A plane wave travelling along +z has E = 4sin(kz − ωt)x̂ + 4cos(kz − ωt)ŷ V/m. Identify its polarisation state and handedness. What would change if the ŷ amplitude were 2 instead of 4? Step 1: Rewrite the ŷ term: 4cos(kz − ωt) = 4sin(kz − ωt + π/2). So E₀ₓ = E₀ᵧ = 4 V/m with phase difference π/2 — equal amplitudes, 90° apart. Step 2: Equal amplitudes + 90° phase difference = circular polarisation. At fixed z, as t increases the phase (kz − ωt) decreases; tracking the vector: at phase 0, E = (0, 4); slightly later, phase −ε: E ≈ (−4ε... let's just note x = 4sin(−ε) ≈ −4ε < 0, y = 4cos(−ε) ≈ 4. The tip moves from (0,4) toward (−ε, 4), i.e. toward −x from +y — anticlockwise when viewed from +z looking back toward the source... Standard convention: looking into the oncoming wave (from +z toward −z), this rotation is right-circular. (Handedness conventions vary; the key identification is circular.) Step 3: If the ŷ amplitude were 2, amplitudes unequal with 90° phase → elliptical polarisation (ellipse with axes 4 and 2).