Multiple Wavelengths and Fringe Coincidence
At a common bright fringe, m1 lambda1 = m2 lambda2 for integer orders. A common dark fringe additionally requires odd-half-integer conditions for both wavelengths and may not exist for every wavelength ratio.
Why this shows up in the exam
Spectral line comparison · Multiwavelength interferometry · White-light fringe analysis
Learn the idea
Fringes of two wavelengths coincide when their optical path differences equal integer multiples of both wavelengths. Two stripe rulers with different spacing line up again only after a common multiple of their spacings.
🧠 Memory hook: Find the smallest common optical path, not just a common order number.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- m1 lambda1 = m2 lambda2 — condition for coincident bright fringes
- (2m1+1) lambda1 = (2m2+1) lambda2 — condition for coincident dark fringes
- y_common = D Delta_common/d — screen location from common path difference
How to approach it
- 1Reduce the wavelength ratio
- 2Choose the smallest valid integer pair
- 3Convert the common path difference to position
Common slip-ups that cost marks
- •Equating fringe orders instead of path differences
- •Using an LCM directly on decimal wavelengths without reducing their ratio
- •Assuming bright and dark coincidence obey the same integer rule
🌟 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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