JEE Main 2025 — Wave Optics Question with Solution

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:

$E_1=E_0\sin \omega t$

$E_2=E_0\sin \left(\omega t+\frac{\pi }{3}\right)$

The general formula for the amplitude $E$ of two superimposed waves $E_1$ and $E_2$ with a phase difference is:

$E=\sqrt{E_1^2+E_2^2+2E_1{E}_2\cos \varphi }$

where $\varphi$ is the phase difference between the two waves. In this scenario, $\varphi =\frac{\pi }{3}$.

Substituting the given expressions and phase difference into the formula, we get:

$E=\sqrt{(E_0{)}^2+(E_0{)}^2+2(E_0)(E_0)\cos \left(\frac{\pi }{3}\right)}$

Given that $\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$, this simplifies to:

$E=\sqrt{2E_0^2+E_0^2}=\sqrt{3}{E}_0$

Thus, the amplitude of the resultant wave is $\sqrt{3}{E}_0\approx 1.73E_0$.