In the Young's double slit experiment,the int | vedclass

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

Question

(B) The intensity at any point on the screen is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_0$ is the intensity of each individual wave and $\phi$ is

Vedclass · Accepted Answer

$K/2$

Vedclass · Answer

(B) The intensity at any point on the screen is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference.The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = (2\pi / \lambda) 	imes \Delta x$.Case $1$: When path difference $\Delta x = \lambda$,the phase difference is $\phi = (2\pi / \lambda) 	imes \lambda = 2\pi$.Substituting this into the intensity formula: $I = 4I_0 \cos^2(2\pi / 2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Given this intensity is $K$,we have $K = 4I_0$.Case $2$: When path difference $\Delta x = \lambda / 4$,the phase difference is $\phi = (2\pi / \lambda) 	imes (\lambda / 4) = \pi / 2$.Substituting this into the intensity formula: $I' = 4I_0 \cos^2((\pi / 2) / 2) = 4I_0 \cos^2(\pi / 4) = 4I_0 (1 / \sqrt{2})^2 = 4I_0 (1 / 2) = 2I_0$.Since $K = 4I_0$,then $2I_0 = K / 2$. Therefore,the intensity is $K / 2$.

In the Young's double slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda / 4$ will be:

Similar Questions

In Young's double-slit experiment using monochromatic light of wavelength $\lambda$,the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ units. What is the intensity of light at a point where the path difference is $\lambda / 3$?

In Young's double-slit experiment,the ratio of intensities of bright and dark fringes is $9$. What is the ratio of the intensities of the sources?

Waves emitted from two identical sources produce an intensity of $K$ units at a point on the screen. If the path difference between these two waves is $\lambda$,calculate the intensity at a point on the screen where the path difference is $\lambda/4$.

In a Young's double slit experiment,the intensity at a point is $\left(\frac{1}{4}\right)^{\text{th}}$ of the maximum intensity. The minimum distance of the point from the central maximum is . . . . . . . . $\mu m$.
(Given: $\lambda = 600 \ nm, d = 1.0 \ mm, D = 1.0 \ m$)

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda / 6$ is $I'$. If $I_0$ denotes the maximum intensity,then $I'/I_0$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module