AS 91523 · 2023 paper · diffraction glasses
Question 3(d) Merit
Practises Diffraction & refraction
At a fair, children buy “Rainbow Glasses” made of diffraction gratings in a cardboard frame. Steve shines a laser pointer of wavelength 643 nm (6.43 × 10⁻⁷ m) through one of the gratings onto a wall. The pattern is a bright red spot at the centre, with slightly dimmer red spots either side. The wall is 1.43 m from the grating, and the distance from the central bright spot to the second dimmer spot is 1.75 m.
| λ | 643 nm (6.43 × 10⁻⁷ m) |
|---|---|
| grating to wall | 1.43 m |
| centre to 2nd spot | 1.75 m |
Looking at a spot of white light through the glasses, the children see the white spot with spectra on either side (“rainbows”). Give an in-depth explanation: describe where the spectra occur, explain the position of the colours in them and why they are there, and include a labelled sketch.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Where is the light white, and why?
Straight through, at the centre: at θ = 0 every colour has zero path difference, so they are all bright together and mix into white. -
2 For the first order, what path difference does each colour need?
One of its own wavelengths: violet the least (about 400 nm), red the most (about 700 nm). -
3 So at what angle is each colour bright?
sin θ = nλ / d: the shorter the wavelength, the smaller the angle. Violet is bright closest to the centre, red furthest out, and the other colours in between. -
4 Where do the spectra appear, and what goes in the sketch?
On both sides of the white centre, one complete spectrum for each order (n = 1, 2), each with violet on the inside and red on the outside. Sketch the white central spot labelled n = 0, then on each side the n = 1 and n = 2 spectra, each running violet to red outwards. With these glasses (d = 1.66 × 10⁻⁶ m) the first-order spectrum runs from about 14° (violet) to 25° (red).
What the marker wanted
- Achieved One idea: a complete spectrum is seen, violet inside and red outside; a whole-wavelength path difference puts the waves in phase; a whole-wavelength path difference gives a maximum; or waves in phase give a maximum.
- Merit The right reasoning for violet inside and red outside: red’s longer wavelength needs a larger path difference, so a larger angle in nλ = d sin θ.
In our words, from NZQA’s 2023 marking schedule.
✕Common mistake
Putting red on the inside. NZQA’s 2023 report lists spectra drawn the wrong way round among answers that did not reach Achieved. It is the reverse of a prism: with a grating, the shortest wavelength is nearest the centre.How did it go?
Got it moves it further out. Review brings it back tomorrow.