Question

If two electromagnetic waves with electric fields given by \[ \vec{E_1} = E_0 \sin (kx - \omega t) \hat{j} \] and \[ \vec{E_2} = E_0 \sin (kx - \omega t + \pi) \hat{j} \] - interfere, then the peak value of the electric field of the resultant wave is

• A. \(E_0\)
• B. \(\frac{E_0}{2}\)
• C. \(\frac{E_0}{\sqrt{2}}\)
• D. Zero

Hint:

When two waves with equal amplitude and opposite phase (\(\pi\) phase difference) interfere, they cancel each other completely, leading to a resultant electric field of zero.

Answer 1

The Correct Option is D

Step 1: Understanding Wave Interference When two electromagnetic waves superimpose, their resultant electric field is given by: \[ \vec{E} = \vec{E_1} + \vec{E_2} \] Given the wave equations: \[ \vec{E_1} = E_0 \sin (kx - \omega t) \hat{j} \] \[ \vec{E_2} = E_0 \sin (kx - \omega t + \pi) \hat{j} \] Since \(\sin(\theta + \pi) = -\sin(\theta)\), we rewrite: \[ E_2 = E_0 \sin (kx - \omega t + \pi) = -E_0 \sin (kx - \omega t) \] Step 2: Adding the Fields \[ E_{\text{resultant}} = E_0 \sin (kx - \omega t) + (-E_0 \sin (kx - \omega t)) \] \[ E_{\text{resultant}} = 0 \] Thus, the resultant electric field is zero, meaning complete destructive interference has occurred. Thus, the correct answer is \( \mathbf{(4)} \) Zero.