There is a destructive interference between t | vedclass

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

Question

(B) For destructive interference,the path difference between two waves is given by $\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, ...$. F

Vedclass · Accepted Answer

$\frac{\lambda}{2}$

Vedclass · Answer

(B) For destructive interference,the path difference between two waves is given by $\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, ...$. For the first case of destructive interference $(n=0)$,the path difference is $\Delta x = \frac{\lambda}{2}$.To achieve constructive interference,the path difference must be an integer multiple of the wavelength,i.e.,$\Delta x' = n\lambda$. The condition for constructive interference is $\Delta x' = m\lambda$ (where $m$ is an integer).If we increase the path of one wave by $\frac{\lambda}{2}$,the new path difference becomes $\Delta x_{new} = \Delta x + \frac{\lambda}{2} = \frac{\lambda}{2} + \frac{\lambda}{2} = \lambda$. Since $\lambda$ is an integer multiple of the wavelength $(m=1)$,constructive interference occurs.

There is a destructive interference between two waves of wavelength $\lambda$ coming from two different paths at a point. To get maximum sound or constructive interference at that point,the path of one wave is to be increased by

Similar Questions

Equations of motion for two waves traveling in the same direction are given by $y_1 = 2a \sin(\omega t - kx)$ and $y_2 = 2a \sin(\omega t - kx - \theta)$. The resultant amplitude of the medium particle will be:

Three harmonic waves having equal frequency $v$ and same intensity $I_{0}$,have phase angles $0, \frac{\pi}{4}$ and $-\frac{\pi}{4}$ respectively. When they are superimposed,the intensity of the resultant wave is close to

Two waves each of intensity $I$ interfere with a phase difference of $120^o$. The resultant intensity of the waves will be:

State the principle of superposition and explain it.

Two sound waves each of wavelength $\lambda$ and same amplitude $A$ interfere at point $Q$. If the path difference is $\frac{\lambda}{4}$, the amplitude of the resultant wave at point $Q$ is $[\sin \frac{\pi}{2} = 1, \cos \frac{\pi}{2} = 0]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module