problem3 - 600 = 4 d(1 50 450-2(18 d = 600nm(19 Therefore d...

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Problem 3 ray 1 ray 2 d n1 = 1.00 n2 = 1.50 Figure 1: Interference with plastic film Since ray of light passes through a plastic film twice Δ x = 2 d (1) Where d is the thickness of the plastic film. So extra time ray 2 travels respect to ray 1 is. Δ t = Δ x v = 2 dn 2 c (2) Therefore phase difference due to path difference is Δ φ t = ω Δ t = ω 2 dn 2 c (3) There is also phase change from reflection for ray 1. Δ φ total = Δ φ t - Δ φ r (4) Δ φ total = ω 2 dn 2 c - π (5) Δ φ total = 4 πdn 2 λ - π (6) To have minima, Δ φ total has to be odd multiple of π . So we have 2 equations m 360 π = 4 πdn 2 360 - π (7) m 450 π = 4 πdn 2 450 - π (8) and m 600 π = 4 πdn 2 600 - π (9) 1
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also we have condition m 450 = m 360 - 2 (10) 4 πdn 2 450 - π = 4 πdn 2 360 - π - 2 π (11) 4 πdn 2 450 = 4 πdn 2 360 - 2 π (12) 4 d (1 . 50) 450 = 4 d (1 . 50) 360 - 2 (13) d = 600nm (14) also relationship between 450 and 600 yield same result m 600 = m 450 - 2 (15) 4 πdn 2 600 - π = 4 πdn 2 450 - π - 2 π (16) 4 πdn 2 600 = 4 πdn 2 450 - 2 π (17) 4 d (1 . 50)
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Unformatted text preview: 600 = 4 d (1 . 50) 450-2 (18) d = 600nm (19) Therefore d = 600nm (20) I get m 360 = 9 , m 450 = 7 and m 600 = 5 . To get bright spot, I can ask when is light constructively interfering? Since we want to look for visible spectrum, its safe to pick m = 6 (since its between 450 and 600). Then 6 = 4 600(1 . 50) - (21) 7 = 4(600)(1 . 50) (22) = 514 (23) Thus = 514nm (24) Other m might work, but its getting close to edge of visible spectrum. Last one requires no calculation since n 3 = 1 . 6 would only cause extra phase shift of to ray 2. Which means old maxima is new minima, and old minima is now new maxima. 2...
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problem3 - 600 = 4 d(1 50 450-2(18 d = 600nm(19 Therefore d...

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