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

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

Question

(C) Given: $d = 0.3 	ext{ mm} = 3 	imes 10^{-4} 	ext{ m}$,$D = 1 	ext{ m}$,$\lambda = 600 	ext{ nm} = 6 	imes 10^{-7} 	ext{ m}$,$PO = y = 11

Vedclass · Accepted Answer

For $\alpha = \frac{0.36}{\pi}$ degree,there will be destructive interference at point $P$.

Vedclass · Answer

(C) Given: $d = 0.3 	ext{ mm} = 3 	imes 10^{-4} 	ext{ m}$,$D = 1 	ext{ m}$,$\lambda = 600 	ext{ nm} = 6 	imes 10^{-7} 	ext{ m}$,$PO = y = 11 	ext{ mm} = 1.1 	imes 10^{-2} 	ext{ m}$.$(1)$ At point $O$ $(y=0)$,the path difference is $\Delta x = d \sin \alpha \approx d \alpha$ (for small $\alpha$).Given $\alpha = \frac{0.36}{\pi} 	ext{ degrees} = \frac{0.36}{\pi} 	imes \frac{\pi}{180} 	ext{ radians} = 2 	imes 10^{-3} 	ext{ rad}$.Path difference $\Delta x = (3 	imes 10^{-4} 	ext{ m}) 	imes (2 	imes 10^{-3} 	ext{ rad}) = 6 	imes 10^{-7} 	ext{ m} = \lambda$.Since $\Delta x = n\lambda$ (where $n=1$),there is constructive interference at $O$. Thus,statement $(A)$ is incorrect.$(2)$ Fringe spacing $\beta = \frac{D\lambda}{d}$. This depends only on $D, \lambda, d$,not on $\alpha$. Thus,statement $(B)$ is incorrect.$(3)$ At point $P$,path difference $\Delta x_P = d \sin \alpha + \frac{dy}{D} \approx d \alpha + \frac{dy}{D}$.$\Delta x_P = (3 	imes 10^{-4})(2 	imes 10^{-3}) + \frac{(3 	imes 10^{-4})(1.1 	imes 10^{-2})}{1} = 6 	imes 10^{-7} + 33 	imes 10^{-7} = 39 	imes 10^{-7} 	ext{ m}$.$\frac{\Delta x_P}{\lambda} = \frac{39 	imes 10^{-7}}{6 	imes 10^{-7}} = 6.5$. Since it is a half-integer multiple,there is destructive interference at $P$. Thus,statement $(C)$ is correct.$(4)$ For $\alpha = 0$,$\Delta x_P = \frac{dy}{D} = 33 	imes 10^{-7} 	ext{ m}$.$\frac{\Delta x_P}{\lambda} = \frac{33 	imes 10^{-7}}{6 	imes 10^{-7}} = 5.5$. Since it is a half-integer multiple,there is destructive interference at $P$. Thus,statement $(D)$ is incorrect.

Similar Questions

Two sources produce an interference pattern which is observed on a screen,at a distance $D$ from the sources. The fringe width is $2w$. If the distance $D$ is now doubled,the fringe width will be:

In Young's double slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be (in $:1$)

In a Young's double slit experiment,the slit separation is $0.2 \ cm$,the distance between the screen and slit is $1 \ m$. The wavelength of the light used is $5000 \ \mathring{A}$. The distance between two consecutive dark fringes (in $mm$) is

In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module