C.3.12 (HL)—Single-slit diffraction

Recognize the pattern

A monochromatic wave passing through a narrow rectangular slit spreads and forms a broad central maximum with weaker side maxima separated by minima. The pattern results from interference between contributions across the slit.

Use the first-minimum condition

For slit width $b$ , the first minimum satisfies $\theta\approx\lambda/b$ for small angles. On a screen a distance $x$ away, the central maximum width is approximately $2x\lambda/b$ .

Check the trends

A narrower slit or longer wavelength produces greater angular spreading and a wider central maximum. A wider slit or shorter wavelength produces a narrower pattern. The syllabus treatment is monochromatic light and rectangular slits at normal incidence.

Common trap

Do not confuse the distance from the central maximum to the first minimum with the full central-maximum width; the latter is twice the first-minimum distance on the screen.

How it is assessed

Questions ask you to find wavelength from an intensity graph or choose the width of the central maximum. The evidence rewards λ = bθ and 2xλ/b, with the factor of two handled correctly.

Command terms

Calculate / What is

What earns marks

Read the first intensity minimum, use θ ≈ λ/b, and multiply by the slit-to-screen distance when the question asks for central-maximum width. Check whether the requested width is one-sided or the full width.

Watch for

Using λ/b as the full central-maximum width on the screen and omitting the factor 2x.

Representative question

[Maximum number: 2]

The graph shows the variation with diffraction angle $\theta$ of the intensity I on the screen.

\(I / \mathrm{Wm

The slit width is $1.3 \times 10^{-5} \mathrm{~m}$ . Calculate the wavelength of the light.

First diffraction minimum at / value use of $\theta=0.04 \mathrm{rad}$ $\theta=\frac{\lambda}{b} \Rightarrow \lambda=b \theta=1.3 \times 10^{-5} \times 0.04=5.2 \times 10^{-7} \mathrm{~m}$

Marking guidance:

Award full marks for a CNA

Allow ECF from MP1

[2]

The HL extension is secure when you can model diffraction as interference and read the limits of the pattern.

Single-slit diffraction: b sinθ = nλ for minima
Diffraction envelopes in multiple-slit patterns
Diffraction grating maxima: nλ=d sinθ
Allowed orders satisfy nλ≤d

C.3.12 (HL)—Single-slit diffraction

Model Single-Slit Diffraction

C.3.12 (HL) Exam Analysis

Assessment in practice

Retrieve the HL C.3 Wave Phenomena Model

Question 1