Resultant Amplitude of Two Polarized Waves — 90° Phase Trick in 59 Sec 🔥
❓ Question Two plane-polarized light waves combine at a point. Their electric field components are E 1 = E 0 sin ( ω t ) , E 2 = E 0 sin ( ω t + π 2 ) . E_1 = E_0 \sin(\omega t), \qquad E_2 = E_0 \sin\!\left(\omega t + \frac{\pi}{2}\right). Find the amplitude of the resultant wave . 🖼️ Question Image ✍️ Short Solution This is a classic JEE superposition + phase difference problem. No trigonometric expansion marathon needed — phasor (vector) method makes it instant 🔥 🔹 Step 1 — Identify Phase Difference (MOST IMPORTANT 💯)** Given: E 1 = E 0 sin ( ω t ) , E 2 = E 0 sin ( ω t + π 2 ) E_1 = E_0 \sin(\omega t), \quad E_2 = E_0 \sin(\omega t + \tfrac{\pi}{2}) So the phase difference : Δ ϕ = π 2 = 90 ∘ \Delta\phi = \frac{\pi}{2} = 90^\circ 📌 This means the two waves are in quadrature . 🔹 Step 2 — Use Resultant Amplitude Formula For two waves of amplitudes E 0 E_0 and E 0 E_0 with phase difference Δ ϕ \Delta\phi : E res = E 0 2 + E 0 2 + 2 E 0...
