Explain why the pattern of maxima and minima observed in single-slit diffraction differs from that observed in double-slit interference, even when the slit separation and wavelength are the same.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
In double‑slit interference the two slits are essentially point sources, so the path difference between the two waves at a screen point is
d = dsin heta
where d is the slit separation. Constructive interference occurs when dsin heta = mlambda, giving equally spaced bright fringes of equal intensity.
In single‑slit diffraction the slit has a finite width a. Huygens’ principle shows that each point across the slit acts as a secondary source. The path difference between the extreme parts of the slit is asin heta. Destructive interference occurs when asin heta = mlambda (m≠0), so the minima are at the same angles as the double‑slit condition, but the intensity between minima is not constant. The central maximum is twice as wide as the secondary maxima, and the intensity of successive maxima decreases because the contributions from different parts of the slit add with varying phase.
Thus the single‑slit pattern is a diffraction envelope that modulates the double‑slit interference fringes. Where the envelope has a minimum, the interference fringes are suppressed or absent, and the remaining fringes have reduced intensity.
d = dsin heta
where d is the slit separation. Constructive interference occurs when dsin heta = mlambda, giving equally spaced bright fringes of equal intensity.
In single‑slit diffraction the slit has a finite width a. Huygens’ principle shows that each point across the slit acts as a secondary source. The path difference between the extreme parts of the slit is asin heta. Destructive interference occurs when asin heta = mlambda (m≠0), so the minima are at the same angles as the double‑slit condition, but the intensity between minima is not constant. The central maximum is twice as wide as the secondary maxima, and the intensity of successive maxima decreases because the contributions from different parts of the slit add with varying phase.
Thus the single‑slit pattern is a diffraction envelope that modulates the double‑slit interference fringes. Where the envelope has a minimum, the interference fringes are suppressed or absent, and the remaining fringes have reduced intensity.
Examiner tips
- Use the equations for path difference in both cases; show the difference in source geometry.
- Explain that the single‑slit envelope comes from Huygens’ wavelets across the slit width.
- Mention that the central maximum is twice as wide and that fringe intensity falls off.
- Show that interference fringes are modulated by the diffraction envelope, leading to missing or dimmer fringes at the edges.
Common mistakes
- Confusing the conditions for minima in the two patterns; forgetting that asinθ=mλ in single‑slit. Misstating that the double‑slit fringes are of equal intensity; they are modulated by the envelope. Failing to note that the central maximum is twice as wide in single‑slit diffraction.
Mark scheme (4 marks)
- In double-slit interference, the two slits act as point (coherent) sources, producing equally spaced maxima of equal intensity.
- In single-slit diffraction, path differences arise between wavelets from different parts of the same slit (Huygens' wavelets / secondary sources across the slit width), causing the maxima to have decreasing intensity away from the centre.
- The central maximum in single-slit diffraction is twice as wide as the secondary maxima (minima occur at angles where the path difference across the full slit equals a whole number of wavelengths).
- In the double-slit pattern the single-slit diffraction envelope modulates the intensity of the interference fringes, so fringes near the edges of the envelope are dimmer and some interference maxima may be missing where they coincide with single-slit minima.
Key terms in this question
Related
- All IB DP Physics Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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