The big idea: Shine one colour of light through two narrow slits that are close together.
The light from the two slits overlaps on a screen and makes a row of equally spaced bright and dark bands — called fringes.
Bright = the two waves arrive in step (add up); dark = they arrive out of step (cancel).
New words: Fringe = one of the bright or dark bands on the screen.
Fringe spacing s = the gap from one bright fringe to the next bright fringe.
Coherent = the two slits give light of the same wavelength with a fixed phase relationship — needed for a steady pattern.
Spot it: The fringes are evenly spaced and the central one is the brightest. The spacing s is the same gap all the way across.
The fringe spacing depends on three things: the wavelength of the light, how far away the screen is, and how close the two slits are. They combine into one equation that is given in the data booklet.
- fringe spacing — gap between neighbouring bright fringes (m)
- wavelength of the light (m)
- distance from the slits to the screen (m)
- separation of the two slits (m)
Which way each one pushes the fringes: Bigger λ or bigger D → wider fringes (s up).
Bigger slit separation d → narrower fringes (s down), because d is on the bottom.
[Diagram: phys-formula-triangle] - Available in full study mode
Worked example — find the wavelength
Two slits 0.50 mm apart are lit by a laser. On a screen 2.0 m away the bright fringes are 2.4 mm apart. Find the wavelength of the light.
Solution
- Start with the given formula:
- Rearrange for the wavelength λ:
- Put in the numbers in metres (s = 2.4×10⁻³, d = 0.50×10⁻³, D = 2.0):
- Work it out — keep the unit:
Final answer
wavelength λ = 6.0 × 10⁻⁷ m (600 nm) — orange-red light.
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How this is tested: Double-slit questions come in two flavours.
- Paper 2: plug into s = λD/d to find a missing quantity, OR find the angular separation of the fringes (an angle in radians, using θ ≈ λ/d for small angles). - Paper 1A / short answer: a suggest/explain — e.g. why a dark fringe (zero energy) still obeys energy conservation.
Classic trap: mixing units — get every length into metres (mm = 10⁻³ m, nm = 10⁻⁹ m) before substituting.
Fringe angle in radians: Each bright fringe sits at an angle θ from the slits where d sin θ = nλ.
For the small angles in these experiments, sin θ ≈ θ (in radians), so neighbouring maxima are about λ/d apart, and the nearest minima either side of the centre are also λ/d apart in total.
[Diagram: phys-formula-triangle] - Available in full study mode
IB-style question — angular separation of the nearest minima
In a double-slit experiment the slit separation is 0.25 mm and the wavelength is 5.0 × 10⁻⁷ m. Calculate, in radians, the angular separation of the two nearest minima either side of the central maximum.
Solution
- Maxima are spaced by λ/d (radians, small angle); the minima sit halfway between, so the two nearest minima are also λ/d apart in total:
- Put in the numbers in metres (λ = 5.0×10⁻⁷, d = 0.25×10⁻³):
- Work it out — radians have no unit:
Final answer
angular separation of the nearest minima Δθ = 2.0 × 10⁻³ rad.