Single-slit diffraction, diffraction gratings and measuring wavelength
AQA A-Level Physics 3
〰️Describe single-slit diffraction patterns
📏Explain how slit width and wavelength affect diffraction
🎚️Describe the diffraction grating and its construction
📐Use d sin θ = nλ to find angles and wavelengths
🔬Carry out the required practical with a diffraction grating
🌐State applications of diffraction gratings
Single-Slit Diffraction
Diffraction: The spreading of waves as they pass through a gap or around an obstacle. Diffraction is a wave property — it is evidence that light behaves as a wave.
When monochromatic light passes through a single narrow slit, it diffracts and produces a pattern on a screen:
A central bright maximum (widest and most intense)
Alternating dark minima and dimmer secondary maxima on either side
All secondary maxima are roughly half the width of the central maximum
Diffraction is greatest when the slit width is approximately equal to the wavelength of light (slit width ≈ λ). A wider slit produces less diffraction (narrower central maximum, more like geometrical shadow). A narrower slit produces more diffraction (wider, more spread out pattern).
Unlike a double-slit (which produces evenly spaced, equally bright fringes), a single-slit produces a pattern with a dominant bright central fringe and rapidly diminishing outer fringes.
The Diffraction Grating
Diffraction grating: A plate with a very large number of equally spaced parallel slits (typically 300–1200 lines per mm). Each slit diffracts incoming light, and the diffracted beams from all slits interfere with each other.
Because there are many slits (thousands), the interference maxima are extremely sharp and bright (the sharpness increases with number of slits). This makes diffraction gratings far more accurate for wavelength measurement than a double-slit.
Diffraction grating equation:
d sin θ = nλ
d = grating spacing (m) = 1/(lines per metre)
θ = angle of nth order maximum from straight-ahead (degrees)
n = order (0, 1, 2, 3...)
λ = wavelength (m)
Grating spacing d: If the grating has N lines per metre, then d = 1/N. For example, a grating with 500 lines/mm has d = 1/(500 × 10³) = 2.0 × 10⁻⁶ m = 2.0 μm.
The zero-order maximum (n = 0) is always at θ = 0 (straight ahead). First-order maxima (n = ±1) are at equal angles on either side. Maximum possible order: n_max = d/λ (since sin θ cannot exceed 1).
Orders of Diffraction
The diffraction grating produces sharp maxima at specific angles given by d sin θ = nλ. The order n tells you how many complete wavelengths of path difference there are between adjacent slits.
Order n
Path difference
Position
0
0
Centre (straight-ahead)
±1
±λ
First-order maxima, either side
±2
±2λ
Second-order maxima
±3
±3λ
Third-order (if d sin θ ≤ d)
Maximum order: n_max = d/λ (rounded down to nearest integer). For d = 2.0 μm and λ = 600 nm: n_max = 2.0×10⁻⁶ / 6.0×10⁻⁷ = 3.3 → maximum order is 3.
Using white light with a diffraction grating produces a spectrum for each order (n ≥ 1). The zero-order beam remains white. The 1st order spectrum shows violet at small angles and red at larger angles (opposite to a prism).
Applications of Diffraction Gratings
Spectroscopy: Analysing the spectrum of starlight or laboratory sources to identify elements (emission spectra) or missing wavelengths (absorption spectra). The grating produces sharp lines → high resolution.
X-ray crystallography: Crystal planes act as a diffraction grating for X-rays (spacing comparable to λ_X-ray ≈ 10⁻¹⁰ m). Bragg's law n_λ = 2d sin θ is used to determine crystal structure.
CD/DVD reading: The track spacing on a CD (~1.6 μm) acts as a reflection grating for laser light.
Monochromators: Select a specific wavelength from a broad-spectrum source for use in chemistry or physics experiments.
A diffraction grating has 400 lines per mm. Light of wavelength 589 nm is incident on it. Calculate the angle of the first and second order maxima.
A grating has 600 lines/mm. What is the maximum order visible for red light of wavelength 700 nm?
1d = 1/(600 × 10³) = 1.667 × 10⁻⁶ m
2Maximum order when sin θ = 1: n_max = d/λ = 1.667 × 10⁻⁶ / 700 × 10⁻⁹ = 2.38
3Round down to nearest integer: n_max = 2
Maximum visible order = 2 (0th, ±1st, ±2nd order only)
White light (400–700 nm) falls on a diffraction grating (d = 2.00 × 10⁻⁶ m). Do the 1st and 2nd order spectra overlap?
11st order red: sin θ = λ/d = 700×10⁻⁹/2.00×10⁻⁶ = 0.350 → θ = 20.5°
22nd order violet: sin θ = 2×400×10⁻⁹/2.00×10⁻⁶ = 0.400 → θ = 23.6°
31st order ends at 20.5°; 2nd order starts (violet) at 23.6° → no overlap between 1st and 2nd orders for this grating
No overlap (1st order 20.5° → 2nd order violet starts at 23.6°). For gratings with larger d, orders can overlap.
1. A diffraction grating has 500 lines/mm. What is the grating spacing d?
d = 1/(500 lines mm⁻¹) = 1/(500×10³ m⁻¹) = 2.0×10⁻⁶ m = 2.0 μm.
2. For a diffraction grating, increasing the wavelength of light (at constant d) causes the diffraction angles to:
d sin θ = nλ → sin θ = nλ/d. Larger λ → larger sin θ → larger θ. Red light diffracts more than violet light with a diffraction grating.
3. A grating produces a 2nd order maximum for 450 nm light at 26.7°. What is the grating spacing?
d = nλ/sin θ = (2 × 450×10⁻⁹)/sin(26.7°) = 9.00×10⁻⁷/0.449 = 2.00×10⁻⁶ m ≈ 2.01×10⁻⁶ m.
4. How does a diffraction grating produce sharper maxima than a double-slit?
With thousands of slits, even a tiny deviation from the exact angle for a maximum leads to destructive interference from some combination of slits → almost complete cancellation between maxima → extremely sharp, bright maxima.
5. A 300 lines/mm grating is used with sodium light (λ = 589 nm). Calculate the angle of the first-order maximum to 1 d.p.
1. A diffraction grating spectrometer measures the 3rd order maximum for an unknown wavelength at θ = 48.2°. The grating has 600 lines/mm. Identify the colour of the light.
d = 1/(600×10³) = 1.667×10⁻⁶ m. λ = d sin θ / n = 1.667×10⁻⁶ × sin(48.2°) / 3 = 1.667×10⁻⁶ × 0.7455 / 3 = 4.14×10⁻⁷ m = 414 nm. This is violet light (visible range ~400–420 nm is violet/indigo).
2. Explain why single-slit diffraction is maximal when the slit width equals the wavelength, and what happens when the slit is much wider than the wavelength.
When the slit width ≈ λ, each point across the slit acts as a secondary source (Huygens' principle). The path differences between these sources to any given point cover a full range of phases, but the spread of wavelets in all directions is significant. Constructive interference occurs mainly in a wide central region — maximum diffraction spreading. When the slit width >> λ: rays from opposite sides of the slit arrive nearly in phase even at significant angles, so destructive interference only occurs at very large angles → the light continues mostly in a straight line (geometric shadow). This is why doors don't diffract light but do diffract sound (sound λ ≈ 0.1–3 m, door width ≈ 1 m).
3. A diffraction grating with 750 lines/mm is used with white light. (a) Calculate the angular width of the 1st order spectrum (angle spanned from 400 nm to 700 nm). (b) Show that the 2nd order violet overlaps with the 1st order red.
(a) d = 1/(750×10³) = 1.333×10⁻⁶ m. 1st order violet (400nm): sin θ = 400×10⁻⁹/1.333×10⁻⁶ = 0.300 → θ_v = 17.5°. 1st order red (700nm): sin θ = 700×10⁻⁹/1.333×10⁻⁶ = 0.525 → θ_r = 31.7°. Angular width = 31.7 − 17.5 = 14.2°. (b) 2nd order violet: sin θ = 2×400×10⁻⁹/1.333×10⁻⁶ = 0.600 → θ = 36.9°. 1st order red ends at 31.7°, 2nd order violet starts at 36.9° → they do NOT overlap for this grating. The question's premise is only true for gratings with larger d. Let me verify: actually for d = 1.333 μm, 2nd violet at 36.9° vs 1st red at 31.7° — no overlap. For d = 2 μm (500 lines/mm): 1st red = arcsin(0.350) = 20.5°; 2nd violet = arcsin(0.400) = 23.6° — also no overlap. The 2nd violet overlaps 1st red when d is larger; e.g. d = 2.8 μm (357 lines/mm): 1st red: sin = 0.25 → 14.5°; 2nd violet: sin = 0.286 → 16.6°. Still no. In general: overlap occurs when 2λ_v/d > λ_r/d, i.e. 2×400 > 700 → 800 > 700. So 2nd violet ALWAYS begins after 1st red ends — they always overlap IF 2λ_min > λ_max: 2×400 = 800 > 700 ✓. Therefore 2nd order violet (800/d) > 1st order red (700/d) → the 2nd order violet maximum is always at a larger angle than the 1st order red — so they DO overlap (2nd order spectrum starts before 1st order ends). Re-examining: 1st red = arcsin(λ_r/d); 2nd violet = arcsin(2λ_v/d). Since 2×400 = 800 > 700: 2nd violet angle > 1st red angle → the 2nd order spectrum overlaps with the end of the 1st order spectrum. For the 750 lines/mm grating: 1st red at 31.7°, 2nd violet at 36.9°. The 2nd order continues to longer wavelengths from there, but violet (400nm) in 2nd order is at 36.9°, which is already beyond 1st order red at 31.7°, so the 2nd order spectrum (starting at violet 36.9°) doesn't overlap with 1st order (ending at red 31.7°). There is a gap. So the overlap principle 2λ_v > λ_r applies to whether angles interleave — the 2nd order starts where it does. The angular positions don't overlap here; the spectrum ranges 17.5°–31.7° (1st) and 36.9°–beyond (2nd).
Objective: Use a diffraction grating to measure the wavelength of a laser beam.
Equipment
Laser (red He-Ne at ~632 nm, or similar)
Diffraction grating (e.g. 300 lines/mm)
Grating holder and optical bench
Screen or wall 1–2 m away
Metre rule or tape measure
Pencil to mark spot positions
Method
Clamp the grating perpendicular to the laser beam. Place the screen at distance D (measure accurately with a metre rule).
Turn on the laser. Observe the zero-order spot (central) and multiple first-order (and possibly second-order) spots on the screen.
Mark the positions of the 1st order spots on each side. Measure the distance x from the zero-order spot to one 1st-order spot.
Calculate θ: tan θ = x/D → θ = arctan(x/D).
Apply d sin θ = nλ → λ = d sin θ/n. Use n = 1 for the 1st order.
Repeat for the other 1st order spot and for 2nd order if visible. Calculate mean λ.
Safety
NEVER look into the laser or at its specular reflections. Keep the beam at bench level, away from eyes. Post a warning notice on the laboratory door.
Analysis and Uncertainties
Measurement
Typical uncertainty
How to reduce
Distance D to screen
±0.5 cm
Use a large D (2 m); measure to front face of screen
Position of spot x
±1 mm
Mark centre of spot carefully; repeat both sides
Grating spacing d
Printed on grating (exact)
Use a verified grating
Calculate % uncertainty in θ → % uncertainty in λ using the diffraction grating equation.
Expected Result
For a 300 lines/mm grating (d = 3.33 μm) and red laser (λ = 632 nm): sin θ = λ/d = 632×10⁻⁹/3.33×10⁻⁶ = 0.1898 → θ = 10.9°. At D = 1.50 m: x = D tan θ = 1.50 × tan(10.9°) = 0.289 m ≈ 29 cm from the zero-order spot.