To solve both questions, we need to use the equation for the position of the bright fringes in a double-slit experiment:
d sinθ = mλ,
where d is the slit separation distance, θ is the angle between the central axis and the line connecting the bright fringe to the center of the double-slit, m is the order of the bright fringe, and λ is the wavelength of the light.
The distance between the central bright fringe and the 3rd dark fringe can be found by first calculating the position of the 3rd bright fringe, and then subtracting the position of the central bright fringe:
For the central bright fringe, m=0, so we have:
d sinθ = 0
sinθ = 0
θ = 0
For the 3rd bright fringe, m=3, so we have:
d sinθ = 3λ
sinθ = 3λ/d
θ = sin^(-1)(3λ/d)
The distance between the central bright fringe and the 3rd dark fringe is the distance between the central axis and the line connecting the 3rd bright fringe to the center of the double-slit. This distance is simply the distance from the double-slit to the viewing screen times the tangent of the angle θ:
distance = L tanθ
where L is the distance from the double-slit to the viewing screen.
Plugging in the values given in the problem, we get:
θ = sin^(-1)(3(600 nm)/(3.55 μm)) = 0.317 radians
distance = (1.50 m) tan(0.317) = 0.816 m
Therefore, the distance from the central bright fringe to the 3rd dark fringe is 0.816 meters.
The distance between the 2nd and 3rd dark fringes can be found by calculating the position of the 2nd and 3rd dark fringes, and then subtracting the two positions:
For the central bright fringe, m=0, so we have:
d sinθ = 0
sinθ = 0
θ = 0
For the 2nd dark fringe, m=1, so we have:
d sinθ = λ
sinθ = λ/d
θ = sin^(-1)(λ/d)
For the 3rd dark fringe, m=2, so we have:
d sinθ = 2λ
sinθ = 2λ/d
θ = sin^(-1)(2λ/d)
The distance between the 2nd and 3rd dark fringes is the distance between the lines connecting the two fringes to the central axis. This distance is simply the distance from the double-slit to the viewing screen times the difference between the tangents of the angles θ for the two fringes:
distance = L (tanθ2 - tanθ1)
where L is the distance from the double-slit to the viewing screen, θ1 is the angle for the 2nd dark fringe, and θ2 is the angle for the 3rd dark fringe.
Plugging in the values given in the problem, we get:
θ1 = sin^(-1)(600 nm/3.55 μm) = 0.170 radians
θ2 = sin^(-1)(2(600 nm)/3.55 μm) = 0.332 radians
distance = (1.50 m) (tan(0.332) - tan(0.170)) = 0.207 m = 20.7 cm
Therefore, the distance between the 2nd and 3rd dark fringes is 20.7 cm.