problem36_67 - 36.67: a)  d sin θ = mλ. Place 1st...

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Unformatted text preview: 36.67: a)  d sin θ = mλ. Place 1st maximum at ∞ or θ = 90 . d = λ. If d < λ, this puts the first maximum " beyond ∞." Thus, if d < λ, there is only a single principal maximum. d sin θ . This just scales 2π radians by the fraction the wavelength is b)  Φ path = 2π λ of the path difference between adjacent sources. If we add a relative phase δ between sources, we still must maintain a total phase difference of zero to keep our principal maximum. 2πd sin θ δλ Φ path ± δ = 0 ⇒ = ±δ or θ = sin −1 λ 2πd 0.280 m = 0.0200 m (count the number of spaces between 15 points). c)  d = 14 Let θ = 45. Also recall fλ = c, so δ max = ± 2π (0.0200 m)(8.800 × 10 9 Hz) sin 45o = ±2.61 radians. (3.00 × 10 8 m s) ...
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This note was uploaded on 01/24/2010 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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