Question

Superposition of two electromagnetic waves of same frequency and same amplitude with phase difference $180^\circ$ have resultant of amplitude

• A. half of the single wave amplitude
• B. same as the single wave amplitude
• C. zero
• D. doubles the single wave amplitude

Answer 1

Answer: (C)

zero

Answer 2

The problem describes the superposition of two electromagnetic waves. Let the amplitude of each single wave be $A$. Let the frequency of each wave be $\omega$. The phase difference between the two waves is given as $180^\circ$ or $\pi$ radians.

We can represent the two waves mathematically as:

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

$E_2 = A \sin(\omega t + \phi)$

Given $\phi = 180^\circ = \pi$ radians, the second wave becomes:

$E_2 = A \sin(\omega t + \pi)$

Using the trigonometric identity $\sin(x + \pi) = -\sin(x)$, we get:

$E_2 = -A \sin(\omega t)$

When these two waves superpose, the resultant wave $E_R$ is the sum of the individual waves:

$E_R = E_1 + E_2$

$E_R = A \sin(\omega t) + (-A \sin(\omega t))$

$E_R = A \sin(\omega t) - A \sin(\omega t)$

$E_R = 0$

The resultant amplitude is 0.

Alternatively, using the formula for the resultant amplitude of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$:

$A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos\phi}$

Given $A_1 = A$, $A_2 = A$, and $\phi = 180^\circ$:

$A_R = \sqrt{A^2 + A^2 + 2(A)(A) \cos(180^\circ)}$

Since $\cos(180^\circ) = -1$:

$A_R = \sqrt{2A^2 + 2A^2 (-1)}$

$A_R = \sqrt{2A^2 - 2A^2}$

$A_R = \sqrt{0}$

$A_R = 0$

This phenomenon is known as destructive interference, where the waves cancel each other out completely.