Diffraction — practice questions

Step 1 — Grating spacing: 400 lines per mm means $d = \dfrac{1 \times 10^{-3}}{400} = 2.50 \times 10^{-6}\ \text{m}$.

Step 2 — Grating equation: $d \sin\theta = n\lambda$ with $n = 2$ and $\lambda = 550 \times 10^{-9}\ \text{m}$.

Step 3 — Solve for the angle: $\sin\theta = \dfrac{2 \times 550 \times 10^{-9}}{2.50 \times 10^{-6}} = 0.440$, so $\theta = \sin^{-1}(0.440) = 26.1^{\circ}$.

Check: the first-order maximum is at $\sin^{-1}(0.220) = 12.7^{\circ}$; the second order is further out, as expected, but not at double the angle — the relationship is in $\sin\theta$, not $\theta$.

(a) Slit spacing. Rearrange $d \sin\theta = n\lambda$ with $n = 1$:

$d = \dfrac{\lambda}{\sin\theta} = \dfrac{680 \times 10^{-9}}{\sin 19.5^{\circ}}$

$d = \dfrac{680 \times 10^{-9}}{0.334} = 2.04 \times 10^{-6}\ \text{m}$

(b) Number of maxima. The largest possible angle is $90^{\circ}$, where $\sin\theta = 1$, so the highest order satisfies $n \le \dfrac{d}{\lambda}$:

$\dfrac{d}{\lambda} = \dfrac{2.04 \times 10^{-6}}{680 \times 10^{-9}} = 2.996$

$n$ must be a whole number, so the highest observable order is $n = 2$ (a third order would need $\sin\theta > 1$, which is impossible).

Maxima are seen at $n = 0$ and at $n = 1, 2$ on each side of the centre: $2 \times 2 + 1 =$ 5 maxima.

Common error: rounding $2.996$ up to 3. Always round the order down — and remember to count the zero-order maximum and both sides.