In YDSE setup,light of wavelength 640 , nm is | vedclass

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

Question

(A) The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$.Substituting the values: $\beta = \frac{640 	imes 10^{-9} \, m 	imes 1 \, m}{

Vedclass · Accepted Answer

Intensity at $y = 0.4 \, mm$ is zero

Vedclass · Answer

(A) The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$.Substituting the values: $\beta = \frac{640 	imes 10^{-9} \, m 	imes 1 \, m}{0.8 	imes 10^{-3} \, m} = 800 	imes 10^{-6} \, m = 0.8 \, mm$.The path difference $\Delta x$ at position $y$ is given by $\Delta x = \frac{yd}{D}$.For $y = 0.4 \, mm$,$\Delta x = \frac{0.4 	imes 10^{-3} 	imes 0.8 	imes 10^{-3}}{1} = 0.32 	imes 10^{-6} \, m = 320 \, nm$.The phase difference $\phi$ is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{640 \, nm} 	imes 320 \, nm = \pi$.The intensity at any point is $I = I_0 \cos^2(\frac{\phi}{2})$.Substituting $\phi = \pi$,$I = I_0 \cos^2(\frac{\pi}{2}) = 0$.

In $YDSE$ setup,light of wavelength $640 \, nm$ is used with $d = 0.8 \, mm$ and $D = 1 \, m$. If intensity at central maximum is $I_0$ and its position is $y = 0$,then:

Similar Questions

In a Young's double slit experiment,$15$ fringes are observed on a small portion of the screen when light of wavelength $500 \; nm$ is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in $nm$):

In a Young's double-slit experiment,the wavelength of light used is $6000 \, \mathring A$. The phase difference between the central bright fringe and the third bright fringe is .......

The width of one of the two slits in a Young's double slit experiment is $4$ times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (in $: 1$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module