Monochromatic light (500 nm) passes through two slits and onto a screen 2.5 m away. The second-order fringes are located 12 mm away from the central fringe. What is the distance between the two slits?
Question
Monochromatic light (500 nm) passes through two slits and onto a screen 2.5 m away. The second-order fringes are located 12 mm away from the central fringe. What is the distance between the two slits?
Solution
To solve this problem, we can use the formula for double-slit interference:
dsin(θ) = mλ
where:
- d is the distance between the two slits,
- θ is the angle between the central maximum and the mth order fringe,
- m is the order of the fringe, and
- λ is the wavelength of the light.
First, we need to convert the given values to the correct units. The wavelength λ is given as 500 nm, which is 500 x 10^-9 m. The distance from the slits to the screen is 2.5 m, and the distance from the central fringe to the second-order fringe is 12 mm, which is 0.012 m.
We can find the angle θ using the small angle approximation tan(θ) ≈ sin(θ) ≈ y/L, where y is the distance from the central fringe to the mth order fringe, and L is the distance from the slits to the screen. So, θ = y/L = 0.012 m / 2.5 m = 0.0048 rad.
Substituting the known values into the formula, we get:
dsin(θ) = mλ dsin(0.0048 rad) = 2(500 x 10^-9 m) d = 2*(500 x 10^-9 m) / sin(0.0048 rad)
Solving for d, we find that the distance between the two slits is approximately 0.000208 m, or 0.208 mm.
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