PHY 102: Lecture Linear Superposition 11.2 Young’s Two Slit Experiment 11.3 Thin Films 11.4 Diffraction 11.5 Resolving Power 11.6 Diffraction Grating

PHY 102: Lecture 11 Interference Wave Nature of Light 11.1 Linear Superposition

Linear Superposition - 1 Several sound waves are present at the same place at the same time The pressure disturbance that results is governed by the principle of linear superposition The resultant disturbance is the sum of the disturbances from the individual waves Light is also a wave, an electromagnetic wave, and it too obeys the superposition principle

Linear Superposition - 2 When two or more light waves pass through a given point, their electric fields combine according to the principle of linear superposition and produce a resultant electric field Interference alters the brightness of light, just as it affects the loudness of sound

Linear Superposition - 3 Two identical waves arrive at the point P in phase – crest-to-crest and trough-to-trough By the principle of linear superposition, the waves reinforce each other and constructive interference occurs The resulting total wave at P has amplitude that is twice the amplitude of either individual wave

Linear Superposition - 4 The waves start out in phase and are in phase at P because the distance l 1 and l 2 between P and sources of waves differ by one wavelength These distances are l 1 = 2.25 wavelengths and l 2 = 3.25 wavelengths When the waves start out in phase, constructive interference will result at P whenever the distance are the same or differ by integer number of wavelengths Assume l 2 is larger distance l 2 – l 1 = m, m=0, 1, 2, 3, …

Linear Superposition - 5 Two identical waves arrive at the point P out of phase with one another - crest-to-trough The waves mutually cancel, according to the principle of linear superposition Destructive interference results With light waves this would mean that there is no brightness

Linear Superposition - 6 Waves begin with same phase but are out of phase at P because the distance through which they travel in reaching this spot differ by one-half of a wavelength ( l 1 = 2.75 wavelengths and l 2 = 3.25 wavelengths) When waves start out in phase, destructive interference will take place at P whenever the distances differ by any odd integer number of half-wavelengths Assume l 2 is larger distance l 2 – l 1 =(m+ ½), m=0, 1, 2, 3, …

Coherent Light Source If constructive or destructive interference is to continue occurring at a point, the sources of the waves must be coherent sources two sources are coherent if the waves they emit maintain a constant phase relation This means that the waves do not shift relative to one another as time passes Lasers are coherent sources of light Incandescent light bulbs and fluorescent bulbs are incoherent sources

PHY 102: Lecture 11 Interference Wave Nature of Light 11.2 Young’s Two Slit Experiment

Two Slit Experiment - 1 This experiment demonstrated the wave nature of light by showing that two overlapping light waves interfered with each other This experiment also permitted the determination of the wavelength of light

Two Slit Experiment - 2 Light of a single wavelength passes through a single narrow slit and falls on two closely spaced, narrow slits S 1 and S 2 The two slits act as coherent sources of light waves that interfere constructively and destructively at different points on the screen

Two Slit Experiment - 3 The purpose the single slit is to ensure that only light from one direction falls on the double slit The slits S 1 and S 2 act as coherent sources of light waves because the light from each originates from the same primary source

Two-Slit Explanation - 1 (a) Illustrates how a bright fringe arises directly opposite the mid point between the two slits Waves from each slit travel to the midpoint on the screen At this location, distance l 1 and l 2 to the slits are equal Each contain the same number of wavelengths.

Two-Slit Explanation - 2 (b) constructive interference produces another bright fringe on one side of the midpoint when the distance l 2 is larger than l 1 by exactly one wavelength Constructive interference produces additional bright fringes on both sides of the middle where the difference between l 1 and l 2 is an integer number of wavelengths

Two-Slit Explanation - 2 (c) shows how the first dark fringe arises Distance l 2 is large than l 1 by exactly one-half wavelength The waves interfere destructively, giving the dark fringe Destructive interference creates additional dark fringes on both side of center wherever the difference between l 1 and l 2 equal an odd integer number of ½ wavelengths

Photograph of Two-Slit Experiment

Derivation of Two Slit Equations Bright fringes of a double slit sin  = m /dm = 0, 1, 2, 3, … Dark fringes of a double slit sin  = (m + ½ ) /dm = 0, 1, 2, 3, …

Problem Red light ( = 664 nm in vacuum) is used in Young’s experiment with the slits separated by a distance d = 1.20 x m The screen is located at a distance of L = 2.75 m from the slits Find the distance y on the screen between the central bright fringe and the third-order bright fringe

Problem sin  = m /d = 3 x 6.64 x /1.20 x sin  =  = sin -1 (0.0166) = tan  = y/L tan(0.951) = y / 2.75 y = (2.75)tan(0.951) = m

PHY 102: Lecture 11 Interference Wave Nature of Light 11.3 Thin Films

Thin-Film Interference - 1 Figure shows a thin film Assume that the film has constant thickness Monochromatic light strikes the film nearly perpendicular At top surface reflection occurs This is Ray 1

Thin-Film Interference - 2 Refraction also occurs Some light enters the film Part of this light reflects from the bottom surface of the film and passes back up through the film Eventually reentering air Second wave is Ray 2 Ray 2 travels further than Ray 1

Thin-Film Interference - 3 Ray 1 can interfere with Ray 2 The wavelength that is important for thin-film interference is the wavelength within the film, not the wavelength in vacuum  film = vacuum / n film

Phase Change on Reflection - 1 Light travels through a material with smaller refractive index toward a material with larger refractive index Reflection at the boundary occurs along with a phase change of ½ wavelength

Phase Change on Reflection - 2 Light travels through a material with larger refractive index toward a material with smaller refractive index There is no phase change upon reflection at the boundary

Problem A thin film of gasoline floats on a puddle of water Sunlight falls almost perpendicularly on the film and reflects into your eyes Sunlight is white since it contains all colors The film looks yellow because destructive interference eliminates color blue ( vacuum =469 nm) from the reflected light The refractive indices of the blue light in gasoline and in water are 1.40 and 1.33 Determine minimum nonzero thickness t of film

Problem Phase change for wave 1 is ½ wavelength Wave 1 travels from a smaller refractive index (n air = 1.00) toward a larger refractive index (n gasoline = 1.40) No phase change when ray 2 reflects from the gasoline-water interface Net phase change between ray 1 and ray 2 is ½ wavelength

Problem t + ½ film =½ film, 3/2 film, 5/2 film t = m film m = 0, 1, 2, 3, …  film = vacuum /n = 4.69 x /1.40  film = 3.35 x m t = 1(3.35 x )/2 = 1.68 x m

PHY 102: Lecture 11 Interference Wave Nature of Light 11.4 Diffraction

Try Yourself Very simple demonstration of diffraction Holding your hand in front of a light source Slowly close two fingers while observing the light transmitted between them As the fingers approach each other and come very close together, a series of dark lines parallel to the fingers begins to appear The parallel lines are an example of diffraction patterns

Corona Coronae are produced by the diffraction of light by tiny cloud droplets or sometimes small ice crystalsdiffraction

Pollen Corona

Spider Web Diffraction

Diffraction - 1 Diffraction is the bending of waves around obstacle or the edges of an opening Sound waves are leaving a room through an open doorway The exiting sound waves bend, or diffract, around the edges of the opening A listener outside the room can hear the sound even when standing around the corner from the doorway

Diffraction - 2 Diffraction is an interference effect Huygen’s principle explains diffraction Every point on a wave front acts as a source of tiny wavelets that move forward with the same speed as the wave The wave front at a later instant is the surface that is tangent to the wavelets

Diffraction - 3 Wavelets emitted by the Huygens sources in the slit can also interfere destructively on the screen (a) Light rays directed from each source toward the first dark fringe  gives the position of this dark fringe relative to the line between the midpoint of the slit and the midpoint of the central bright fringe

Diffraction - 4 Screen is very far from slit Rays from each Huygens source are nearly parallel and oriented at nearly the same angle  Wavelet from source 1 travels shortest distance to screen Wavelet from source 5 travels the farthest

Diffraction - 5 Destructive interference creates the first dark fringe The extra distance traveled by the wavelet from source 5 is exactly one wavelength Under this condition, the extra distance traveled by the wavelet from source 3 at the center of the slit is exactly ½ wavelength

Diffraction - 6 Wavelets from sources 1 and 3 are exactly out of phase and interfere destructively at the screen Each wavelet from the upper half of the slit cancels a corresponding wavelet from the lower half and no light reaches screen sin  = / W W is width of slit

Diffraction - 7 Similar argument is used to find the second dark fringe

Diffraction Equation Dark fringes for single-slit diffraction sin  = m( / W)m = 1, 2, 3, … W is the width of the slit

Diffraction Pattern - 1 Between each pair of dark fringes there is a bright fringe Brightness of the fringes is related to the light intensity The width of the central fringe provides some indication of the extent of the diffraction

Problem Light passes through a slit and shines on a flat screen that is L = 0.40 m away The wavelength of the light in a vacuum is = 410 nm The distance between the midpoint of the central bright fringe and the first dark fringe is y. Determine the width 2y of the central bright fringe when the width of the slit is (a) W = 5.0 x m (b) W = 2.5 x m

Problem (a) W = 5.0 x m  = sin -1 ( /W)  =sin -1 (4.10x10 -7 /5.0x10 -6 )  = tan  =y/L 2y = 2Ltan  2y = 2(0.40)tan4.7 2y = m

Problem (B) W = 2.5 x m  = sin -1 ( /W)  =sin -1 (4.10x10 -7 /2.5x10 -6 )  = tan  =y/L 2y = 2Ltan  2y = 2(0.40)tan9.4 2y = 0.13 m

PHY 102: Lecture 11 Interference Wave Nature of Light 11.5 Resolving Power

Diffraction - Resolving Power - 1 Resolving Power –ability of an optical instrument to distinguish (or resolve) two closely-spaced objects Look at the headlights of a car as it backs away from you to a far distance When the car is close, it is easy to distinguish two separate headlights As it gets farther away, it’s harder to resolve the two headlights There is a certain point that we can’t distinguish the two headlights clearly This inability to resolve two closely- spaced objects is due to diffraction

Diffraction - Resolving Power - 2 Light passes through openings, eyes, telescope, microscope Diffraction through openings limits resolution Screen to right shows diffraction pattern for light passing through small circular opening There is a central bright fringe and alternating bright and dark fringes Θ locates angle from central bright fringe to first dark fringe If the screen distance is much larger than the width of the circular aperture (D), then –sinθ = 1.22 / D

Diffraction - Resolving Power - 3 Two objects create diffraction patterns on screen Can distinguish the two objects, since their diffraction patterns are widely separated Move them closer together Their diffraction patterns overlap Unable to distinguish two separate objects

Diffraction - Resolving Power - 4 Criterion for judging whether or not two objects are resolved Rayleigh Criterion for Resolution –Two closely-spaced objects are resolved when the first dark fringe of one image falls on the central bright fringe of other object

Diffraction - Resolving Power - 5 Image of two objects are just resolved The first dark fringe of one image is right at the edge of the central bright fringe of the other This sets a condition on the minimum angle between the two objects being resolved If θ < θ min, then we won’t be able to resolve the two objects Since θ min is small, then sin(  min ) = θ min Rayleigh Criterion for Resolution is  min = 1.22( / D)

Problem 3 Headlights on a car are s = 1.5 meters apart The pupil of eye is D = 2.0 x meters At what distance will the headlights appear as a single object? The light is yellow ( air = 580 x m) Index of refraction of the eye is 1.36 –Wavelength in eye is less than in air – eye = air / n eye = = 580 x / 1.36 = 426 x m –s = r  min, where r is distance to car –  min = 1.22 eye / D = s/r –r = sD / 1.22 eye = 1.5 x 2.0 x / 1.22 x 426 x –r = 5772 meters = 3.6 miles

PHY 102: Lecture 11 Interference Wave Nature of Light 11.6 Diffraction Grating

Types of Diffraction Gratings Transmission Grating –An opaque plate with many closely and regularly spaced slits Reflection Grating –Many closely and regularly spaced reflecting surfaces

How Many Lines (Slits) Diffraction grating can be made with up to 4,000 lines (slits) per millimeter

How Diffraction Gratings Work - 1 Light waves coming from the slits interfere constructively or destructively

How Diffraction Gratings Work - 2 Maximums (bright spots) occur when path length difference is 0, 1, 2, … You then have constructive interference from the light coming from each slit

Diffraction Grating Exact Mathematics Bright Spots (Maxima) sin  = m / dm = 0, 1, 2, 3, …

Problem 5 (m = 1) Grating is 1000 lines/cm d = 1/1000 lines/cm d = cm/line  x m = 380 x cm sin  = m / d = 380 x /0.001  = sin -1 (0.038) =  x m = 380 x cm sin  = m / d = 750 x /0.001  = sin -1 (0.075) = ColorWavelength violet380–450 nm blue450–495 nm green495–570 nm yellow570–590 nm orange590–620 nm red620–750 nm

Problem 5 (m = 2) Grating is 1000 lines/cm d = 1/1000 lines/cm d = cm/line  x m = 380 x cm sin  = m / d = 2x380x10 -7 /0.001  = sin -1 (0.076) =  x m = 380 x cm sin  = m / d = 2x750x10 -7 /0.001  = sin -1 (0.150) = ColorWavelength violet380–450 nm blue450–495 nm green495–570 nm yellow570–590 nm orange590–620 nm red620–750 nm

Problem 5 (m = 3) Grating is 1000 lines/cm d = 1/1000 lines/cm d = cm/line  x m = 380 x cm sin  = m / d = 3x380x10 -7 /0.001  = sin -1 (0.114) =  x m = 380 x cm sin  = m / d = 3x750x10 -7 /0.001  = sin -1 (0.225) = ColorWavelength violet380–450 nm blue450–495 nm green495–570 nm yellow570–590 nm orange590–620 nm red620–750 nm

Problem 5 Summary m-order maximumm = 1m = 2m = 3 Violet Angle Red Angle

Problem 5 Results First-order and second-order lines do not overlap Second-order and third-order lines overlap