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Answers to selected problems from Jackson's Classical by van Wijk K.

By van Wijk K.

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21) And ık · nψ(x )|z =0 = ık cos θE0 . 22) So what we are left with is an integral we solved in part a: ψ(x) = = ıkE0 eıkr (1 + cos θ) e−ık·x da 4πr S1 ıE0 eıkr (1 + cos θ) ka sin θ cos φ sin 2 2 πrk sin θ cos φ sin φ sin kb sin θ sin φ 2 . 23) sin2 kb sin θ sin φ 2 . 14. DIFFRACTION FROM A RECTANGULAR OPENING 45 c. A Comparison For the case: b = a, β = π/4, φ → 0 and ka/2 = kb/2 = 2π the power per unit angle for the vector and the scalar approximation is respectively dP dω = × = Using that 2E02 sin2 k 2 π 2 sin θ cos2 φ sin2 φ sin2 θ sin2 (φ − β) + cos2 θ 4 ka sin θ cos φ 2 2E02 sin2 (2π sin θ) sin2 (2π sin θφ) k 2 π 2 sin4 θφ2 sin2 kb sin θ sin φ 2 × sin2 θ + cos2 θ .

14. DIFFRACTION FROM A RECTANGULAR OPENING 45 c. A Comparison For the case: b = a, β = π/4, φ → 0 and ka/2 = kb/2 = 2π the power per unit angle for the vector and the scalar approximation is respectively dP dω = × = Using that 2E02 sin2 k 2 π 2 sin θ cos2 φ sin2 φ sin2 θ sin2 (φ − β) + cos2 θ 4 ka sin θ cos φ 2 2E02 sin2 (2π sin θ) sin2 (2π sin θφ) k 2 π 2 sin4 θφ2 sin2 kb sin θ sin φ 2 × sin2 θ + cos2 θ . 26) lim we get for the vector approximation of the power per unit angle that dP 8E 2 = 2 02 sin2 (2π sin θ) φ→0 dω k sin θ lim 1 cos2 θ + 2 2 = 4E02 sin2 (2π sin θ) 1 + cos2 θ k 2 sin2 θ .

In other words, what is t? 15) 1 . 16) 1 − β2 The twin in the rocket ship feels a force but does not move in its reference frame: x = 0, so v c β= and γ= dt = γdt . 17) I wrote this in differentials because γ is function of velocity and the velocity of the space ship as seen from earth increases with time. We’ll find this velocity via the acceleration a. We know a = g and that a=a x ˆ= 1+ gt = 1− v 1− v2 c2 3 vu x ˆ=g 1− v2 c2 3/2 x ˆ⇔ c2 x ˆ⇔ −3/2 v2 c2 3/2 3/2 dv v2 =g 1− 2 dt c gdt = v2 c2 1− a dv ⇔ .

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