Diffraction Envelope and Interference Fringes
For slit width a and centre separation d, interference maxima satisfy d sin(theta)=m lambda while the single-slit envelope has minima at a sin(theta)=p lambda. An interference order is missing when both conditions occur at the same angle.
Why this shows up in the exam
Diffraction gratings · Aperture-array design · Counting fringes within the central envelope
Learn the idea
Finite double-slit widths place fine interference fringes inside a broad single-slit envelope. Two narrow openings create regular stripes, but each opening also diffracts; the broad envelope decides which stripes are bright enough to appear or go missing.
🧠 Memory hook: Fine fringes obey d; the broad envelope obeys a.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- d sin(theta) = m lambda — two-slit interference maxima
- a sin(theta) = p lambda — single-slit envelope minima
- m = p d/a — missing-order condition when this is an integer
How to approach it
- 1Find the central-envelope limits
- 2List interference orders inside those limits
- 3Exclude any missing orders
Common slip-ups that cost marks
- •Using slit separation for diffraction minima
- •Counting boundary fringes that coincide with envelope minima
- •Assuming all interference maxima have equal brightness
🌟 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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