 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-resit/5/solution

d sin θ

d sin θ = mλ

holds. Thus the groove spacing is

d = \frac{3 ( 589 \times 1 0 ^{- 9} m )}{s i n 3 2 ^{\circ}} = 3.334 \times 1 0^{- 6} m, N = \frac{1 0 ^{- 3} m}{d} = 299.9 lines per mm ≃ 300 lines per mm .

An allowed propagating diffraction order must have

∣ sin θ ∣ \leq 1

. Therefore

∣ m ∣ \leq \frac{d}{λ} = \frac{3}{s i n 3 2 ^{\circ}} = 5.6615, ∣ m ∣_{m a x} = 5.

The fifth diffraction order is the highest geometrically permitted order for this wavelength. Its actual brightness depends on groove shape and the single-slit diffraction envelope, which the source does not specify; geometric permission alone does not guarantee a bright observable maximum.

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