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Diffraction Physics 202 Professor Lee Carkner Lecture 26
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PAL #25 Interference Applications Wavelength of laser D = d = 0.25 mm = 0.00025 m y = 1.5 cm = 0.015 m (between any 2 maxima) = yd/D = (0.015)(2.5X10 -4 )/(5.5) = 682 nm Is this reasonable?
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Diffraction When light passes though a small aperture it spreads out This effect also occurs when light passes by an obstacle The pattern consists of minima and maxima of decreasing intensity as you move away from the center
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Diffraction and Optics Real point sources of light always experience diffraction and so we never see true point images Diffraction limits even the best optics Degree of diffraction (and blurriness) depends on aperture size and wavelength
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Diffraction and Interference Light rays from different parts of the same aperture can also produce interference Instead of two rays from two slits, we have a continuum of rays emerging from one slit
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Path Length Difference Minima (dark fringes) should occur at the point where half of the rays are out of phase with the other half If we assume that the distance to the screen (D) is much larger than the slit width (a) then the path difference is where d is the distance between the origin points of the two rays We will pair up the rays, and find the path length for which each pair cancels out
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Single Slit Diffraction
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Location of the Minima Since: L /d = sin How far apart can a pair of rays get? For the first minima L must equal /2: a sin =
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Diffraction Patterns Since a sin = for m=1, we can say for the more general case: a sin = m (min) The maxima are located halfway between the minima Since waves from the top and bottom half cancel Smaller slit means more flaring
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Rays from Slit
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Intensity Intensity of maxima decrease with increasing We will use which is half the phase difference between the top and bottom of the slit = ½ = ( a/ ) sin I = I m (sin / ) 2
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Diffraction and Interference
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Intensity Variations The intensity falls off rapidly with linear distance y Remember tan = y/D The narrower the slit the broader the maximum Remember: m = 1.5, 2.5, 3.5 … maxima
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Diffraction and Circular Apertures The location of the minima depend on the wavelength and the diameter instead of slit width: For m = 1 The minima and maxima appear as concentric circles
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Resolution Since virtually all imaging devices have apertures, virtually all images are blurry If you view two point sources that are very close together, you may not be able to distinguish them
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Resolution and Circular Aperture
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Rayleigh’s Criterion This will be true if their angular separation satisfies the diffraction formula R = 1.22 /d This is called Rayleigh’s criterion
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