AS 91523 · 2025 paper · interference patterns
Question 3(a) Achieved
Practises Wave basics & superposition
A laser shines green light onto a double slit of separation 1.25 × 10⁻⁴ m. Each slit acts as a light source in phase with the other, giving an interference pattern of closely spaced fringes on a screen. With the screen 3.40 m from the slits, the distance from the central maximum to the 3rd-order maximum is 4.20 × 10⁻² m.
| d | 1.25×10⁻⁴ m |
|---|---|
| L | 3.40 m |
| x (centre to 3rd order) | 4.20×10⁻² m |
Show that the laser has a wavelength of 515 nm.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 Which formula links the slit separation, the screen distance, and the distance from the centre to a bright fringe?
nλ = dx / L, from the resource booklet, where x runs from the central maximum to the nth bright fringe. It is a small-angle form, and these angles are tiny. -
2 Rearrange for λ and put the values in. It is a “show that”, so keep every digit.
λ = dx / (nL) = (1.25 × 10⁻⁴ × 4.20 × 10⁻²) / (3 × 3.40) = 5.25 × 10⁻⁶ / 10.2 = 5.147 × 10⁻⁷ m -
3 Was the small-angle form safe to use?
tan θ = 0.0420 / 3.40 gives θ = 0.708°, far too small for sin θ and tan θ to differ. Working through d sin θ = nλ with that angle gives 514.67 nm instead of 514.71 nm: the same to three figures.
What the marker wanted
- Achieved The right unrounded value, 514.7 nm, from nλ = dx / L or worked through the angle (0.708°).
In our words, from NZQA’s 2025 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.