Chp42_5 - = Thus(assuming small angles θ<< 1 we get D 22 1 min ≈ This is only an approximation(the actual resolution is usually worse than

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PHY2061 R. D. Field Department of Physics chp42_5.doc University of Florida Resolving Power The fact that round holes produce a diffraction patter in which the light spreads out on the screen is important when we wish to distinguish ) two distant objects whose angular separation is small. for resolvability states that the minimum angle θ min such that two objects can be distinguished as two separate objects occurs when the center of the central bright spot of object 1 is located at the first diffraction minimum of object 2 (or visa-versa). For angles less that θ min the two objects will appear as one. For round hole diffraction the first diffraction minimum of object 2 occurs at D λ θ 22 . 1 sin
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Unformatted text preview: = . Thus (assuming small angles, θ << 1), we get D 22 . 1 min ≈ This is only an approximation (the actual resolution is usually worse than this), but it allows us to make calculations. Example Problem: A hang glider is flying at an altitude of 120 m . Green light ( λ = 555 nm ) enters the pilot²s eye through a pupil that has a diameter D = 2.5 mm . If the average index of refraction of the material in the eye is n = 1.36 , determine how far apart two point objects on the ground must be if the pilot is to have any hope of distinguishing between them. Answer: 2.4 cm Object 1 Diffraction Patterns on Screen Round hole diameter D Object 2 Intensity θ min Rayleigh&s Criterion for resolution...
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This note was uploaded on 05/31/2011 for the course PHY 2061 taught by Professor Fry during the Spring '08 term at University of Florida.

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