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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 = ?f (25-4) v = c / n (26-10) My versions for Section 28-1: destructive ?l = (2m+1)?/2 constructive ?l = m? d sin? = m? , m = 0, ±1, ±2, ... (28-1 ) Energy conversion: 1.0 eV = 1.6×10 -19 J d sin? = (m ? ½)? , m = 1, 2, 3,... and d sin? = (m + ½)? , m = -1, -2, -3,... (28-2) My versions (no ? signs): d sin? = m? , and d sin? = (m + ½)? , m = 0, 1, 2, 3,... in either case y = L tan? (28-3) ? n = ?/n (28-4) ½ + 2d/? = m (air, constructive) (28-5) ½ + 2d/? = m + ½ (air, destructive) (28-6) l eff,1 / ? = ½ (28-7) l eff,2 /? n = 2t /? n = 2nt /? vacuum (28-8) difference in phase changes = 2nt /? vacuum ? ½ (28-9) My version: 2t = m?/n , bright with no reflection effect, dark with reflection effect. 2nt /? vacuum = m, m = 0, 1, 2, ...(destr) (28-10) 2nt /? vacuum ?½ = m, m = 0, 1, 2, ...(const) (28-11) W sin? = m? (28-12) Approx angular width central fringe = 2?/W (28-13) (My version: a sin? = m?) sin ? = 1.22 ? / D (28-14) ? min = 1.22 ? / D (28-15) d sin? = m? , m = 0, ±1, ±2, ... (diff grating) (28-16) My version: d sin? = m? , m = 0, 1, 2, ... ?t 0 = 2d /c (29-1) ?t = ?t 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 ?s (meters): ?s 2 = [?x 2 +?y 2 +?z 2 ?c 2 ?t 2 ] gamma ? = 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 / ? (30-11) hf = hf? + K (30-12) h /? = (h /??) cos ? + p e cos ? (30-13) 0 = (h /??) sin ? ? p e sin ? (30-14) ?? = ?? ? ? = [h / m e c] (1 ?cos ?) (30-15) ? = h / p (30-16) sin ? = ? / W (30-18) 2d sin ? = m? , m = 1, 2, 3,... (30-17) ?p y ?y ? h /(2?) (30-19) ?E ?t ? h /(2?) (30-20) pedigo Numbered Equations from Walker, Physics, 3rd edition, as modified by R

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