Two waves of equal amplitude and frequency in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The resultant intensity $I$ of two interfering waves of equal intensity $I_0$ is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase diff

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

(C) The resultant intensity $I$ of two interfering waves of equal intensity $I_0$ is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference between the waves.For waves arriving in phase,the phase difference $\phi_1 = 0^{\circ}$.Thus,$I_1 = 4I_0 \cos^2(0/2) = 4I_0 \cos^2(0) = 4I_0$.For waves arriving $90^{\circ}$ out of phase,the phase difference $\phi_2 = 90^{\circ}$.Thus,$I_2 = 4I_0 \cos^2(90^{\circ}/2) = 4I_0 \cos^2(45^{\circ}) = 4I_0 	imes (1/\sqrt{2})^2 = 4I_0 	imes (1/2) = 2I_0$.The ratio of the intensities is $\frac{I_1}{I_2} = \frac{4I_0}{2I_0} = \frac{2}{1}$.

Two waves of equal amplitude and frequency interfere with each other. The ratio of intensity when the two waves arrive in phase to that when they arrive $90^{\circ}$ out of phase is:

Similar Questions

The resultant amplitude due to the superposition of two waves $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$ is:

Four harmonic waves of equal frequencies and equal intensities $I_0$ have phase angles $0, \pi / 3, 2 \pi / 3$ and $\pi$. When they are superposed,the intensity of the resulting wave is $nI_0$. The value of $n$ is

If the phase difference between two waves is $2\pi$ during superposition,then the resultant amplitude is

The similarity between sound and light waves is:

Two sinusoidal waves given below are superposed:
$y_1 = A \sin \left(kx - \omega t + \frac{\pi}{6}\right), \quad y_2 = A \sin \left(kx - \omega t - \frac{\pi}{6}\right)$
The equation of the resultant wave is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module