ch36-p028 - 28 From Fig 36-44(a we find the diameter D′...

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Unformatted text preview: 28. From Fig. 36-44(a), we find the diameter D′ on the retina to be D′ = D L′ 2.00 cm = (2.00 mm) = 0.0889 mm . L 45.0 cm Next, using Fig. 36-44(b), the angle from the axis is ⎛ D′ / 2 ⎞ −1 ⎛ 0.0889 mm / 2 ⎞ ⎟ = tan ⎜ ⎟ = 0.424° . ⎝x⎠ ⎝ 6.00 mm ⎠ θ = tan −1 ⎜ Since the angle corresponds to the first minimum in the diffraction pattern, we have sin θ = 1.22λ / d , where λ is the wavelength and d is the diameter of the defect. With λ = 550 nm, we obtain d= 1.22λ 1.22(550 nm) = = 9.06 × 10−5 m ≈ 91µ m . sin θ sin(0.424°) ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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