Respuesta :
Answer:
3.17 mm
Explanation:
Given from the question
wavelength 1 λ₁= 660 nm = 6.6 x 10^-7 m
wavelength 2 λ₂= 470 nm = 4.7 x 10^-7 m
Distance d = 0.300 mm = 3.0 x 10^-4 m
interference L = 5.00 m
use young's slit formula :
y = k (λ L)/d . . . . . (k = order of bright fringe)
for : λ₁= 6.6 x 10^-7 m (the first-order k = 1)
y₁= 1 {(6.6 x 10^-7) (5.00)} / 3.0 x 10^-4
y₁= 11.0 x 10^-3 m = 11 mm
for : λ₂= 4.7 x 10^-7 m . . . . . . (the first-order, k = 1)
y₂= 1 {(4.7 x 10^-7 ) (5.00)} / 3.0 x 10^-4
y₂= 7.83 x 10^-3 m = 7.83 mm
so, the distance on the screen between the first-order bright fringe for each wavelength Is given by
∆y = y₁- y₂
∆y = 11mm - 7.83mm = 3.17 mm
Answer:
Distance on the screen between thefirst-order bright fringes for the two wavelengths = 0.0032 m
Explanation:
The formula for a first order bright fringe is given by:
[tex]y_{n} = R(n \lambda /d)[/tex]
ym=R(mλ/d)
Distance between the screen and the slits, R = 5 m
Separation of the two narrow slits, d = 0.300 mm = 0.0003 m
The fringe is a first order bright fringe:
Order of the bright fringe, n = 1
Wavelength for the red light, λ =660 nm = 660*10⁻⁹ m
Wavelength for the blue light, λ =470 nm =470*10⁻⁹ m
For the first order bright fringe for red light:
[tex]y_{r} = 5(1 * (660 * 10^{-9} /(3 * 10^{-4} )\\y_{r} = 0.011 m[/tex]
For the first order bright fringe for blue light:
[tex]y_{b} = 5(1 * (470 * 10^{-9} /(3 * 10^{-4} )\\y_{b} = 0.0078 m[/tex]
Distance between the first order bright fringes for the two wavelengths
[tex]y_{n} = y_{r} - y_{b} \\y_{n} = 0.011-0.0078\\y_{n} = 0.0032[/tex]