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handin07

# handin07 - Physics 212 Classical and Modern Physics Spring...

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Physics 212 Classical and Modern Physics Spring 2008 Solutions for Hand-In Set #7 A47 – Resolution I paced off 184 steps to the point where I could just barely resolve the numerals on the clock, so my dis- tance was 184 paces × 0 . 75 m 1 pace + 30 ft × 0 . 304 m 1 ft = 147 m The diameter of my pupil in bright sunlight is about 0 . 3 cm = 3 × 10 - 3 m, so for λ = 500 nm, my eye’s reso- lution is θ = 1 . 22 λ d = 2 . 0 × 10 - 4 rad . For the angular size of the separation between numbers to be the same as my resolution, then tan (2 . 0 × 10 - 4 rad) = x 147 m , where x is the linear separation between the numerals. So, x = 147 tan (1 . 1 × 10 - 4 rad) = 3 × 10 - 2 m = 3 cm. A88 In this two-slit problem, we’re looking for the angle be- tween the centeral maximum and the third side maxi- mum. third side max central max θ At the angle of the third side maximum, Δ φ = 6 π , so Δ r = Δ φ 2 π λ = 3 λ. But this path length difference is related to the angle θ by Δ r = d sin θ where d is the separation between the slits. So, sin θ = 3 λ d = 3(633 × 10 - 9 ) 3 × 10 - 6 = 0 . 633 and θ = 39 . 3 . A90 We can calculate the path lengths for all three waves directly from the geometry of the problem: 3 m 3 m P 5 m 5 m 4 m The path length difference between the waves from the top and middle speakers is Δ r = 1 m, and we get the same value for the path length difference between waves from the middle and bottom speakers. This gives a phase difference of Δ φ = 2 π Δ r λ = π.

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handin07 - Physics 212 Classical and Modern Physics Spring...

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