A double slit apparatus is held 1.2 m from a screen.
(a) When red light (λ = 600 nm) is sent through the double slit, the interference pattern on the screen shows a distance of 12.5 cm between the first and tenth dark fringes. What is the separation of the slits?
(b) What will be the difference in path length for the waves travelling from each slit to the tenth nodal line?
(a) We can use the formula for the distance between consecutive dark fringes in a double-slit interference pattern:
d sinθ = mλ
where d is the slit separation, θ is the angle between the central axis and the location of the mth dark fringe, λ is the wavelength of the light, and m is an integer that represents the order of the fringe.
For the first dark fringe (m = 1), sinθ ≈ θ = 12.5 cm / 1.2 m = 0.1042. We can assume that this angle is small and use the approximation sinθ ≈ θ. Substituting this into the formula gives:
d × 0.1042 ≈ 1 × 600 nm
d ≈ 6.09 μm
Therefore, the separation of the slits is approximately 6.09 μm.
(b) The path difference between the waves from the two slits at the tenth nodal line can be found using the formula:
Δl = d sinθ
where d is the slit separation and θ is the angle between the central axis and the location of the tenth nodal line. From part (a), we know that d ≈ 6.09 μm. For the tenth nodal line (m = 10), sinθ ≈ θ = (10 × λ) / d = (10 × 600 nm) / 6.09 μm = 0.98. Substituting these values into the formula gives:
Δl ≈ 6.09 μm × 0.98 ≈ 5.98 μm
Therefore, the difference in path length for the waves travelling from each slit to the tenth nodal line is approximately 5.98 μm.