JEE Physics 2025 Question with Solution

Given: Two electric field components are

$$E_1 = E_0 \sin(\omega t)$$

and

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

So the two waves have equal amplitudes $E_0$ and phase difference $\phi = \frac{\pi}{3}$.

Find: The amplitude of the resultant wave.

For two waves of amplitudes $E_1$ and $E_2$ with phase difference $\phi$, the resultant amplitude is

$$E_{\text{res}} = \sqrt{E_1^2 + E_2^2 + 2E_1E_2\cos\phi}$$

Substituting $E_1 = E_2 = E_0$ and $\phi = \frac{\pi}{3}$,

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

Since

$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

we get

$$E_{\text{res}} = \sqrt{E_0^2 + E_0^2 + E_0^2} = \sqrt{3E_0^2} = \sqrt{3}E_0$$

Numerically,

$$\sqrt{3}E_0 \approx 1.7E_0$$

Therefore, the amplitude of the resultant wave is $\sqrt{3}E_0 \approx 1.7E_0$, so the correct option is C.

Given: Two equal-magnitude fields $E_0$ make an angle $60^\circ = \frac{\pi}{3}$ with each other.

Find: The magnitude of the resultant electric field.

Place $\vec{E}_1$ along the $x$-axis:

$$\vec{E}_1 = \langle E_0, 0 \rangle$$

Resolve $\vec{E}_2$ into components:

$$\vec{E}_2 = \left\langle E_0 \cos 60^\circ, E_0 \sin 60^\circ \right\rangle = \left\langle \frac{E_0}{2}, \frac{\sqrt{3}}{2}E_0 \right\rangle$$

Then the resultant is

$$\vec{E} = \vec{E}_1 + \vec{E}_2 = \left\langle E_0 + \frac{E_0}{2}, \frac{\sqrt{3}}{2}E_0 \right\rangle = \left\langle \frac{3E_0}{2}, \frac{\sqrt{3}}{2}E_0 \right\rangle$$

Its magnitude is

$$E = \sqrt{\left(\frac{3E_0}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}E_0\right)^2}$$ $$E = \sqrt{\frac{9E_0^2}{4} + \frac{3E_0^2}{4}} = \sqrt{\frac{12E_0^2}{4}} = \sqrt{3}E_0$$

Thus the resultant magnitude is $\sqrt{3}E_0$, which is approximately $1.73E_0$. Hence the correct option is C.