Explain why a single narrow slit produces a diffraction pattern with a bright central maximum flanked by alternating dark and bright fringes of decreasing intensity.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Monochromatic light of wavelength 550 nm is directed at a single narrow slit. A screen placed far from the slit shows a diffraction pattern consisting of a bright central maximum and several secondary maxima of diminishing brightness on either side.
Model answer (4 marks)
1. Each point on the slit acts as a source of secondary wavelets (Huygens’ principle). 2. The wavelets interfere beyond the slit; at the centre of the screen the path difference between all wavelets is zero, giving constructive interference and a bright central maximum. 3. Dark fringes occur when the path difference between wavelets from opposite edges of the slit is a half‑wavelength (or odd multiple), producing complete destructive interference. 4. Secondary maxima are less bright because only a portion of the wavelets interfere constructively; the constructive fraction decreases for higher‑order fringes, so the intensity falls off.
Examiner tips
- Use Huygens’ principle and path‑difference arguments; mention constructive vs destructive interference; explain why central maximum is brightest; note intensity decreases with order.
Common mistakes
- Confusing the condition for minima (half‑wavelength) with maxima; forgetting that the central maximum is due to zero path difference; not explaining why higher maxima are dimmer.
Mark scheme (4 marks)
- Each point on the wavefront within the slit acts as a source of secondary wavelets (Huygens' principle), and these wavelets superpose / interfere beyond the slit.
- The central maximum is bright because wavelets from across the entire slit arrive at the centre of the screen with zero path difference, so they interfere constructively.
- Dark fringes (minima) occur at angles where wavelets from different parts of the slit can be paired so that each pair has a path difference of half a wavelength, resulting in complete destructive interference.
- Secondary maxima are less bright than the central maximum because only a fraction of the slit's wavelets contribute constructively (the remaining wavelets cancel), and the fraction contributing constructively decreases for higher-order maxima.
Key terms in this question
Related
- All IB DP Physics Standard Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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