1049_Physics ProblemsTechnical Physics

# 1049_Physics ProblemsTechnical Physics - 390 Interference...

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390 Interference of Light Waves (c) E i i j i j R E = + °+ ° + °+ ° 0 120 120 240 240 ± ± cos ± sin ± cos ± sin e j e j E i j R P E E = + = = 0 0 0 0 0 ± ± E 1 E 2 E 3 y x FIG. P37.22(c) (d) E i i j i j R E = + + F H G I K J + + L N M O Q P 0 3 2 3 2 3 3 ± ± cos ± sin ± cos ± sin π π π π e j E i j R P E E E E E t = = °= = + F H G I K J 0 0 0 0 0 1 00 3 2 3 2 ± . ± sin at 270 at rad π ω π E 1 E 2 E 3 y x FIG. P37.22(d) P37.23 E i j R = + = + F H G I K J 6 00 8 00 6 00 8 00 8 00 6 00 2 2 1 . ± . ± . . tan . . a f a f at E R P E t = °= = + 10 0 53 1 10 0 0 927 10 0 100 0 927 . . . . . sin . at at rad π b g 6.00 8.00 y x E R π 2 α FIG. P37.23 P37.24 If E E t 1 01 = sin ω and E E t 2 02 = + sin ω φ b g , then by phasor addition, the amplitude of E is E E E E E E E E 0 01 02 2 02 2 01 2 01 02 02 2 2 = + + = + + cos sin cos φ φ φ b g b g and the phase angle is found from sin sin θ φ = E E 02 0 . E 01 E 02 E 0 x y θ φ FIG. P37.24 P37.25 E i i j R = + ° + ° 12 0 18 0 60 0 18 0 60 0 . ± . cos . ± . sin . ± e j E i j R R E t = + = ° = + ° 21 0 15 6 26 2 26 2 36 6 . ± . ± . . sin . at 36.6 ω
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