Among the two interfering monochromatic sourc | vedclass

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

Question

(A) The total phase difference at point $P$ is the sum of the initial phase difference and the phase difference caused by the path difference.Initial

Vedclass · Accepted Answer

$156$

Vedclass · Answer

(A) The total phase difference at point $P$ is the sum of the initial phase difference and the phase difference caused by the path difference.Initial phase difference $\phi_i = 66^\circ$ (where $A$ is ahead of $B$).Path difference $\Delta x = PB - PA = \lambda / 4$.The phase difference due to path difference is given by $\phi_p = \frac{2\pi}{\lambda} 	imes \Delta x = \frac{360^\circ}{\lambda} 	imes \frac{\lambda}{4} = 90^\circ$.Since $A$ is ahead of $B$ by $66^\circ$,and the path $PB$ is longer than $PA$ by $\lambda/4$,the wave from $B$ lags behind the wave from $A$ due to the path difference.The total phase difference $\Delta \phi = 66^\circ + 90^\circ = 156^\circ$.

Among the two interfering monochromatic sources $A$ and $B$; $A$ is ahead of $B$ in phase by $66^\circ$. If the observation is taken from point $P$,such that $PB - PA = \lambda / 4$. Then the phase difference between the waves from $A$ and $B$ reaching $P$ is.....$^\circ$.

Similar Questions

When a light wave goes from air into water,the quality that remains unchanged is its

Two sources are said to be coherent if they produce waves

If the wavelength of light is $4000 \, Å$, then the number of waves in a length of $1 \, mm$ will be:

Two plane polarized light waves combine at a certain point whose electric field components are $E_1 = E_0 \sin(\omega t)$ and $E_2 = E_0 \sin(\omega t + \frac{\pi}{3})$. Find the amplitude of the resultant wave.

Two light waves of intensities $I_0$ and $9I_0$ superpose at a point to produce a resultant intensity of $7I_0$. Calculate the phase difference between the light waves.

Vedclass Test Series

Exam Paper Generator

Online Exam Module