Respuesta :
Answer:
2.316e-3 and 3.47e-3
Explanation:
Now, at that angle, we look at the bright spots on the screen.
tan θ = x / L (x is the horizontal distance from the centre of the screen, L is distance to screen)
For small angles, we can approximate that tan θ = sin θ.
nλ / d = x / L so then x = n λ L / d
First bright fringe, n = 1
x = (1) (641*10^-9 m) (3.65 m) / (1.01*10^-3 m) = 2.316e-3
For destructive interference (dark fringes), equation becomes:
x = (n - 0.5) λ L / d
Second dark fringe, n = 1.5
x = 1.5 (641*10^-9 m) (3.65 m) / (1.01*10^-3 m) = 3.47e-3
The distance of the the first bright fringe and the second dark fringe from the central bright fringe 2.316*10^-3 and 3.47*10^-3
What is double slit experiment?
The double-slit experiment is a demonstration that light and matter can display characteristics of both classically defined waves and particles;
moreover, it displays the fundamentally probabilistic nature of quantum mechanical phenomena.
Now, at that angle, we look at the bright spots on the screen.
[tex]tan\theta = \dfrac{x} { L}[/tex]
(x is the horizontal distance from the centre of the screen, L is distance to screen)
For small angles, we can approximate that tan θ = sin θ.
[tex]\dfrac{n\lambda }{ d} = \dfrac{x} { L}[/tex]
so then
[tex]x = \dfrac{n \lambda L} { d}[/tex]
First bright fringe, n = 1
[tex]x = \dfrac{(1) (641\times 10^{-9} ) (3.65 m)} { (1.01\times 10^{-3} ) }= 2.316\times 10^{-3}[/tex]
For destructive interference (dark fringes), equation becomes:
[tex]x = \dfrac{(n - 0.5) \lambda L} { d}[/tex]
Second dark fringe, n = 1.5
[tex]x = \dfrac{1.5 (641\times 10^{-9} ) (3.65 m) }{ (1.01\times 10^{-3} )} = 3.47\times 10^{-3}[/tex]
Hence the distance of the the first bright fringe and the second dark fringe from the central bright fringe 2.316*10^-3 and 3.47*10^-3
To know more about double slit experiment follow
https://brainly.com/question/16392311
