Lista Interferência - Resolvida física 3-d

1) λ = 389.5 nm ΔC = λf ΔC = f λ f0 = 3.103 0.850.10 H z 5.003.10 H z λ= 389.5 nm m= 1,52 m = C s / c t C1 = λf0 c1 = C s / C1 C1 = C s 3.10 15 5,1.10 10 C = λf C = λf 399,5.15.10 5,1.10 10 C1 = 1,57.10 mm 2) ΔN = L (m2 - m1) λ L = ΔN. λ (m2 - m1) ans: N de complementos de onda x = 0.9 com 5.6 mmed = 13 med = 18 de onda 2nd = 1, 3 I = I0 - I0 = I0 (5,66) I = 0,903 I0 I = I0 (0,37) 3) a) ΔN = 3.5.10 - 6 (1,5 - 1,6) 1,7 (continua de ondas) d) ΔN = 3.9.10 - 6 (1,62-1,72) = 1,7 λ= 2.7 1.7 = x x = 10^(-7) c) ΔN = 3,25.10 - 6 (1,70-1,69) = 1.3 ΔN = - 6.23.10 1 = 12.5 med I0 = 0.34 3) θ = 5.5.*10 - 7 ( 1.5 - 16) 5.10 = 0.216 med 12,59 m = 5.10 4 d) θ = 0.5 = λ ΔN = L = 0.5.6.12 3 ( 1.6-1.65) 1,55*10 λ = 0.5.6 | | 5.5.10 θ = mλ m = 0.5 12 θ = p 1 = esp= 5.5.1 =.2 (ESP / => θ = ∆ 1,5.10 - 6 0,12D0 h g = D - D x/d = l θ = z 3.5.1 3,85.10 3 = 3,85 5.05 I= 64 T eθ = 10(10) = 13) 9) δ = 0.681 I = 5.5.0 I δ / x = 100* = 3.5.103 med 5.k=A = 2k.5 Au k m = n θ= = 12 0.01 med 2.5 m = 8.33= 38.5 (36 = I ) 0.05. 10.5 = 13 ΔCo = Y - x = λ/3 TOTAL CONSTRUTIVA 3 MÁXIMO \n dt + x = (x + mλ)² \n x = λ(c² - m²)² \n (2mλ)²: \n y / m = 1, 2, 3. \n\n (16) \n m2 = m1: \n 𝒟𝑡 = λ/2 \n 2L m2 = m1λ \n 2L m2 = (m1 / 2)? \n \n 𝒟n2 = 0 \n 2.121.10⁶.133 = \n 5.85.10⁷ \n\n (17) \n L = (m + 1/2) λ. m0 = m1 = 2 λ = 117nm \n L = Lm: 351.5mm \n L = 1.17.10⁻⁹ \n L = 3/2 \n 6.26.10⁻²: \n 2.173 \n = 3.51.8.10⁹ \n\n (18) \n 𝒟𝑡 = 𝒟𝑡1 + 𝒟𝑡2 + 𝒟𝑡3 \n PRECISA SEM NÚMERO DECIMAL \n = λ/3 CONSUMO \n L = (m + 1/2) λ / 2m \n L = λ.1 \n L = λ/5 \n L = \n\n (19) \n L = 0.410-10 \n m3 = 1.5 \n 𝑚3 = 0.1; 1.0 \n λ0 = 2460mm N VISÍVEL \n VISÍVEL