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Unformatted text preview: Recap of last lecture Revisit 2 slit interference: actually diffraction pattern from each slit interfere to produce the pattern seen which is a little different from the idealized 2 slit pattern we saw in the last chapter. 2 2 /2 /2 sin 2 cos I I = ( 2 p / l )asin q f = (2 p/l) dsin q meters) ( 100 1 lines/cm, are there if : Note ce interferen slit  2 as formula Same , 2 , 1 , , ) sin( : fringes Bright n d n m d m l q 1.22 sin n D l q q The angular width of the central intensity maximum is 2 q where q =1.22 l/ D is given by the above Rayleigh formula. Light passing through a circular aperture of diameter D is spread into a wider circular beam with alternating bright and dark rings D l q 22 . 1 Holograms Problem 36.46 A wildlife photographer uses a moderate telephoto lens of focal length 135mm and maximum aperture f/4 to photograph a bear that is 11.5 m away. Assume the wavelength is 550 nm. a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? b) If, the photographer opens the aperture only to f/22 what would be the width of the smallest feature on the bear? Raleighs criterion is sin q q = 1.22 l /D. (a) Using Rayleighs criterion sin q q = 1.22 l /D = (1.22)(550 nm)/(135/4 mm) = 1.99 10 5 rad On the bear this angle subtends a distance x then q = x/R and...
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This note was uploaded on 01/27/2012 for the course PH 2233 taught by Professor Dipinkardutta during the Spring '11 term at Mississippi State.
 Spring '11
 DipinkarDutta
 Physics, Diffraction

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