Answer:
69.7 cm
Explanation:
What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.
Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m
Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ
So, d/mλ = x/D for a second order bright fringe, m = 2.
So, d/2λ = x/D
x = dD/2λ
So, x =
For a dark fringe, we have
d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.
For a second-order dark fringe, m = 2. So,
d/(2 + 1/2)λ = x'/D
d/(5/2)λ = x'/D
2d/5λ = x'/D
x' = 2dD/5λ
So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ
x" = dD/10λ
Substituting the values of the variables into the equation, we have
x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)
x" = 1.15/165 × 10² m
x" = 0.00697 × 10² m
x" = 0.697 m
x" = 69.7 cm
