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1047_Physics ProblemsTechnical Physics

# 1047_Physics ProblemsTechnical Physics - 388 P37.19...

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388 Interference of Light Waves P37.19 (a) From Equation 37.8, φ π λ θ π λ φ π λ π = = + = × × × = 2 2 2 2 0 850 10 2 50 10 600 10 2 80 7 95 2 2 3 3 9 d d y y D yd D sin . . . . m m m m rad e je j e j a f (b) I I d d m max max cos sin cos sin cos cos = = 2 2 2 2 2 π λ θ π λ θ φ π b g b g b g I I max cos cos . . = = F H G I K J = 2 2 2 7 95 0 453 φ rad 2 P37.20 (a) The resultant amplitude is E E t E t E t r = + + + + 0 0 0 2 sin sin ω ω φ ω φ b g b g , where φ π λ θ = 2 d sin . E E t t t t t E E t E t E E t t E t r r r = + + + + = + + + + = + + = + + 0 0 2 0 0 0 2 2 1 2 1 2 1 2 1 2 sin sin cos cos sin sin cos cos sin sin cos cos cos sin sin cos cos sin cos cos sin cos sin ω ω φ ω φ ω φ ω φ ω φ φ ω φ φ φ φ ω φ ω φ φ ω φ b g b g e j b gb g b gb g b g b g Then the intensity is I E E r = + F H G I K J 2 0 2 2 1 2 1 2 cos φ b g where the time average of sin 2 ω φ t + b g is 1 2 . From one slit alone we would get intensity I E max F H G I K J 0 2 1 2 so I I d = + F H G I K J L N M O Q P max cos
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