Two plane polarized light waves combine at a certain point whose electric field components are

$$\begin{aligned} & E_1=E_0 \operatorname{Sin} \omega t \\ & E_2=E_0 \operatorname{Sin}\left(\omega t+\frac{\pi}{3}\right) \end{aligned}$$

Find the amplitude of the resultant wave.

<p>To find the amplitude of the resultant wave for the given electric field components, we use the formula for the resultant amplitude of two combining waves. These components are:</p> <p><p>$ E_1 = E_0 \sin \omega t $</p></p> <p><p>$ E_2 = E_0 \sin \left(\omega t + \frac{\pi}{3}\right) $</p></p> <p>The general formula for the amplitude $ E $ of two superimposed waves $ E_1 $ and $ E_2 $ with a phase difference is:</p> <p>$ E = \sqrt{E_1^2 + E_2^2 + 2E_1E_2 \cos\phi} $</p> <p>where $ \phi $ is the phase difference between the two waves. In this scenario, $ \phi = \frac{\pi}{3} $.</p> <p>Substituting the given expressions and phase difference into the formula, we get:</p> <p>$ E = \sqrt{(E_0)^2 + (E_0)^2 + 2(E_0)(E_0)\cos\left(\frac{\pi}{3}\right)} $</p> <p>Given that $\cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$, this simplifies to:</p> <p>$ E = \sqrt{2E_0^2 + E_0^2} = \sqrt{3}E_0 $</p> <p>Thus, the amplitude of the resultant wave is $\sqrt{3}E_0 \approx 1.73E_0$.</p>