Phasors & Superposition (IPhO)

Phasors are a bookkeeping tool: they turn same-frequency sinusoids into vectors so you can add them without expanding trig identities. In waves/optics problems, this usually means: reduce everything to a phase difference, add phasors, then square the amplitude if the question asks for intensity.

Practice feedback

• For beats, two phasors rotate at slightly different angular speeds. Their relative angle changes slowly, so the resultant amplitude cycles between reinforcement and cancellation at the difference frequency.

1. Definitions (Must Know)

• Sinusoid (one frequency): x(t) = A cos(ω t + φ) where ω = 2π f.
• Phase difference: Δφ = φ₂-φ₁ (only differences matter).
• Phasor / complex amplitude: represent x(t) as the real part of x tilde e^(iω t) with x tilde = Ae^iφ.
• Superposition (same ω): x tilde ₜₒₜ = ∑ⱼ x tilde ⱼ and the physical amplitude is |x tilde ₜₒₜ|.
• Path and time to phase: Δφ = ωΔ t = kΔ x with k = 2π/λ (in one medium).
• Beats (close frequencies): a fast oscillation with a slow amplitude envelope; beat frequency f_b = |f₁-f₂|.

2. Key Ideas (What Earns Marks)

• Pick a reference: set one phasor along the real axis, then every other term is defined by a phase difference.
• Add phasors as vectors (or complex numbers), then take a magnitude for the resultant amplitude.
• Use limiting checks: Δφ = 0 gives A₁ + A₂, Δφ = π gives |A₁-A₂|.
• Convert geometry to phase early: Δφ = 2π(Δ x/λ), and only then do the addition.
• For many equal steps in phase, recognize a geometric series (same trick behind gratings).

3. Detailed Explanations

3.1 The phasor model (one frequency)

Write

The time factor e^(iω t) is common to all same-frequency terms, so superposition reduces to adding the complex amplitudes x tilde.

3.2 Adding two sinusoids with the same ω

Let

Factor out a reference phase by setting φ₁ = 0 (you are free to choose the time origin), so the phasor sum is

The resultant amplitude is the magnitude:

The resultant phase relative to the reference is

For equal amplitudes A₁ = A₂ = A, this collapses to

3.3 Many phasors with equal phase steps (the grating shape)

If you add N equal phasors, each with amplitude A and a constant phase step δ:

then the magnitude is

This one result explains why adding more equally spaced sources gives narrow, strong principal maxima.

3.4 Beats as “phasors with slightly different angular speeds”

For equal amplitudes,

can be rewritten as

where ω bar = (ω₁ + ω₂)/2 and Δω = ω₂-ω₁. The envelope oscillates at f_b = |f₁-f₂|.

4. Common Mistakes

• Treating amplitudes like scalars (adding A₁ + A₂) when there is a phase difference.
• Mixing sine and cosine conventions mid-solution (pick one and stick to it).
• Losing track of degrees versus radians when evaluating trig for phases.
• Forgetting the physics step: many optics questions ask for intensity, so use I ∝ A² after you find the resultant amplitude.
• Using a path difference Δ x but forgetting the 2π when converting to phase.

5. Exam Tips

• Draw one clean phasor diagram and label Δφ; it often replaces a page of algebra.
• If the algebra feels messy, reset the reference phase so one term is purely real.
• Always sanity-check special phases: in-phase, opposite phase, and quarter-cycle.
• When asked for a ratio of intensities, compute amplitude ratio first, then square once at the end.

6. Worked Examples

7. Mind Stretchers

• Visibility with unequal amplitudes: two beams with amplitudes A₁ and A₂ interfere. Show that Iₘₐₓ ∝ (A₁ + A₂)² and Iₘᵢₙ ∝ (A₁-A₂)², then write the fringe visibility in terms of A₁/A₂.
• Random phases: add N equal phasors with phases uniformly spread from 0 to 2π. Estimate the typical size of the resultant (think “2D random walk”).
• Standing wave via phasors: at fixed x, add an incident and reflected wave and predict where the amplitude is always zero.

8. Practice