Resultant amplitude of two waves
For two waves of the same frequency and equal amplitude A meeting with phase difference Δφ, the resultant amplitude is 2A cos(Δφ/2): it is 2A when Δφ = 0 and zero when Δφ = π. If the amplitudes differ, the resultant runs from A₁ + A₂ (in phase) down to |A₁ − A₂| (antiphase).
A_R = 2A cos(Δφ/2) · in phase: A₁ + A₂ · antiphase: |A₁ − A₂| · I ∝ A_R²What this actually means
Amplitudes add, intensities do not. Two equal sources in phase give amplitude 2A, so intensity 4I, not 2I. Examiners love this because the factor of four catches people every time.
Work in amplitude first, then square at the very end to get intensity. Going the other way (adding intensities) is the single most common numerical error in this topic.
The cos(Δφ/2) form is not on the formula sheet and you are not asked to derive it, but knowing it explains why intensity fades smoothly between a bright fringe and a dark fringe rather than switching abruptly.
Unequal amplitudes are why contrast matters. If A₁ = 3A₂, the minima have amplitude 2A₂ rather than zero, so the dark fringes are grey rather than black and the pattern looks washed out.
For the ratio-of-intensities questions, write I_max/I_min = (A₁ + A₂)²/(A₁ − A₂)². That one line usually earns the whole calculation.
Adding intensities instead of amplitudes, giving 2I at a maximum instead of 4I.
Prove it — watch it be true
- Switch to the two-sines mode and set both amplitudes equal.
- Sweep the Δφ slider from 0 to 2π and watch the resultant amplitude readout track 2A cos(Δφ/2).
- Stop at Δφ = π and confirm the resultant is a flat line, then at Δφ = 2π and confirm it is back to 2A.