In a standard YDSE setup,two sources S_1 and | vedclass

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

Question

(B) The intensity distribution in a $YDSE$ experiment is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$.From the graph,we observe that t

Vedclass · Accepted Answer

$S_1$ and $S_2$ have a constant phase difference.

Vedclass · Answer

(B) The intensity distribution in a $YDSE$ experiment is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$.From the graph,we observe that the minimum intensity $(I_{min})$ is zero. Since $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$,$I_{min} = 0$ implies that $I_1 = I_2$. Thus,the sources must have the same intensity.Furthermore,for a stable interference pattern to be observed on the screen,the sources $S_1$ and $S_2$ must be coherent. Coherent sources are defined as sources that maintain a constant phase difference over time. Therefore,the statement that $S_1$ and $S_2$ have a constant phase difference is a fundamental requirement for the observed interference pattern.

In a standard $YDSE$ setup,two sources $S_1$ and $S_2$ of intensities $I_1$ and $I_2$ are placed in front of a screen (figure). The pattern of intensity distribution seen in the central portion is given by the graph. In this case,which of the following statements is true?

Similar Questions

In Young's double slit experiment,the central bright fringe can be identified

In a Young's double-slit experiment,the separation of the two slits is doubled. To keep the same spacing of fringes,the distance $D$ of the screen from the slits should be made

In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module