Lecture%2040%202220%20Spring%202009

Lecture%2040%202220%20Spring%202009 - Reminders and...

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1 Reminders and Announcements • HW 22 is due today ; HW 23 is due next Tuesday; HW 24 will be optional (extra credit) and will be due a week from Friday • Help Room Hours (Physics 117): – M 4-8 pm, T 3-6:30, W 3-6:30, R 4:30-8, F 2-6 Young’s Double-Slit Experiment: Schematic • Thomas Young first demonstrated interference in light waves from two sources in 1801 • The narrow slits S 1 and S 2 act as sources of coherent waves Resulting Interference Pattern • The light from the two slits forms a visible pattern on a screen The pattern consists of a series The pattern consists of a series of bright and dark parallel bands called fringes Constructive interference occurs where a bright fringe occurs Destructive interference results in a dark fringe Young’s Double-Slit Experiment: Geometry • The path difference, δ , is found from the tan triangle δ = r 2 r 1 = d sin θ – This assumes the paths are parallel – Not exactly true, but a very good approximation if L is much greater than d Uses for Young’s Double-Slit Experiment • Young’s double-slit experiment provides a method for measuring wavelength of the light • This experiment gave the wave model of light a great deal of credibility – It was inconceivable that particles of light could cancel each other in a way that would explain the dark fringes Intensity Distribution: Double- Slit Interference Pattern • Assumptions: – The two slits represent coherent sources of sinusoidal waves – The waves from the slits have the same angular frequency, ω – The waves have a constant phase difference, φ • The total magnitude of the electric field at any point on the screen is the superposition of the two waves
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2 Intensity Distribution, Electric Fields • The magnitude of each wave at point P can be found E 1 = E o sin ω t E 2 = E o sin ( ω t + φ ) – Both waves have the same amplitude, E o Intensity Distribution, Phase Relationships • The phase difference between the two waves at P depends on their path difference δ = r 2 r 1 = d sin θ • A path difference of λ corresponds to a phase difference of 2 π rad • A path difference of is the same fraction of λ
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Lecture%2040%202220%20Spring%202009 - Reminders and...

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