Single-Slit Diffraction Minima and Central Maximum
For Fraunhofer diffraction by a slit of width a, minima satisfy a sin(theta) = m lambda for nonzero integer m. The central maximum lies between the first minima and is twice the approximate width of a secondary maximum.
Why this shows up in the exam
Slit-width measurement · Spectrometer aperture design · Microwave diffraction demonstrations
Learn the idea
A finite slit spreads light, with minima set by destructive interference across the aperture. Think of the slit as many tiny sources. At special angles, contributions pair off across the slit and cancel, producing dark bands.
🧠 Memory hook: First dark on each side sets the central width.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- a sin(theta_m) = m lambda — single-slit minima
- angular central width = 2 sin⁻¹(lambda/a) — exact full angular width when first minima exist
- linear central width approximately 2 lambda D/a — small-angle screen width
- linear focal width approximately 2 f lambda/a — pattern in a lens focal plane
How to approach it
- 1Mark the relevant minimum orders
- 2Use a sin(theta)=m lambda
- 3Double the first-minimum distance only for full central width
Common slip-ups that cost marks
- •Using m=0 as a minimum
- •Calling lambda D/a the full central width
- •Using small-angle formulas when lambda/a is not small
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A real object is placed 30 cm from a converging lens of focal length 10 cm. Find the real image distance.
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