In YDSE,the intensity of the central bright f | vedclass

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

Question

(B) The intensity at any point in $YDSE$ is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $I_{max}$ is the intensity of the central bright fring

Vedclass · Accepted Answer

$6 \, mW/m^2$

Vedclass · Answer

(B) The intensity at any point in $YDSE$ is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $I_{max}$ is the intensity of the central bright fringe.Given $I_{max} = 8 \, mW/m^2$ and path difference $\Delta x = \frac{\lambda}{6}$.The phase difference $\phi$ is calculated as $\phi = \frac{2\pi}{\lambda} 	imes \Delta x = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}$.Substituting the values into the intensity formula:$I = 8 \cos^2(\frac{\pi/3}{2}) = 8 \cos^2(\frac{\pi}{6})$.Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have $\cos^2(\frac{\pi}{6}) = \frac{3}{4}$.Therefore,$I = 8 	imes \frac{3}{4} = 6 \, mW/m^2$.

In $YDSE$,the intensity of the central bright fringe is $8 \, mW/m^2$. What will be the intensity at a path difference of $\frac{\lambda}{6}$?

Similar Questions

Two coherent light sources having intensity in the ratio $2x$ produce an interference pattern. Then the value of $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,the ratio $\frac{I_0}{I}$ is equal to:

The ratio of intensities of two waves is $9 : 1$. They are producing interference. The ratio of maximum and minimum intensities will be

In a $YDSE$ setup,the intensity due to two coherent beams differs from each other by $1\%$. If one of the beams has intensity $I$,then the intensity of the minima is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module