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1036_PartUniversity Physics Solution

1036_PartUniversity Physics Solution - Diffraction 36.72...

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Diffraction 36-19 36.72. I DENTIFY : Apply sin 1.22 D λ θ = . S ET U P : θ is small, so sin x R θ Δ , where x Δ is the size of the details and R is the distance to the earth. 15 1 ly 9.41 10 m = × . E XECUTE : (a) 6 5 17 5 (6.00 10 m)(2.50 10 m) 1.23 10 m 13.1ly 1.22 (1.22)(1.0 10 m) D x R λ Δ × × = = = × = × (b) 5 15 8 1.22 (1.22)(1.0 10 m)(4.22 ly)(9.41 10 m ly) 4.84 10 km 1.0 m R x D λ × × Δ = = = × . This is about 10,000 times the diameter of the earth! Not enough resolution to see an earth-like planet! x Δ is about 3 times the distance from the earth to the sun. (c) 5 15 6 6 (1.22)(1.0 10 m)(59 ly)(9.41 10 m ly) 1.13 10 m 1130 km. 6.00 10 m x × × Δ = = × = × 3 5 planet 1130 km 8.19 10 ; 1.38 10 km x x D Δ = = × Δ × is small compared to the size of the planet. E VALUATE : The very large diameter of Planet Imager allows it to resolve planet-sized detail at great distances. 36.73. I DENTIFY and S ET U P : Follow the steps specified in the problem. E XECUTE : (a) From the segment , dy the fraction of the amplitude of 0 E that gets through is 0 0 sin( ). dy dy E dE E kx t a a ω = (b) The path difference between each little piece is 0 sin ( sin ) sin( ( sin ) ). E dy y kx k D y dE k D y t a θ θ θ ω = = This can be rewritten as 0 (sin( )cos( sin ) sin( sin )cos( )). E dy dE kD t ky ky kD t a ω θ θ ω = + (c) So the total amplitude is given by the integral over the slit of the above. 2 2 0 2 2 (sin( ) cos( sin ) sin( sin ) a a a a E E dE dy kD t ky ky a ω θ θ = = + cos( )). kD t ω But the second term integrates to zero, so we have: [ ] [ ] 2 2 0 0 2 2 0 0 0 sin( sin ) sin( ) (cos( sin )) sin ( ) sin 2 sin( (sin ) 2) sin( (sin ) ) sin( ) sin( ) . (sin ) 2 (sin ) sin ... At 0, 1 sin( ... a a a a E ky E kD t dy ky E kD t a ka ka a E E kD t E kD t ka a E E θ ω θ ω θ θ π θ λ ω ω θ π θ λ θ = = = = = = = ). kD t ω (d) Since 2 2 2 0 0 sin( (sin )/2) sin( 2) , (sin )/2 2 ka I E I I I ka θ β θ β = = where we have used 2 2 0 0 sin ( ). I E kx t ω = E VALUATE : The same result for ( ) I θ is obtained as was obtained using phasors. 36.74. I DENTIFY and S ET U P : Follow the steps specified in the problem. E XECUTE : (a) Each source can be thought of as a traveling wave evaluated at x R = with a maximum amplitude of 0 . E However, each successive source will pick up an extra phase from its respective pathlength to point sin 2 d P . θ φ π λ = which is just 2 , π the maximum phase, scaled by whatever fraction the path difference, sin , d θ is of the wavelength, λ . By adding up the contributions from each source (including the accumulating phase difference) this gives the expression provided. (b) ( ) cos( ) sin( ). i kR t n e kR t n i kR t n ω φ ω φ ω φ + = + + + The real part is just cos ( ). kR t n ω φ + So, 1 1 ( ) 0 0 0 0 Re e cos( ).
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