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Chapter 37 - Interference and Diffraction A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007

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Objectives: After completing this module, you should be able to: Define and apply concepts of constructive interference, destructive interference, diffraction, and resolving power.Define and apply concepts of constructive interference, destructive interference, diffraction, and resolving power. Describe Youngs experiment and be able to predict the location of dark and bright fringes formed from the interference of light waves.Describe Youngs experiment and be able to predict the location of dark and bright fringes formed from the interference of light waves. Discuss the use of a diffraction grating, derive the grating equation, and apply it to the solution of optical problems.Discuss the use of a diffraction grating, derive the grating equation, and apply it to the solution of optical problems.

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Diffraction of Light Diffraction is the ability of light waves to bend around obstacles placed in their path. OceanBeach Water waves easily bend around obstacles, but light waves also bend, as evidenced by the lack of a sharp shadow on the wall. Fuzzy Shadow Light rays

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Water Waves A wave generator sends periodic water waves into a barrier with a small gap, as shown below. A new set of waves is observed emerging from the gap to the wall.

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Interference of Water Waves An interference pattern is set up by water waves leaving two slits at the same instant.

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Youngs Experiment In Youngs experiment, light from a monochromatic source falls on two slits, setting up an interference pattern analogous to that with water waves. Light source S1S1 S2S2

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The Superposition Principle The resultant displacement of two simul- taneous waves (blue and green) is the algebraic sum of the two displacements.The resultant displacement of two simul- taneous waves (blue and green) is the algebraic sum of the two displacements. The superposition of two coherent light waves results in light and dark fringes on a screen. The composite wave is shown in yellow.The composite wave is shown in yellow. Constructive Interference Destructive Interference

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Youngs Interference Pattern s1s1 s2s2 s1s1 s2s2 s1s1 s2s2 Constructive Bright fringe Dark fringe Destructive

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Conditions for Bright Fringes Bright fringes occur when the difference in path p is an integral multiple of one wave length. p1p1p1p1 p2p2p2p2 p3p3p3p3 p4p4p4p4 Path difference p = 0,, 2, 3, Bright fringes: p = n, n = 0, 1, 2,...

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Conditions for Dark Fringes Dark fringes occur when the difference in path p is an odd multiple of one-half of a wave length . p1p1p1p1 p2p2p2p2 p3p3p3p3 p3p3p3p3 n = odd n = 1,3,5 Dark fringes:

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Analytical Methods for Fringes x y d sin s1s1 s2s2 d p1p1 p2p2 Bright fringes: d sin = n, n = 0, 1, 2, 3,... Dark fringes: d sin = n , n = 1, 3, 5,... p = p 1 p 2 p = d sin Path difference determines light and dark pattern.

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Analytical Methods (Cont.) x y d sin s1s1 s2s2 d p1p1 p2p2 From geometry, we recall that: Bright fringes:Dark fringes: So that...

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Example 1: Two slits are 0.08 mm apart, and the screen is 2 m away. How far is the third dark fringe located from the central maximum if light of wavelength 600 nm is used? x = 2 m; d = 0.08 mm = 600 nm; y = ? = 600 nm; y = ? The third dark fringe occurs when n = 5 x y d sin s1s1 s2s2 n = 1, 3, 5 Dark fringes: d sin = 5( /2)

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Example 1 (Cont.): Two slits are 0.08 mm apart, and the screen is 2 m away. How far is the third dark fringe located from the central maximum if = 600 nm? x = 2 m; d = 0.08 mm = 600 nm; y = ? = 600 nm; y = ? x y d sin s1s1 s2s2 n = 1, 3, 5 y = 3.75 cm

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The Diffraction Grating A diffraction grating consists of thousands of parallel slits etched on glass so that brighter and sharper patterns can be observed than with Youngs experiment. Equation is similar. d sin d d sin n n = 1, 2, 3,

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The Grating Equation The grating equation:d = slit width (spacing) = wavelength of light = angular deviation n = order of fringe 1 st order 2 nd order

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Example 2: Light (600 nm) strikes a grating ruled with 300 lines/mm. What is the angular deviation of the 2 nd order bright fringe? 300 lines/mm n = 2 To find slit separation, we take reciprocal of 300 lines/mm: Lines/mm mm/line

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Example (Cont.) 2: A grating is ruled with 300 lines/mm. What is the angular deviation of the 2 nd order bright fringe? 2 = 21.1 0 Angular deviation of second order fringe is: 300 lines/mm n = 2 = 600 nm

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A compact disk acts as a diffraction grating. The colors and intensity of the reflected light depend on the orientation of the disc relative to the eye.

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Interference From Single Slit Pattern Exaggerated When monochromatic light strikes a single slit, diffraction from the edges produces an interference pattern as illustrated. Relative intensity The interference results from the fact that not all paths of light travel the same distance some arrive out of phase.

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Single Slit Interference Pattern a/2 a 1 2 4 3 5 Each point inside slit acts as a source. For rays 1 and 3 and for 2 and 4: First dark fringe: For every ray there is another ray that differs by this path and therefore interferes destructively.

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Single Slit Interference Pattern a/2 a 1 2 4 3 5 First dark fringe: Other dark fringes occur for integral multiples of this fraction /a.

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Example 3: Monochromatic light shines on a single slit of width 0.45 mm. On a screen 1.5 m away, the first dark fringe is displaced 2 mm from the central maximum. What is the wavelength of the light? x = 1.5 m y a = 0.35 mm = ? = 600 nm

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Diffraction for a Circular Opening Circular diffraction D The diffraction of light passing through a circular opening produces circular interference fringes that often blur images. For optical instruments, the problem increases with larger diameters D.

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Resolution of Images Consider light through a pinhole. As two objects get closer the interference fringes overlap, making it difficult to distinguish separate images. d2d2 Separate images barely seen d1d1 Clear image of each object

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Resolution Limit d2d2 Resolution limit Images are just resolved when central maximum of one pattern coincides with first dark fringe of the other pattern. Resolution Limit Separate images

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Resolving Power of Instruments The resolving power of an instrument is a measure of its ability to produce well-defined separate images. Limiting angle of resolution: For small angles, sin , and the limiting angle of resolution for a circular opening is: Limiting angle D

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Resolution and Distance Limiting Angle of Resolution: sosososop D Limiting angle o

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Example 4: The tail lights ( = 632 nm) of an auto are 1.2 m apart and the pupil of the eye is around 2 mm in diameter. How far away can the tail lights be resolved as separate images? sosososopEye D Tail lights p = 3.11 km

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Summary Bright fringes:Dark fringes: Youngs Experiment: Monochromatic light falls on two slits, producing interference fringes on a screen. x y d sin s1s1 s2s2 d p1p1 p2p2

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Summary (Cont.) The grating equation: d = slit width (spacing) = wavelength of light = angular deviation n = order of fringe

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Summary (Cont.) Pattern Exaggerated Relative Intensity Interference from a single slit of width a :

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Summary (cont.) Limiting Angle of Resolution: sosososop D Limiting angle o The resolving power of instruments.

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CONCLUSION: Chapter 37 Interference and Diffraction