What is the Wavelength of Light Used in a Double-Slit Interference Experiment?

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SUMMARY

The wavelength of light used in Thomas Young's double-slit experiment is determined to be 6.1 x 10-7 m through three distinct calculations. The parameters include an angle to the eighth maximum of 1.12 degrees, a distance from the slits to the screen of 3.02 m, and a distance between the slits of 0.00025 m. The calculations utilize the formulas Sinθ = 8λ/d, Δx = λL/d, and λ = Δxd / 4L, all yielding the same wavelength result. The mentor notes that the second and third methods are fundamentally similar in approach.

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student07
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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]
 
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2 and 3 look very similar, but there is no fundamentally different thing you can do.
 
thank you
 

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