From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of
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Answer:
the wavelength λ of the light when it is traveling in air = 560 nm
the smallest thickness t of the air film = 140 nm
Explanation:
From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)
Now for constructive interference;
Δx=
replacing ;
Δx = 2t ; we have:
2t =
Given that thickness t = 700 nm
Then
2× 700 = --- equation (1)
For thickness t = 980 nm that is next to constructive interference
2× 980 = ----- equation (2)
Equating the difference of equation (2) and equation (1); we have:'
λ = (2 × 980) - ( 2× 700 )
λ = 1960 - 1400
λ = 560 nm
Thus; the wavelength λ of the light when it is traveling in air = 560 nm
b)
For the smallest thickness
∴
Thus, the smallest thickness t of the air film = 140 nm