Question

Two plane polarized light waves combine at a certain point whose electric field components are $ E_1 = E_0 \sin(\omega t) $ $ E_2 = E_0 \sin(\omega t + \frac{\pi}{3}) $ Find the amplitude of the resultant wave.

• A. \( 0.9 E \)
• B. \( E_0 \)
• C. \( 1.7 E_0 \)
• D. \( 3.4 E_0 \)

Hint:

To find the resultant amplitude of two waves, use the formula for the resultant of two waves with a phase difference. Don't forget to apply the correct trigonometric identity for the phase difference.

Answer 1

The Correct Option is C

We are given two electric field components of plane polarized light: \[ E_1 = E_0 \sin(\omega t) \] \[ E_2 = E_0 \sin\left( \omega t + \frac{\pi}{3} \right) \] The amplitude of the resultant wave is given by the formula: \[ E_{\text{res}} = \sqrt{E_1^2 + E_2^2 + 2E_1 E_2 \cos(\phi)} \] where \( \phi = \frac{\pi}{3} \) is the phase difference between the two waves. Substitute the values: \[ E_{\text{res}} = \sqrt{E_0^2 + E_0^2 + 2E_0^2 \cos\left( \frac{\pi}{3} \right)} \] Since \( \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \), we get: \[ E_{\text{res}} = \sqrt{E_0^2 + E_0^2 + E_0^2} = \sqrt{3E_0^2} = \sqrt{3} E_0 \]
Thus, the amplitude of the resultant wave is \( \sqrt{3} E_0 \approx 1.7 E_0 \).

Answer 2

Given two equal-magnitude fields \(E_0\) making an angle \(60^\circ=\dfrac{\pi}{3}\) with each other. We need the magnitude of the resultant \(E\).

<Method 1: Law of Cosines (Parallelogram Law)>
For two vectors \(\vec{E}_1\) and \(\vec{E}_2\) with angle \(\theta\) between them, \[ E=\lVert\vec{E}_1+\vec{E}_2\rVert=\sqrt{E_1^2+E_2^2+2E_1E_2\cos\theta}. \] Here \(E_1=E_2=E_0\) and \(\theta=\dfrac{\pi}{3}\). Hence, \[ E=\sqrt{E_0^2+E_0^2+2E_0E_0\cos\frac{\pi}{3}} =\sqrt{2E_0^2+2E_0^2\cdot\frac{1}{2}} =\sqrt{2E_0^2+E_0^2} =\sqrt{3E_0^2} =\sqrt{3}\,E_0. \] Therefore \(E\approx1.732\,E_0\).

<Method 2: Component (Resolution) Method>
Place \(\vec{E}_1\) along the \(x\)-axis: \(\vec{E}_1=\langle E_0,\,0\rangle\). Let \(\vec{E}_2\) make \(60^\circ\) with \(\vec{E}_1\): \[ \vec{E}_2=\langle E_0\cos60^\circ,\,E_0\sin60^\circ\rangle =\left\langle \tfrac{E_0}{2},\,\tfrac{\sqrt{3}}{2}E_0\right\rangle. \] Then \[ \vec{E}=\vec{E}_1+\vec{E}_2 =\left\langle E_0+\tfrac{E_0}{2},\,\tfrac{\sqrt{3}}{2}E_0\right\rangle =\left\langle \tfrac{3E_0}{2},\,\tfrac{\sqrt{3}}{2}E_0\right\rangle. \] Magnitude: \[ E=\sqrt{\left(\tfrac{3E_0}{2}\right)^2+\left(\tfrac{\sqrt{3}}{2}E_0\right)^2} =\sqrt{\tfrac{9E_0^2}{4}+\tfrac{3E_0^2}{4}} =\sqrt{\tfrac{12E_0^2}{4}} =\sqrt{3}\,E_0. \] Direction (optional): the angle \(\phi\) that \(\vec{E}\) makes with \(\vec{E}_1\) satisfies \[ \tan\phi=\frac{E_y}{E_x} =\frac{\tfrac{\sqrt{3}}{2}E_0}{\tfrac{3}{2}E_0} =\frac{\sqrt{3}}{3} \;\Rightarrow\; \phi=30^\circ. \] Thus the resultant has magnitude \(\boxed{\sqrt{3}\,E_0\approx1.73\,E_0}\) and lies \(30^\circ\) from either vector toward the other (bisecting the angle since magnitudes are equal).