AS 91523 · 2025 paper · interference patterns
Question 3(d) Merit to Excellence
Practises Diffraction & refraction
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 |
In another experiment, a green 515 nm laser is shone through a different diffraction grating and 7 maxima are seen on the screen. The screen is 0.900 m wide, centred on the straight-through beam, and 1.20 m from the grating.
Calculate the possible range of values for the slit separation of the grating.
| λ | 515 nm |
|---|---|
| screen width | 0.900 m |
| grating to screen | 1.20 m |
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Seven maxima: which orders is that?
The central maximum (n = 0) and three either side of it. So orders 1, 2 and 3 are on the screen, and order 4 is not. -
2 The screen is 0.900 m wide and centred. At what angle is its edge?
Each edge is 0.450 m from the centre and 1.20 m from the grating: tan θ = 0.450 / 1.20, so θ = 20.556° and sin θ = 0.3511. -
3 For the 3rd order to land on the screen, what is the smallest d can be?
Its angle can be at most 20.556°. From d sin θ = 3λ: d ≥ 3 × 515 × 10⁻⁹ / sin 20.556° = 4.400 × 10⁻⁶ m. Slits any closer would push the 3rd order off the edge. -
4 For the 4th order to miss the screen, what is the largest d can be?
Its angle must be more than 20.556°: d < 4 × 515 × 10⁻⁹ / sin 20.556° = 5.867 × 10⁻⁶ m. Slits any further apart would bring the 4th order onto the screen. -
5 Why not use nλ = dx / L, as in part (a)?
That form assumes small angles, where sin θ ≈ tan θ. At 20.6° they differ by about 7%, and nλ = dx / L gives 4.12 × 10⁻⁶ and 5.49 × 10⁻⁶ m, both off. Find the angle from the screen first, then use d sin θ = nλ.
What the marker wanted
- Achieved One of: n = 3 is the outermost order on the screen; or the angle to the screen’s edge, 20.556° (0.359 rad).
- Merit The lower limit, 4.40 × 10⁻⁶ m, from n = 3; or both limits found with the small-angle form nλ = dx / L (4.12 and 5.49 × 10⁻⁶ m).
- Excellence The right range, with both its lower and upper limits.
In our words, from NZQA’s 2025 marking schedule.
✕Common mistake
Using nλ = dx / L for both limits. It gives 4.12 × 10⁻⁶ and 5.49 × 10⁻⁶ m, which the schedule credits at Merit only: at 20.6° the small-angle form is not accurate enough.How did it go?
Got it moves it further out. Review brings it back tomorrow.