Maxima, Minima, and Fringe Visibility
For two coherent beams, maximum intensity occurs at phase difference 2m pi and minimum at (2m+1) pi. Visibility compares the modulation depth with the mean intensity.
Why this shows up in the exam
Balancing interferometer arms · Assessing coherence quality · Optical contrast measurement
Learn the idea
Bright-dark contrast is controlled by the amplitude balance of the interfering waves. Perfect darkness needs equal waves to cancel. If one beam is stronger, subtraction leaves a residual field and the dark fringe is not black.
🧠 Memory hook: Equal amplitudes make the darkest dark.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Imax = (sqrt(I1) + sqrt(I2))² — maximum resultant intensity
- Imin = (sqrt(I1) - sqrt(I2))² — minimum resultant intensity
- V = (Imax - Imin)/(Imax + Imin) — fringe visibility
How to approach it
- 1Take square roots to get amplitude ratio
- 2Form sum and difference amplitudes
- 3Square before taking the requested ratio
Common slip-ups that cost marks
- •Using the intensity ratio as the amplitude ratio
- •Assuming every minimum is zero
- •Confusing Imax/Imin with visibility
🌟 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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