Why one slit makes a pattern: Light passing through a single narrow slit of width b spreads out and produces a pattern of a broad central bright fringe flanked by much dimmer side fringes. This happens because waves from different parts of the same slit interfere — the wider the slit relative to λ, the less the spread.
The HL result you need is the position of the first dark fringe (first minimum). For small angles it sits at θ = λ/b. This single formula tells you the half-width of the central maximum — make the slit narrower (smaller b) or use a longer wavelength (larger λ) and the pattern spreads wider.
- angle from the centre to the first minimum (rad)
- wavelength of the light (m)
- width of the slit (m)
Worked example — first minimum of a single slit
Monochromatic light of wavelength 600 nm passes through a single slit of width 0.12 mm. Find the angle from the centre of the pattern to the first minimum.
Solution
- Write the given formula first:
- Convert to metres: λ = 600 nm = 6.0×10⁻⁷ m, b = 0.12 mm = 1.2×10⁻⁴ m. Substitute:
- Work it out — small angle, answer in radians:
Final answer
θ = 5.0×10⁻³ rad (≈ 0.29°). The full central maximum is twice this wide.
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Narrower slit ⇒ wider pattern: Because θ = λ/b, the spread of the pattern is inversely proportional to the slit width and directly proportional to the wavelength:
- Halve the slit width b → the pattern is twice as wide. - Use red light (longer λ) instead of blue → the pattern is wider.
Make θ BIGGER (wider pattern)
- Narrower slit (smaller b)
- Longer wavelength (red light)
Make θ SMALLER (narrower pattern)
- Wider slit (larger b)
- Shorter wavelength (blue light)
It limits resolution too: Single-slit spreading is why every lens or telescope has a diffraction limit: a smaller aperture (smaller b) gives a wider diffraction blur, so finer detail can no longer be resolved.
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Many slits → razor-sharp maxima: A diffraction grating has thousands of equally spaced slits. Where the path difference between adjacent slits is a whole number of wavelengths, all the slits add in phase, giving a very sharp, bright maximum. The more slits per millimetre, the sharper and more widely separated these maxima become.
- slit spacing = distance between adjacent slits (m)
- angle of the nth bright maximum (degrees or rad)
- order of the maximum (0, 1, 2, …)
- wavelength of the light (m)
First find d from the lines per mm: Gratings are quoted as lines per mm (or per metre). The slit spacing d is the reciprocal of the number of lines per metre:
d = 1 ÷ (lines per metre). Convert lines-per-mm to lines-per-metre by ×10³ first.
Worked example — first-order angle from a grating
A diffraction grating has 500 lines per mm. Light of wavelength 600 nm is shone on it at normal incidence. Find the angle of the first-order (n = 1) maximum.
Solution
- Find the slit spacing d (reciprocal of lines per metre). 500 lines/mm = 500×10³ lines/m:
- Write the given grating equation and make sinθ the subject:
- Substitute λ = 6.0×10⁻⁷ m, n = 1, d = 2.0×10⁻⁶ m:
- Take the inverse sine — keep the unit:
Final answer
θ = 17° for the first-order maximum.
All three involve the same physics of superposition, but they look very different on the screen. Keep the formulas straight — a grating uses d sinθ = nλ, a single slit uses θ = λ/b for its first minimum.
| Set-up | Pattern on screen | Key relation |
|---|---|---|
| Single slit (width b) | Broad central bright fringe, faint sides | First minimum at θ = λ/b |
| Double slit | Many evenly spaced fringes of similar brightness | Path difference = nλ for bright |
| Diffraction grating | A few very sharp, bright, well-separated maxima | d sinθ = nλ |
Why gratings beat two slits: Adding more slits makes each bright maximum narrower and brighter while keeping their angular positions fixed (still d sinθ = nλ). That sharpness is what makes a grating a precise tool for measuring wavelengths (spectroscopy).
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Where it shows up: Single-slit diffraction and the grating are HL only (C.3):
- Paper 1A — 'which change widens the central maximum?', 'what is the slit spacing of a 600 lines/mm grating?', or 'how does θ change with λ?'. - Paper 2 — determine a grating angle or an unknown wavelength from a measured angle, or find the central-maximum width from θ = λ/b.
Three easy marks: (1) For a grating, get d from the lines per metre first (d = 1 ÷ lines/m). (2) Use the whole grating equation d sinθ = nλ — don't drop the sin. (3) For a single slit, θ = λ/b gives the first minimum, and the central maximum is twice that wide.
IB-style question — measuring a wavelength with a grating
A grating with 300 lines per mm is used with a laser. The second-order (n = 2) maximum is seen at an angle of 23°. Determine the wavelength of the laser light.
Solution
- Find the slit spacing. 300 lines/mm = 300×10³ lines/m:
- Rearrange the given grating equation for λ:
- Substitute n = 2, θ = 23°, sin 23° = 0.391:
- Work it out — keep the unit:
Final answer
λ = 6.5×10⁻⁷ m = 650 nm (red light).