1036_PartUniversity Physics Solution

1036_PartUniversity Physics Solution - Diffraction 36.72....

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Diffraction 36-19 36.72. IDENTIFY: Apply sin 1.22 D λ θ = . SET UP: is small, so sin x R 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 . EXECUTE: (a) 65 17 5 (6.00 10 m)(2.50 10 m) 1.23 10 m 13.1ly 1.22 (1.22)(1.0 10 m) Dx R Δ× × == = ×= × (b) 51 5 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 ×× 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 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 = × = × 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. EVALUATE: The very large diameter of Planet Imager allows it to resolve planet-sized detail at great distances. 36.73. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) From the segment , dy the fraction of the amplitude of 0 E that gets through is 00 sin( ). dy dy Ed E E k x t aa ω ′′ ⎛⎞ =− ⎜⎟ ⎝⎠ (b) The path difference between each little piece is 0 sin ( sin ) sin( ( sin ) ). Edy yk x k D y d E k D y t a θθ This can be rewritten as 0 (sin( )cos( sin ) sin( sin )cos( )). dE kD t ky ky kD t a ωθ + (c) So the total amplitude is given by the integral over the slit of the above. 22 0 (sin( ) cos( sin ) sin( sin ) E E dE dy kD t ky ky a −− + ∫∫ cos( )). kD t But the second term integrates to zero, so we have: [] 2 2 0 0 2 2 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 Ek y E k Dt d y k y E k Dt ak a ka a E E kD t E kD t ka a EE θπ ωω = = − = ). kD t (d) Since 2 sin( (sin )/2) sin( 2) , (sin )/2 2 ka IE II I ka θβ where we have used sin ( ). IE k xt EVALUATE: The same result for () I is obtained as was obtained using phasors. 36.74. IDENTIFY and SET UP: Follow the steps specified in the problem. EXECUTE: (a) Each source can be thought of as a traveling wave evaluated at xR = 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. φπ = which is just 2, π the maximum phase, scaled by whatever fraction the path difference, sin , d is of the wavelength, . By adding up the contributions from each source (including the accumulating phase difference) this gives the expression provided. (b) () cos( ) sin( ). ikR t n ek R t n i k R t n ωφ φ −+ + + + The real part is just cos . kR t n So, 11 Re e cos( ). NN nn k R t n ⎡⎤ + ⎢⎥ ⎣⎦ ∑∑ (Note: Re means “the real part of . . . .”). But this is just 0 0 cos( ) cos( ) cos( 2 ) cos( ( 1) ) R t R t R t R t N +− + + + + "
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36-20 Chapter 36 (c) 11 1 () ( ) 00 0 0 ee e e e e . NN N ikR t n i t i kR i n t i n nn n EE E ω φ −− −+ + == = ∑∑ 1 e( e ) . N in i n φφ ∞− = But recall 1 0 1 1 N N n n x x x = = .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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1036_PartUniversity Physics Solution - Diffraction 36.72....

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