Superposition: Question 3

Worked solution

Part (a): Grating spacing

The grating has $600$ lines per millimetre, so the number of lines per metre is: $600 \times 1000 = 6.00\times10^{5}\text{ lines per metre}$

The spacing $d$ between adjacent lines is the reciprocal of this: $d = \frac{1}{6.00\times10^{5}} = 1.6\overline{6}\times10^{-6}\text{ m} \approx 1.67\times10^{-6}\text{ m}$

Check (independent method): $600$ lines per mm means each line is $\frac{1}{600}\text{ mm} = 0.001\overline{6}\text{ mm}$ from the next; converting to metres, $0.001\overline{6}\text{ mm} = 1.67\times10^{-6}\text{ m}$ (3 s.f.), the same result.

So $d \approx \boxed{1.67\times10^{-6}\text{ m}}$, as required.

Part (b): Angle of the first-order maximum

The diffraction grating equation is: $d\sin\theta = n\lambda$

For the first order, $n=1$, with $\lambda = 450\text{ nm} = 4.50\times10^{-7}\text{ m}$ and $d = 1.6\overline{6}\times10^{-6}\text{ m}$ (kept unrounded for accuracy): $\sin\theta_1 = \frac{n\lambda}{d} = \frac{1 \times 4.50\times10^{-7}}{1.6\overline{6}\times10^{-6}} = 0.270$

$\theta_1 = \arcsin(0.270) = 15.7° \text{ (3 s.f.)}$

Check (recompute independently): using the rounded $d = 1.67\times10^{-6}\text{ m}$, $\sin\theta_1 = 4.50\times10^{-7}/1.67\times10^{-6} = 0.2695 \approx 0.270$, consistent with the unrounded calculation; and $\sin(15.7°) \approx 0.2707 \approx 0.270$, confirming the angle.

So $\theta_1 = \boxed{15.7°}$.

Part (c): Highest observable order

The largest possible value of $\sin\theta$ is $1$, so the highest order $n$ satisfies $n\lambda/d \le 1$, i.e.: $n \le \frac{d}{\lambda} = \frac{1.6\overline{6}\times10^{-6}}{4.50\times10^{-7}} = 3.70 \text{ (3 s.f.)}$

Check (recompute independently): using $\sin\theta_1 = 0.270$ from part (b), $d/\lambda = 1/0.270 = 3.70$, consistent.

Since $n$ must be a whole number, testing each candidate directly against $n\lambda/d \le 1$:

• $n=3$: $3 \times 0.270 = 0.810 \le 1$, valid, $\theta_3 = \arcsin(0.810) \approx 54.1°$, an observable angle.
• $n=4$: $4 \times 0.270 = 1.08 > 1$, invalid, since $\sin\theta$ cannot exceed $1$; there is no real angle $\theta$ for which this maximum could appear.

So the highest order that can actually be observed is: $n_{\max} = \boxed{3}$

Final answers

• (a) $d \approx \boxed{1.67\times10^{-6}\text{ m}}$
• (b) $\theta_1 = \boxed{15.7°}$
• (c) Highest observable order $= \boxed{3}$ (since $n=4$ would require $\sin\theta = 1.08 > 1$, which is impossible)