Physical Optics.pdf

Chapter 37 - Interferenceand DiffractionA PowerPoint Presentation byPaul E. Tippens, Professor of PhysicsSouthern Polytechnic State University© 2007

Objectives: After completing thismodule, you should be able to:• Define and apply concepts of constructiveinterference, destructive interference,diffraction, , and resolving power.• Describe Young’s s experiment and be able topredict the location of dark and bright fringesformed from the interference of light waves.• Discuss the use of a diffraction grating, , derivethe grating equation, and apply it to thesolution of optical problems.

Diffraction of LightDiffraction is is the ability of of light waves to to bendaround obstacles placed in in their path.OceanBeachLight raysFuzzy ShadowWater waves easily bend around obstacles, butlight waves also bend, as evidenced by the lackof a sharp shadow on the wall.

Water WavesA wave generator sends periodic water wavesinto a barrier with a small gap, as shown below.A new set of of waves is is observedemerging from the gap to to the wall.

Interference of Water WavesAn interference pattern is set up by waterwaves leaving two slits at the same instant.

Young’s s ExperimentIn Young’s s experiment, , light from a monochromaticsource falls on two slits, setting up an interferencepattern analogous to that with water waves.Lightsource S 1S 2

The Superposition Principle• The resultant displacement of two simul-taneous waves (blue(and green) ) is thealgebraic sum of the two displacements.• The composite wave is shown in yellow.Constructive InterferenceDestructive InterferenceThe superposition of two coherent light wavesresults in light and dark fringes on a screen.

Young’s s Interference Patterns 1s 2ConstructiveBright fringes 1s 2s 1DestructiveDark fringes 2ConstructiveBright fringe

Conditions for Bright FringesBright fringes occur when the difference in path pis an integral multiple of one wave length .p 1 p 2p 3p 4Path differencep = 0, , 2, 3, …Bright fringes:p = n, n = 0, 1, 2, . . .

Conditions for Dark FringesDark fringes occur when the difference in path pis an odd multiple of one-half of a wave length .p 1p 2p 3p 32pn 2n = oddn =1,3,5 …Dark fringes: p n n1,3, 5, 7, . . .2

Analytical Methods for Fringess 1xd sin s 2d p 1p 2yPath differencedetermines lightand dark pattern.p = p 1 –p 2p = d sin Bright fringes: d sin = n, n = 0, 1, 2, 3, . . .Dark fringes: d sin = n, n = 1, 3, 5, . . .

Analytical Methods (Cont.)s 1xd sin s 2d p 1p 2yFrom geometry,we recall that:sindtanSo that . . .sin dyxyxdyxBright fringes:n, n0, 1, 2, ...dyxDark fringes:n , n1, 3, 5...2

Example 1: Two slits are 0.08 mm apart, andthe screen is 2 m away. How far is the thirddark fringe located from the central maximum iflight of wavelength 600 nm is used?x = 2 m; d = 0.08 mm = 600 nm; y = ?d sin = 5(/2)The third dark fringeoccurs when n = 5s 1s 2xd sin n = 1, 3, 5ydyxDark fringes:n , n1, 3, 5...2dyx52

Example 1 (Cont.): Two slits are 0.08 mmapart, and the screen is 2 m away. How far isthe third dark fringe located from the centralmaximum if = 600 nm?x = 2 m; d = 0.08 mmx = 600 nm; y = ?dyx52s 1s 2d sin n = 1, 3, 5yy-95x5(600 x 10 m)(2 m)-32 2(0.08 x 10 m)dy = 3.75 cm

The Diffraction GratingA diffraction grating consists of of thousands ofofparallel slits etched on glass so that brighter andsharper patterns can be observed than withYoung’s s experiment. Equation is is similar.d sin dd sin nn = 1, 2, 3, …

The Grating EquationThe grating equation:dsin nn1, 2, 3, ...d = slit width (spacing) = wavelength of light = angular deviationn = order of fringe1 storder 2 ndorder

Example 2: Light (600(nm) ) strikes a grating ruledwith 300 lines/mm. . What is the angular deviationof the 2 nd order bright fringe?To find slit separation,we take reciprocal of300 lines/mm:n = 2Lines/mm mm/line300 lines/mmdd 1300 lines/mm3mm 10 m 0.00333 line 1 mm 0.00333 mm/lined -63 x 10 m

Example (Cont.) 2: A grating is ruled with 300lines/mm. . What is the angular deviation of the2 nd order bright fringe?= 600 nmd -63 x 10 mn = 2dsin nn2-922(600 x 10 m) sin 0.360sin ;-6d 3.33 x 10300 lines/mmAngular deviation ofsecond order fringe is: 22 = 21.1 0

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

Interference From Single SlitWhen monochromatic light strikes a single slit,diffraction from the edges produces an interferencepattern as illustrated.Relative intensityPattern ExaggeratedThe interference results from the fact that not allpaths of light travel the same distance somearrive out of phase.

Single Slit Interference Patternaa/2a/2a2sin12345Each point inside slitacts as a source.For rays 1 and 3and for 2 and 4:pasin2First dark fringe:a sin 2 2For every ray there is another ray that differs bythis path and therefore interferes destructively.

Single Slit Interference Patterna2sina sin 2 2aa/2a/212345First dark fringe:sin aOther dark fringes occurfor integral multiples ofthis fraction /a.

Example 3: Monochromatic light shines on asingle slit of width 0.45 mm. . On a screen 1.5 maway, the first dark fringe is displaced 2 mmfrom the central maximum. What is thewavelength of the light? = ?sin ax = 1.5 ma = 0.35 mmy y yasin tan ; ; x x a x (0.002 m)(0.00045 m)1.50 m= 600 nmy

Diffraction for a Circular OpeningDCircular diffractionThe diffraction of of light passing through a circularopening produces circular interference fringesthat often blur images. For optical instruments,the problem increases with larger diameters D.

Resolution of ImagesConsider light through a pinhole. As two objectsget closer the interference fringes overlap,making it it difficult to to distinguish separate images.Clear image ofeach objectd 1Separate imagesbarely seend 2

Resolution LimitImages are just resolvedwhen central maximumof of one pattern coincideswith first dark fringe ofofthe other pattern.Resolutionlimitd 2Separate imagesResolution Limit

Resolving Power of InstrumentsThe resolving power of ofan instrument is is ameasure of of its ability to toproduce well-definedseparate images.DLimiting angleFor small angles, sin , , and the limitingangle of of resolution for a circular opening is:Limiting angle of ofresolution:0 1.22 D

Resolution and Distanceps oDLimiting angle oLimiting Angleof Resolution:0 1.22Ds0p

Example 4: The tail lights ((= 632 nm) ) of anauto are 1.2 m apart and the pupil of the eyeis around 2 mm in diameter. How far away canthe tail lights be resolved as separate images?ps oDTail lights s00 1.22D p(1.2 m)(0.002 m)p -91.22(632 x 10 m)Eyeps0D1.22p = 3.11 km

SummaryYoung’sExperiment:Monochromaticlight falls on twoslits, producinginterference fringeson a screen.s 1xd sin s 2d p 1p 2dyd sin xydyxBright fringes:n, n0, 1, 2, ...dyxDark fringes:n , n1, 3, 5...2

Summary (Cont.)The grating equation:dsin nn1, 2, 3, ...d = slit width (spacing) = wavelength of light = angular deviationn = order of fringe

Summary (Cont.)Interference from a single slit of of width a:Relative IntensityPattern ExaggeratedDark Fringes: sin nn1, 2, 3, . . .a

Summary (cont.)The resolving power of of instruments.ps oDLimiting angle oLimiting Angleof Resolution:0 1.22Ds0p

CONCLUSION: Chapter 37Interference and Diffraction