Two slits separated by a distance of 1 ,mm ar | vedclass

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

Question

(B) Given: Distance between slits $d = 1 \,mm = 10^{-3} \,m$, wavelength $\lambda = 6.5 	imes 10^{-7} \,m$, distance of screen $D = 1 \,m$.For the $n

Vedclass · Accepted Answer

$1.625$

Vedclass · Answer

(B) Given: Distance between slits $d = 1 \,mm = 10^{-3} \,m$, wavelength $\lambda = 6.5 	imes 10^{-7} \,m$, distance of screen $D = 1 \,m$.For the $n^{th}$ dark fringe, the position is given by $y_n = (n - 0.5) \frac{\lambda D}{d}$.For the $3^{rd}$ dark fringe $(n=3)$: $y_3 = (3 - 0.5) \frac{6.5 	imes 10^{-7} 	imes 1}{10^{-3}} = 2.5 	imes 6.5 	imes 10^{-4} = 16.25 	imes 10^{-4} \,m$.For the $n^{th}$ bright fringe, the position is given by $y_n = \frac{n \lambda D}{d}$.For the $5^{th}$ bright fringe $(n=5)$: $y_5 = \frac{5 	imes 6.5 	imes 10^{-7} 	imes 1}{10^{-3}} = 32.5 	imes 10^{-4} \,m$.The distance between the $3^{rd}$ dark fringe and the $5^{th}$ bright fringe is $\Delta y = y_5 - y_3 = 32.5 	imes 10^{-4} - 16.25 	imes 10^{-4} = 16.25 	imes 10^{-4} \,m = 1.625 \,mm$.

Two slits separated by a distance of $1 \,mm$ are illuminated with light of wavelength $6.5 \times 10^{-7} \,m$. The interference fringes are observed on a screen placed at $1 \,m$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (in $\,mm$)

Similar Questions

In a Young's double slit experiment,the source illuminating the slits is changed from blue to violet. The width of the fringes

Assertion: In Young's double slit experiment,if the wavelength of incident monochromatic light is doubled,the number of bright fringes on the screen will increase.
Reason: The maximum number of bright fringes on the screen is directly proportional to the wavelength of light used.

In Young's double-slit experiment using monochromatic light,the shape of the interference pattern on the screen is .....

In a Young's double-slit experiment,the fringe width is $0.8 \, mm$ when light of wavelength $6000 \, \mathring{A}$ is used,and the screen is at a distance of $2.5 \, m$. If the experiment is performed in a liquid with a refractive index of $1.6$,the new fringe width will be.....$mm$.

In Young's double-slit experiment,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

Vedclass Test Series

Exam Paper Generator

Online Exam Module