Lecture22 - Lecture 22: More on Interference Last time we...

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Last time we determined the intensity of a wave that results from the interference of the following two waves: We found that the intensity of the summed wave is: The maximum intensity occurs when cos φ= 1, so we can also write: But it took some algebra and trigonometry to get this answer, so we’d like to have an easier way… I μ ¢ E 2 = 2 E 2 1+ cos f ( ) I I max = 2 E 2 f ( ) 4 E 2 = 1 2 f ( ) = cos 2 f 2 Lecture 22: More on Interference E sin ϖτ ( 29 Εσινϖτ+φ ( 29
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Phasor Addition We can use a geometrical construct to help when adding two waves of any amplitude and phase difference as long as the frequencies are the same! We do this by drawing a phasor for each wave This is an arrow with the following properties: 1. Length of arrow represents amplitude 2. Direction of arrow represents phase Then wave addition can be treated just like vector addition: A 1 sin ϖ t 2 sin( ϖ t φ29 A T sin( ϖ t + ε 29
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We can break each phasor up into its components: From this, we find that: A 1 x = Α 1 Α 2 ξ = Α 2 χοσφ 2 ψ = Α 2 σινφ A T = 1 + 2 ( 29 2 + 2 ( 29 2 τανε = 2 1 + 2
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Three-slit Interference Patterns Using phasors, we can determine what happens when light passes through three equally spaced slits: d θ δ d So now there are three waves being added together at each point on the screen, with phases of 0, 2 πδ/λ , and 4
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The phasor approach tells us that: Using the approximation and the fact that intensity is proportional to E 2
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Lecture22 - Lecture 22: More on Interference Last time we...

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