Unformatted text preview: 28. From Fig. 3644(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. 3644(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.
 Spring '08
 Any
 Physics

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