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- StudyBlue
- Washington
- University of Washington - Seattle Campus
- Physics
- Physics 116
- Pedigo
- Exam3eqns.pdf

Vladimir S.

Numbered Equations from Walker, Physics, 3 rd edition, as modified by R. D. Pedigo Vector symbols in bold (no arrows above these), equation numbers also in bold Misc old equations: c = 3.0×10 8 m/s (25-1) c = λf (25-4) v = c / n (26-10) My versions for Section 28-1: destructive Δl = (2m+1)λ/2 constructive Δl = mλ d sinθ = mλ , m = 0, ±1, ±2, ... (28-1 ) Energy conversion: 1.0 eV = 1.6×10 -19 J d sinθ = (m – ½)λ , m = 1, 2, 3,... and d sinθ = (m + ½)λ , m = -1, -2, -3,... (28-2) My versions (no − signs): d sinθ = mλ , and d sinθ = (m + ½)λ , m = 0, 1, 2, 3,... in either case y = L tanθ (28-3) λ n = λ/n (28-4) ½ + 2d/λ = m (air, constructive) (28-5) ½ + 2d/λ = m + ½ (air, destructive) (28-6) l eff,1 / λ = ½ (28-7) l eff,2 /λ n = 2t /λ n = 2nt /λ vacuum (28-8) difference in phase changes = 2nt /λ vacuum − ½ (28-9) My version: 2t = mλ/n , bright with no reflection effect, dark with reflection effect. 2nt /λ vacuum = m, m = 0, 1, 2, ...(destr) (28-10) 2nt /λ vacuum −½ = m, m = 0, 1, 2, ...(const) (28-11) W sinθ = mλ (28-12) Approx angular width central fringe = 2λ/W (28-13) (My version: a sinθ = mλ) sin θ = 1.22 λ / D (28-14) θ min = 1.22 λ / D (28-15) d sinθ = mλ , m = 0, ±1, ±2, ... (diff grating) (28-16) My version: d sinθ = mλ , m = 0, 1, 2, ... Δt 0 = 2d /c (29-1) Δt = Δt 0 / [1 − v 2 /c 2 ] ½ (29-2) L = L 0 [1 − v 2 /c 2 ] ½ (29-3) v 23 = (v 21 + v 13 ) / [1 + (v 21 v 13 / c 2 )] (29-4) p = mv / [1 − v 2 /c 2 ] ½ (29-5) m = m 0 / [1 − v 2 /c 2 ] ½ (29-6) E = m 0 c 2 / [1 − v 2 /c 2 ] ½ = mc 2 (29-7) E 0 = m 0 c 2 (29-8) K = m 0 c 2 / [1 − v 2 /c 2 ] ½ − m 0 c 2 (29-9) R = 2 G M / c 2 (29-10) G = 6.67×10 -11 (SI units) My additions to Chapter 29: Lorentz transformations for Special Relativity and the metric (matrix of “g” values) for general Relativity (you do not need either of them for the exam). These you might: Invariant 4-D interval Δs (meters): Δs 2 = [Δx 2 +Δy 2 +Δz 2 –c 2 Δt 2 ] gamma γ = 1 / [1 − v 2 /c 2 ] ½ f peak = (5.88×10 10 ) T (SI units, T in Kelvin) (30-1) E n = nhf n = 0, 1, 2, ... (30-2) h = 6.63×10 -34 J•s (30-3) E = hf (30-4) K max = E – W 0 (30-5) f 0 = W 0 / h (30-6) K max = hf – W 0 (30-7) E [1 − v 2 /c 2 ] ½ = m 0 c 2 (30-8) m 0 = 0 (30-9) p [1 − v 2 /c 2 ] ½ = m 0 v (30-10) p = hf / c = h / λ (30-11) hf = hf’ + K (30-12) h /λ = (h /λ’) cos θ + p e cos φ (30-13) 0 = (h /λ’) sin θ − p e sin φ (30-14) Δλ = λ’ − λ = [h / m e c] (1 −cos θ) (30-15) λ = h / p (30-16) sin θ = λ / W (30-18) 2d sin θ = mλ , m = 1, 2, 3,... (30-17) Δp y Δy ≥ h /(2π) (30-19) ΔE Δt ≥ h /(2π) (30-20) pedigo Numbered Equations from Walker, Physics, 3rd edition, as modified by R

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