1 Chapter 37 - Interference and Diffraction A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007
2 Objectives: After completing this module, you should be able to: Define and apply concepts of constructive interference, destructive interference, diffraction,, and resolving power. Describe Young s s 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.
3 Diffraction of Light Diffraction is is the ability of of light waves to to bend around obstacles placed in in their path. Ocean Beach Light rays Fuzzy Shadow Water waves easily bend around obstacles, but light waves also bend, as evidenced by the lack of a sharp shadow on the wall.
4 Water Waves A wave generator sends periodic water waves into a barrier with a small gap, as shown below. A new set of of waves is is observed emerging from the gap to to the wall.
5 Interference of Water Waves An interference pattern is set up by water waves leaving two slits at the same instant.
6 Young s s Experiment In Young s s experiment,, light from a monochromatic source falls on two slits, setting up an interference pattern analogous to that with water waves. Light source S 1 S 2
7 The Superposition Principle The resultant displacement of two simul- taneous waves (blue( and green) ) is the algebraic sum of the two displacements. The composite wave is shown in yellow. Constructive Interference Destructive Interference The superposition of two coherent light waves results in light and dark fringes on a screen.
8 Young s s Interference Pattern s 1 s 2 Constructive Bright fringe s 1 s 2 s 1 Destructive Dark fringe s 2 Constructive Bright fringe
9 Conditions for Bright Fringes Bright fringes occur when the difference in path p is an integral multiple of one wave length. p 1 p 2 p 3 p 4 Path difference p = 0,, 2, 3, Bright fringes: p = n, n = 0, 1, 2,...
10 Conditions for Dark Fringes Dark fringes occur when the difference in path p is an odd multiple of one-half of a wave length. p 1 p 2 p 3 p 3 2 p n 2 n = odd n = 1,3,5 Dark fringes: p n n1, 3, 5, 7,... 2
11 Analytical Methods for Fringes s 1 x d sin s 2 d p 1 p 2 y Path difference determines light and dark pattern. p = p 1 p 2 p = d sin Bright fringes: d sin = n, n = 0, 1, 2, 3,... Dark fringes: d sin = n, n = 1, 3, 5,...
12 Analytical Methods (Cont.) s 1 x d sin s 2 d p 1 p 2 y From geometry, we recall that: sin d tan So that... sin dy x y x dy x Bright fringes: n, n0, 1, 2,... dy x Dark fringes: n, n1, 3,
13 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 =? d sin = 5(/2) The third dark fringe occurs when n = 5 s 1 s 2 x d sin n = 1, 3, 5 y dy x Dark fringes: n, n1, 3, dy x 5 2
14 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 x = 600 nm; y =? dy x 5 2 s 1 s 2 d sin n = 1, 3, 5 y y -9 5x 5(600 x 10 m)(2 m) (0.08 x 10 m) d y = 3.75 cm
15 The Diffraction Grating A diffraction grating consists of of thousands of of parallel slits etched on glass so that brighter and sharper patterns can be observed than with Young s s experiment. Equation is is similar. d sin d d sin n n = 1, 2, 3,
16 The Grating Equation The grating equation: dsin n n1, 2, 3,... d = slit width (spacing) = wavelength of light = angular deviation n = order of fringe 1 st order 2 nd order
17 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? To find slit separation, we take reciprocal of 300 lines/mm: n = 2 Lines/mm mm/line 300 lines/mm d d lines/mm 3 mm 10 m line 1 mm mm/line d -6 3 x 10 m
18 Example (Cont.) 2: A grating is ruled with 300 lines/mm.. What is the angular deviation of the 2 nd order bright fringe? = 600 nm d -6 3 x 10 m n = 2 dsin n n (600 x 10 m) sin sin ; -6 d 3.33 x lines/mm Angular deviation of second order fringe is: 22 =
19 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.
20 Interference From Single Slit When monochromatic light strikes a single slit, diffraction from the edges produces an interference pattern as illustrated. Relative intensity Pattern Exaggerated The interference results from the fact that not all paths of light travel the same distance some arrive out of phase.
21 Single Slit Interference Pattern a a/2 a/2 a 2 sin Each point inside slit acts as a source. For rays 1 and 3 and for 2 and 4: p a sin 2 First dark fringe: a sin 2 2 For every ray there is another ray that differs by this path and therefore interferes destructively.
22 Single Slit Interference Pattern a 2 sin a sin 2 2 a a/2 a/ First dark fringe: sin a Other dark fringes occur for integral multiples of this fraction /a.
23 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? =? sin a x = 1.5 m a = 0.35 mm y y ya sin tan ; ; x x a x (0.002 m)( m) 1.50 m = 600 nm y
24 Diffraction for a Circular Opening D Circular diffraction The diffraction of 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.
25 Resolution of Images Consider light through a pinhole. As two objects get closer the interference fringes overlap, making it it difficult to to distinguish separate images. Clear image of each object Separate images barely seen d 1 d 2
26 Resolution Limit Images are just resolved when central maximum of of one pattern coincides with first dark fringe of of the other pattern. Resolution limit d 2 Separate images Resolution Limit
27 Resolving Power of Instruments The resolving power of of an instrument is is a measure of of its ability to to produce well-defined separate images. D Limiting angle For small angles, sin,, and the limiting angle of of resolution for a circular opening is: Limiting angle of of resolution: D
28 Resolution and Distance p s o D Limiting angle o Limiting Angle of Resolution: D s0 p
29 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? p s o D Tail lights s D p (1.2 m)(0.002 m) p (632 x 10 m) Eye p s0d 1.22 p = 3.11 km
30 Summary Young s Experiment: Monochromatic light falls on two slits, producing interference fringes on a screen. s 1 x d sin s 2 d p 1 p 2 dy d sin x y dy x Bright fringes: n, n0, 1, 2,... dy x Dark fringes: n, n1, 3,
31 Summary (Cont.) The grating equation: dsin n n1, 2, 3,... d = slit width (spacing) = wavelength of light = angular deviation n = order of fringe
32 Summary (Cont.) Interference from a single slit of of width a: Relative Intensity Pattern Exaggerated Dark Fringes: sin n n1, 2, 3,... a
33 Summary (cont.) The resolving power of of instruments. p s o D Limiting angle o Limiting Angle of Resolution: D s0 p
34 CONCLUSION: Chapter 37 Interference and Diffraction