A-Level Physics May/June 2025 Q6(b)(ii): Calculate the number of lines per unit length in the diffraction grating.

1 answer

• ▲0
  accepted ✓

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

  Where:

  • $d$ is the grating spacing
  • $\theta$ is the angle of diffraction ($26^\circ$)
  • $n$ is the order of the maximum (second order, so $n=2$)
  • $\lambda$ is the wavelength of light ($720 \text{ nm} = 720 \times 10^{-9} \text{ m}$)

  First, calculate the grating spacing, $d$: $d = \frac{n\lambda}{\sin \theta}$ $d = \frac{2 \times (720 \times 10^{-9})}{\sin 26^\circ}$ $d = \frac{1.44 \times 10^{-6}}{0.43837...}$ $d = 3.285... \times 10^{-6} \text{ m}$

  The number of lines per unit length, $N$, is the reciprocal of the grating spacing: $N = \frac{1}{d}$ $N = \frac{1}{3.285... \times 10^{-6}}$ $N = 304286... \text{ m}^{-1}$

  Rounding to two significant figures: $N = 3.0 \times 10^5 \text{ m}^{-1}$

  ---

  How the marks are awarded

  • C1 — The first mark is awarded for correctly stating or using the diffraction grating equation, $d \sin \theta = n\lambda$.
  • C1 — The second mark is for the correct substitution of values ($n=2$, $\lambda=720 \times 10^{-9}$ m, $\theta=26^\circ$) into the equation to calculate the grating spacing, $d$.
  • A1 — The final mark is for calculating the number of lines per unit length by taking the reciprocal of the grating spacing ($N=1/d$) and providing a final answer of $3.0 \times 10^5 \text{ m}^{-1}$.

  Common mistakes

  • Forgetting to convert the wavelength from nanometres (nm) to metres (m), i.e., using $\lambda = 720$.
  • Using the wrong order of maximum, for example using $n=1$ instead of $n=2$.
  • Forgetting the final step of calculating the reciprocal ($1/d$) and giving the grating spacing $d$ as the final answer.
  • Having the calculator in radians mode instead of degrees mode when calculating $\sin 26^\circ$.

  Examiner tip: Always be clear about the quantity you are calculating at each step; here, distinguishing between the grating spacing ($d$) and the number of lines per unit length ($1/d$) is crucial.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.