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

$$E_1 = E_0 \sin(\omega t), \qquad E_2 = E_0 \sin\!\left(\omega t + \frac{\pi}{2}\right).$$

Find the amplitude of the resultant wave.

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🖼️ Question Image

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✍️ Short Solution

This is a classic JEE superposition + phase difference problem.
No trigonometric expansion marathon needed — phasor (vector) method makes it instant 🔥

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🔹 Step 1 — Identify Phase Difference (MOST IMPORTANT 💯)**

Given:

$$E_1 = E_0 \sin(\omega t), \quad E_2 = E_0 \sin(\omega t + \tfrac{\pi}{2})$$

So the phase difference:

$$\Delta\phi = \frac{\pi}{2} = 90^\circ$$

📌 This means the two waves are in quadrature.

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🔹 Step 2 — Use Resultant Amplitude Formula

For two waves of amplitudes $E_0$ and $E_0$ with phase difference $\Delta\phi$:

$$E_{\text{res}}=\sqrt{E_0^2+E_0^2+2E_0^2\cos \mathrm{\Delta }\mathrm{\varphi }}$$

This is nothing but vector addition of two phasors.

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🔹 Step 3 — Substitute Phase Difference

Since:

$$\cos\left(\frac{\pi}{2}\right) = 0$$

We get:

$$E_{\text{res}} = \sqrt{E_0^2 + E_0^2} = \sqrt{2E_0^2} = \sqrt{2}\,E_0$$

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🔹 Step 4 — Physical Interpretation (JEE Insight 🧠)**

• Phase difference = 90°

• Vectors are perpendicular

• Resultant magnitude found by Pythagoras

🧠 One-line intuition:

Quadrature waves add like perpendicular vectors

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✅ Final Answer

$$\boxed{E_{\text{resultant}} = \sqrt{2}\,E_0}$$

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⭐ Golden JEE Insight

• Same amplitude + phase difference $\pi/2$
  

• No need to expand sine terms

• Direct phasor addition

📌 Special cases to remember:

• $\Delta\phi = 0$ ⇒ amplitude $= 2E_0$

• $\Delta\phi = \pi$ ⇒ amplitude $= 0$
  

• $\Delta\phi = \pi/2$ ⇒ amplitude $= \sqrt{2}E_0$