Diffraction grating
Waves - OCR A-Level Physics
Key Definition
Diffraction grating
A plate with many equally spaced, parallel slits (typically hundreds per mm). Produces sharp, bright maxima at specific angles where constructive interference occurs.
A plate with many equally spaced, parallel slits (typically hundreds per mm). Produces sharp, bright maxima at specific angles where constructive interference occurs.
$$d\sin\theta = n\lambda$$
- The zero-order maximum ($n = 0)$ is the central bright line at θ = 0°.
- Higher orders (n = 1, 2, 3...) appear at increasing angles either side of the centre.
- The maximum order visible is found from n_max = d/λ (rounded down), because sin θ cannot exceed 1.
- More slits per mm produces sharper, brighter maxima and better resolutionThe smallest change in a quantity that an instrument can detect. For example, a ruler has a resolution of 1 mm..
- Gratings produce much sharper lines than a double slit, making them useful for precise wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). measurement and spectroscopy.
Worked Example [4 marks]
A diffraction gratingAn optical component with many equally spaced parallel slits that produces sharp interference maxima at specific angles. has 600 lines per mm. Monochromatic light of wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). 520 nm is incident on the grating. Calculate the angle of the second-order maximum and the maximum order visible.
Show Solution
1
$d = 1/(600 \times 10^{3}) = 1.667 \times 10⁻⁶\;\text{m}$
[1]2
For n = 2: sin θ = nλ/d = (2 × 520 × 10⁻⁹) / (1.667 × 10⁻⁶) = 0.6240
[1]3
$\theta = sin⁻¹(0.6240) = 38.6°$
[1]4
Maximum order
$n_max = d/\lambda = 1.667 \times 10⁻⁶ / 520 \times 10⁻⁹ = 3.2, so n_max = 3$
[1]Answer
\theta_{2} = 38.6°; maximum order n = 3
Common Mistake
MEDIUM
Students often: Forgetting to convert 'lines per mm' to slit spacing in metres. Using $d = 600$ instead of $d = 1/(600 \times 10^{3})$.
Instead: Make sure to always convert: $d = 1/(N \times 10^{3})$, where N is the number of lines per mm. For 600 lines/mm, d = 1.667 × 10⁻⁶ m.
Instead: Make sure to always convert: $d = 1/(N \times 10^{3})$, where N is the number of lines per mm. For 600 lines/mm, d = 1.667 × 10⁻⁶ m.
Examiner Tips and Tricks
- To find the maximum order, set sin $\theta = 1 (the largest possible value)$ in d sin $\theta = n\lambda$ and solve for n.
- Round down to the nearest whole number.