Daily_004.SOLN - 2011 E. Szarmes PHYS 274 DAILY ASSIGNMENT...

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© 2011 E. Szarmes PHYS 274 DAILY ASSIGNMENT 4 — DUE WEDNESDAY, JAN 26/11 PROF. SZARMES Q2S.2 D = 5.0 km f = 100 MHz λ = c/f = 3(10 8 )m/s/100(10 6 ) Hz = 3.0 m Note that radio waves travel at the speed of light, not the speed of sound! Now, for two antennas emitting in phase, the angle ϑ n of the n th maximum in the interference pattern is where y n is the transverse position of the n th maximum at distance D. Thus, the separation between two adjacent maxima (in the region near the axis where ϑ is small), is For λ = 3 m (i.e. f = 100 MHz), D = 5.0 km, and d = 60 m, we calculate Δ y = 250 m. Q2S.8 The criterion for resolving the headlights (separated by d = 1.4 m at distance D) is where α ~ d / D and ϑ 1 (not shown in the diagram) is the half angle of the diffraction patterns. We assume that the diameter of the eye’s pupil is a = 8 mm and that the average wavelength of the light is λ = 550 nm. Then
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Daily_004.SOLN - 2011 E. Szarmes PHYS 274 DAILY ASSIGNMENT...

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