, 5λ/2, …), so the waves arrive crest-on-trough. d sin θ = (m + ½)λ, with m = 0, 1, 2, … on each side

The whole number m is the order of the bright fringe. The central bright fringe (θ = 0, equal paths) is m = 0. The first bright fringe on either side is m = ±1, where one wave has travelled exactly one wavelength further.

The small-angle form

A fringe a distance y from the centre of the screen has tan θ = y/L. For angles under about 10°, sin θ ≈ tan θ, so d(y/L) = mλ and the distance from the middle of the central bright fringe to the m-th bright fringe is

y_max = mλL/d

• Neighbouring bright fringes are λL/d apart. If you think only about the interference of two waves, the bright fringes are evenly spaced, and the dark fringes are exactly halfway between them.
• The fringes spread out when λ or L increases, or when the slits are moved closer together (smaller d).

Fringes inside a diffraction envelope

The simple rule says every bright fringe is equally bright. A real pattern is not like that, because each slit has a width a. Each slit on its own makes the single-slit pattern of Topic 14.7: a wide central band, with dark minima at a sin θ = mλ. The two-slit fringes can only be as bright as this single-slit pattern allows.

So a real double-slit pattern is interference fringes sitting inside a single-slit diffraction envelope:

• The fringe spacing is set by d (λL/d).
• The envelope is set by a (its central band is 2λL/a wide).
• The slit separation d is always larger than the slit width a (usually several times larger), so the fringes are closer together than the envelope’s minima. Several fringes fit inside the central band of the envelope.
• Where a minimum of the envelope lands exactly on a bright-fringe position, that fringe is missing: one slit alone sends no light there, so there is nothing to interfere.

If you cover one of the two slits, the fringes disappear. What remains is the single-slit envelope on its own, less bright, because only one slit sends light.

Diffraction gratings

A diffraction grating is a large number of evenly spaced, parallel slits (or lines). Each slit diffracts the light on its own, and the grating pattern is what you get when all those diffraction patterns overlap and interfere. Gratings are usually described by lines per millimetre. A grating with N lines per millimetre has slit spacing d = (1/N) mm. For example, 500 lines/mm gives d = 0.002 mm = 2.00 × 10⁻⁶ m.

The bright maxima are at the same angles as for two slits with the same d:

d sin θ = mλ

But they look very different:

• For a bright maximum, the wave from every slit must be in step with every other one. That only happens at angles very close to d sin θ = mλ. A tiny step away from that angle and the waves from hundreds of slits cancel each other almost completely.
• So a grating gives narrow, very bright maxima separated by wide dark regions. This makes the positions easy to measure accurately, which is why gratings are used to measure wavelengths.
• Because d is so small, the angles are large. Do not use the small-angle form for a grating. Use d sin θ = mλ and, if you need positions on a screen, y = L tan θ.
• sin θ cannot exceed 1, so the highest order you can see is the largest whole number m that is less than d/λ. (If d/λ were exactly a whole number, that order would be at 90°, travelling along the grating, so you would not see it.)

White light and a grating

Send white light (all visible wavelengths) at a grating.

• At θ = 0 (m = 0) every wavelength has zero path difference, so every colour is bright there. The central maximum is white.
• In each higher order, sin θ = mλ/d, so a longer wavelength goes to a larger angle. Each order is spread into a spectrum: violet closest to the centre, red farthest from it.
• Higher orders are spread over a wider range of angles, and they can overlap: the start of one order can fall inside the previous one. Figure 3 shows this for the grating in Worked example 2.

Planning a measurement

To find an unknown slit separation (or a wavelength), you can:

1. Fix the slits and a screen a measured distance L away, and shine a laser of known wavelength through the slits.
2. Measure across several fringes, for example from the m = −3 bright fringe to the m = +3 one, and divide by 6 to get the spacing. One spacing alone is too small to measure precisely.
3. Repeat for several values of L.
4. Plot fringe spacing against L. The graph should be a straight line through the origin with gradient λ/d. Then d = λ ÷ gradient.

For a grating, measure the distance between the two first-order spots, halve it to get y, find θ from tan θ = y/L, and use d sin θ = mλ.

Worked example 1: a double slit with an envelope

Question. Light of wavelength 560 nm falls on two slits 0.250 mm apart (centre to centre). Each slit is 0.050 mm wide. The screen is 3.00 m away. Find (a) the fringe spacing, (b) the positions of the third bright fringe and of the first dark fringe, (c) the path difference at the third bright fringe, and (d) how many bright fringes lie inside the central band of the diffraction envelope.

1. Convert: λ = 5.60 × 10⁻⁷ m, d = 2.50 × 10⁻⁴ m, a = 5.0 × 10⁻⁵ m.
2. (a) Spacing = λL/d = (5.60 × 10⁻⁷ m)(3.00 m) ÷ (2.50 × 10⁻⁴ m) = 6.72 × 10⁻³ m = 6.72 mm.
3. (b) Third bright fringe: y = 3 × 6.72 mm = 20.2 mm (20.16 mm). Its angle: sin θ = 3λ/d = 6.72 × 10⁻³, θ = 0.385°, so the small-angle form is fine. The first dark fringe is halfway between m = 0 and m = 1: y = ½ × 6.72 mm = 3.36 mm.
4. (c) At the third bright fringe, ΔL = d sin θ = 3λ = 1.68 × 10⁻⁶ m.
5. (d) The envelope’s first minimum is at y = λL/a = (5.60 × 10⁻⁷)(3.00) ÷ (5.0 × 10⁻⁵) = 33.6 mm. That is 33.6 ÷ 6.72 = 5 fringe spacings, so the m = ±5 fringes are missing. Inside the central band: m = 0, ±1, ±2, ±3, ±4, which is 9 bright fringes.

Check. d/a = 0.250 ÷ 0.050 = 5, which gives the same answer for (d) without any lengths on the screen. This is the pattern in Figure 2.

Worked example 2: white light on a grating

Question. White light containing wavelengths from 400 nm to 700 nm falls on a grating with 600 lines/mm. (a) Find the angles at which the first-order spectrum starts and ends. (b) Do the second and third orders overlap? (c) What is the highest order in which red light (700 nm) appears?

1. d = (1/600) mm = 1.667 × 10⁻⁶ m.
2. (a) Violet: sin θ = (400 × 10⁻⁹) ÷ (1.667 × 10⁻⁶) = 0.240, θ = 13.9°. Red: sin θ = 0.420, θ = 24.8°. The spectrum runs from 13.9° (violet) to 24.8° (red).
3. (b) Second-order red: sin θ = 2 × 0.420 = 0.840, θ = 57.1°. Third-order violet: sin θ = 3 × 0.240 = 0.720, θ = 46.1°. The third order starts at 46.1°, before the second ends at 57.1°, so yes, they overlap.
4. (c) The largest m that is less than d/λ: d/λ = 1.667 × 10⁻⁶ ÷ 7.00 × 10⁻⁷ = 2.38, so red appears only up to m = 2.

Check. If you had used the small-angle form for first-order red on a screen 1.00 m away, you would get y = λL/d = 0.420 m instead of the correct L tan θ = 0.463 m, an error of about 9%. Grating angles are too large for the approximation.

Common misconceptions

• “d sin θ = mλ gives dark fringes.” For two slits or a grating it gives the bright fringes. (For one slit of width a, the same form gives dark fringes.)
• “Moving the slits apart spreads the fringes.” The spacing λL/d gets smaller as d increases.
• “Every double-slit fringe is equally bright.” Only in the interference-only model. Real fringes fade with the single-slit envelope, and some can be missing.
• “Violet is deviated most by a grating.” That is true for a prism (refraction), not for a grating. In each grating order, red is farthest from the centre.
• “The central maximum of a grating is a spectrum.” At m = 0 all colours have zero path difference, so the centre is white.
• Using y = mλL/d for a grating. Grating angles are usually far above 10°; use d sin θ = mλ and tan θ = y/L.
• Forgetting to convert lines/mm. 300 lines/mm means d = 1/300 mm = 3.33 × 10⁻⁶ m, not 300 m or 1/300 m.

Where this leads

Topic 14.9 uses the same path-difference reasoning for light reflected from the two surfaces of a thin film, such as a soap bubble or a layer of oil on water. Continue to Topic 14.9: Thin-Film Interference. First, test yourself with the practice questions, then use the revision notes and the topic checklist. For single-slit diffraction, look back at Topic 14.7: Diffraction.