In Y.D.S.E. using light of wavelength lambda, | vedclass

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

Question

(B) Let the intensity of each individual slit be $I_0$. The resultant intensity is given by $I_R = 4I_0 \cos^2(\frac{\Delta \phi}{2})$.For a path diff

Vedclass · Accepted Answer

$\frac{M}{4}$

Vedclass · Answer

(B) Let the intensity of each individual slit be $I_0$. The resultant intensity is given by $I_R = 4I_0 \cos^2(\frac{\Delta \phi}{2})$.For a path difference $\Delta x = \lambda$,the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.Given $I_R = M$ at $\Delta x = \lambda$,we have $M = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Thus,$I_0 = \frac{M}{4}$.For a path difference $\Delta x = \frac{\lambda}{3}$,the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$.The resultant intensity is $I_R' = 4I_0 \cos^2(\frac{2\pi/3}{2}) = 4I_0 \cos^2(\frac{\pi}{3})$.Substituting $I_0 = \frac{M}{4}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $I_R' = 4(\frac{M}{4}) \cdot (\frac{1}{2})^2 = M \cdot \frac{1}{4} = \frac{M}{4}$.

In $Y.D.S.E.$ using light of wavelength $\lambda$,the intensity of light at a point on the screen with path difference $\lambda$ is $M$ units. Calculate the intensity of light at a point where the path difference is $\frac{\lambda}{3}$.

Similar Questions

The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is $9: 1$. The intensities of the light sources used are in the ratio (in $: 1$)

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $k$. What will be the intensity at a point where the path difference is $\lambda/4$? ($\lambda$ = wavelength of light)

In the Young's double slit experiment,the intensities at two points $P_1$ and $P_2$ on the screen are respectively $I_1$ and $I_2$. If $P_1$ is located at the centre of a bright fringe and $P_2$ is located at a distance equal to a quarter of fringe width from $P_1$,then $\frac{I_1}{I_2}$ is

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