- #1
student07
- 36
- 1
Upon using Thomas Young's double-slit experiment to obtain measurements, the following data were obtained. Use the data to determine the wavelength of light being used to create the interference pattern. Do this in three different ways.
- The angle to the eighth maximum is 1.12 deg.
- The distance from the slits to the screen is 302 cm.
- The distance from the first minimum to the fifth minimum is 2.95 cm.
- The distance between the slits is 0.00025 m.
L = 302 cm = 3.02 m, 4 Δx=2.95 cm, then = 0.74 cm = 0.74 x 10^-2 m, d = 0.00025 m
Attempt at the solution:
1. Sinθ = 8λ/d, then Sinθ d /8=λ = (sin 1.12)(0.00025m) / 8 = 6.1 x 10^-7 m
2. Δx =λL/d, then λ= Δxd/L = (0.74 x 10^-2 m)(0.00025 m) / 3.02 m = 6.1 x 10^-7 m
3. λ = Δxd / 4L =(2.95 x 10^-2 m)(0.00025 m) / 4(3.02 m) = 6.1 x 10^-7 m
Im not sure about the last one, please revise thank you!
[Mentor's Note: Moved from General Physics]
- The angle to the eighth maximum is 1.12 deg.
- The distance from the slits to the screen is 302 cm.
- The distance from the first minimum to the fifth minimum is 2.95 cm.
- The distance between the slits is 0.00025 m.
L = 302 cm = 3.02 m, 4 Δx=2.95 cm, then = 0.74 cm = 0.74 x 10^-2 m, d = 0.00025 m
Attempt at the solution:
1. Sinθ = 8λ/d, then Sinθ d /8=λ = (sin 1.12)(0.00025m) / 8 = 6.1 x 10^-7 m
2. Δx =λL/d, then λ= Δxd/L = (0.74 x 10^-2 m)(0.00025 m) / 3.02 m = 6.1 x 10^-7 m
3. λ = Δxd / 4L =(2.95 x 10^-2 m)(0.00025 m) / 4(3.02 m) = 6.1 x 10^-7 m
Im not sure about the last one, please revise thank you!
[Mentor's Note: Moved from General Physics]
Last edited by a moderator:
