Interference, Diffraction & Polarization PHY232 Remco Zegers zegers@nscl.msu.edu Room W109 cyclotron building http://www.nscl.msu.edu/~zegers/phy232.html
light as waves so far, light has been treated as if it travels in straight lines ray diagrams refraction To describe many optical phenomena, we have to treat light as waves. Just like waves in water, or sound waves, light waves can interact and form interference patterns. remember c=fλ PHY232 - Remco Zegers - interference, diffraction & polarization 2
interference constructive interference destructive interference at any point in time one can construct the total amplitude by adding the individual components PHY232 - Remco Zegers - interference, diffraction & polarization 3
demo: interference λ Interference III λ λ + + λ = = constructive interference waves in phase destructive interference waves ½λ out of phase PHY232 - Remco Zegers - interference, diffraction & polarization 4
Interference in spherical waves maximum of wave r 1 =r 2 minimum of wave r 1 r 2 positive constructive interference negative constructive interference destructive interference if r 2 -r 1 =nλ then constructive interference occurs if r 2 -r 1 =(n+½)λ the destructive interference occurs PHY232 - Remco Zegers - interference, diffraction & polarization 5
PHY232 - Remco Zegers - interference, diffraction & polarization 6
light as waves it works the same as water and sound! PHY232 - Remco Zegers - interference, diffraction & polarization 7
double slit experiment the light from the two sources is incoherent (fixed phase with respect to each other in this case, there is no phase shift between the two sources the two sources of light must have identical wave lengths PHY232 - Remco Zegers - interference, diffraction & polarization 8
Young s interference experiment there is a path difference: depending on its size the waves coming from S 1 or S 2 are in or out of phase PHY232 - Remco Zegers - interference, diffraction & polarization 9
Young s interference experiment If the difference in distance between the screen and each of the two slits is such that the waves are in phase, constructive interference occurs: bright spot difference in distance must be a integer multiple of the wavelength: dsinθ=mλ, m=0,1,2,3 m=0: zeroth order m=1: first order etc path difference if the difference in distance is off by half a wavelength (or one and a half etc), destructive interference occurs (dsinθ=[m+1/2]λ, m=0,1,2,3 ) demo PHY232 - Remco Zegers - interference, diffraction & polarization 10
distance between bright spots tanθ=y/l L if θ is small, then sinθ θ tanθ so: dsinθ=mλ, m=0,1,2,3 converts to dy/l=mλ difference between maximum m and maximum m+1: y m+1 -y m = (m+1)λl/d-mλl/d= λl/d demo y m =mλl/d PHY232 - Remco Zegers - interference, diffraction & polarization 11
do loncapa 1,2,7 from set 9 PHY232 - Remco Zegers - interference, diffraction & polarization 12
question two light sources are put at a distance d from a screen. Each source produces light of the same wavelength, but the sources are out of phase by half a wavelength. On the screen exactly midway between the two sources will occur a) constructive interference b) destructive interference distance is equal so 1/2λ difference: destructive int. +1/2λ PHY232 - Remco Zegers - interference, diffraction & polarization 13
question two narrow slits are illuminated by a laser with a wavelength of 600 nm. the distance between the two slits is 1 cm. a) At what angle from the beam axis does the 3 rd order maximum occur? b) If a screen is put 5 meter away from the slits, what is the distance between the 0 th order and 3 rd order maximum? a) use dsinθ=mλ with m=3 θ=sin -1 (mλ/d)=sin -1 (3x600x10-9 /0.01)=0.0103 0 b) y m =mλl/d m=0: y 0 =0 m=3: y 3 =3x600x10-9 x5/0.01=9x10-4 m =0.9 mm PHY232 - Remco Zegers - interference, diffraction & polarization 14
quiz (extra credit) Two beams of coherent light travel different paths arriving at point P. If constructive interference occurs at point P, the two beams must: a) travel paths that differ by a whole number of wavelengths b)travel paths that differ by an odd number of half wavelengths PHY232 - Remco Zegers - interference, diffraction & polarization 15
other ways of causing interference remember equivalent to: n 1 >n 2 n 1 <n 2 1 2 1 2 PHY232 - Remco Zegers - interference, diffraction & polarization 16
phase changes at boundaries If a light ray travels from medium 1 to medium 2 with n 1 <n 2, the phase of the light ray will change by 1/2λ. This will not happen if n 1 >n 2. n 1 >n 2 1 2 1 2 n 1 <n 2 1/2λ phase change no phase change In a medium with index of refraction n, the wavelength changes (relative to vacuum) to λ/n PHY232 - Remco Zegers - interference, diffraction & polarization 17
thin film interference n=1 n=1.5 The two reflected rays can interfere. To analyze this system, 4 steps are needed: n=1 1. Is there phase inversion at the top surface? 2. Is there phase inversion at the bottom surface 3. What are the conditions for constructive/destructive interference? 4. what should the thickness d be for 3) to happen? PHY232 - Remco Zegers - interference, diffraction & polarization 18
thin film analysis n=1 n=1.5 1 2 1. top surface? 2. bottom surface? 3. conditions? 4. d? n=1 1. top surface: n 1 <n 2 so phase inversion 1/2λ 2. bottom surface: n 1 >n 2 so no phase inversion 3. conditions: 1. constructive: ray 1 and 2 must be in phase 2. destructive: ray 1 and 2 must be out of phase by 1/2λ 4. if phase inversion would not take place at any of the surfaces: constructive: 2d=mλ (difference in path length=integer number of wavelengths) due to phase inversion at top surface: 2d=(m+1/2)λ since the ray travels through film: 2d=(m+1/2)λ film =(m+1/2)λ/n film destructive: 2d=mλ film =mλ/n film PHY232 - Remco Zegers - interference, diffraction & polarization 19
Note The interference is different for light of different wavelengths PHY232 - Remco Zegers - interference, diffraction & polarization 20
question n a =1 n b =1.5 n c =2 Phase inversion will occur at a) top surface b) bottom surface c) top and bottom surface d) neither surface n 1 <n 2 in both cases constructive interference will occur if: a) 2d=(m+1/2)λ/n b b) 2d=mλ/n b c) 2d=(m+1/2)λ/n c d) 2d=mλ/n c note: if destructive 2d=(m+1/2)λ/n b this is used e.g. on sunglasses to reduce reflections PHY232 - Remco Zegers - interference, diffraction & polarization 21
another case 1 2 The air gap in between the plates has varying thickness. Ray 1 is not inverted (n 1 >n 2 ) Ray 2 is inverted (n 1 <n 2 ) where the two glasses touch: no path length difference: dark fringe. if 2t=(m+1/2)λ constructive interference if 2t=mλ destructive interference. PHY232 - Remco Zegers - interference, diffraction & polarization 22
question Given h=1x10-5 m 30 bright fringes are seen, with a dark fringe at the left and the right. What is the wavelength of the light? 2t=mλ destructive interference. m goes from 0 (left) to 30 (right). λ=2t/m=2h/m=2x1x10-5 /30=6.67x10-7 m=667 nm PHY232 - Remco Zegers - interference, diffraction & polarization 23
demo newton s rings spacing not equal PHY232 - Remco Zegers - interference, diffraction & polarization 24
loncapa now do 3,4,5,8 from set 9 PHY232 - Remco Zegers - interference, diffraction & polarization 25
diffraction In Young s experiment, two slits were used to produce an interference pattern. However, interference effects can already occur with a single slit. This is due to diffraction: the capability of light to be deflected by edges/small openings. In fact, every point in the slit opening acts as the source of a new wave front PHY232 - Remco Zegers - interference, diffraction & polarization 26
PHY232 - Remco Zegers - interference, diffraction & polarization 27
interference pattern from a single slit pick two points, 1 and 2, one in the top top half of the slit, one in the bottom half of the slit. Light from these two points interferes destructively if: Δx=(a/2)sinθ=λ/2 so sinθ=λ/a we could also have divided up the slit into 4 pieces: Δx=(a/4)sinθ=λ/2 so sinθ=2λ/a 6 pieces: Δx=(a/6)sinθ=λ/2 so sinθ=3λ/a Minima occur if sinθ=mλ/a m=1,2,3 In between the minima, are maxima: sinθ=(m+1/2)λ/a m=1,2,3 AND sinθ=0 or θ=0 PHY232 - Remco Zegers - interference, diffraction & polarization 28
slit width a a if λ>a sinθ=λ/a > 1 Not possible, so no patterns λ<a : interference pattern is seen if λ<<a sinθ=mλ/a is very small diffraction hardly seen PHY232 - Remco Zegers - interference, diffraction & polarization 29
the diffraction pattern The intensity is not uniform: I=I 0 sin 2 (β)/β 2 β=πa(sinθ)/ λ a a a a a a PHY232 - Remco Zegers - interference, diffraction & polarization 30
question light with a wavelength of 500 nm is used to illuminate a slit of 5μm. At which angle is the 5 th minimum in the diffraction pattern seen? sinθ=mλ/a θ=sin -1 (5x500x10-9 /(5x10-6 ))=30 0 PHY232 - Remco Zegers - interference, diffraction & polarization 31
diffraction from a single hair instead of an slit, we can also use an inverse image, for example a hair! demo PHY232 - Remco Zegers - interference, diffraction & polarization 32
double slit interference revisited The total response from a double slit system is a combination of two single-source slits, combined with a diffraction pattern from each of the slit due to diffraction minima asinθ=mλ, m=1,2,3 maxima asinθ=(m+1/2)λ, m=1,2,3 and θ=0 a: width of individual slit due to 2-slit interference maxima dsinθ=mλ, m=0,1,2,3 minima dsinθ=(m+1/2)λ, m=0,1,2,3 d: distance between two slits PHY232 - Remco Zegers - interference, diffraction & polarization 33
double-slit experiment a d if λ>d, each slit acts as a single source of light and we get a more or less prefect double-slit interference spectrum if λ<d the interference spectrum is folded with the diffraction pattern. PHY232 - Remco Zegers - interference, diffraction & polarization 34
question 7th A person has a double slit plate. He measures the distance between the two slits to be d=1 mm. Next he wants to determine the width of each slit by investigating the interference pattern. He finds that the 7 th order interference maximum lines up with the first diffraction minimum and thus vanishes. What is the width of the slits? 7 th order interference maximum: dsinθ=7λ so sinθ=7λ/d 1 st diffraction minimum: asinθ=1λ so sinθ=λ/a sinθ must be equal for both, so λ/a=7λ/d and a=d/7=1/7 mm PHY232 - Remco Zegers - interference, diffraction & polarization 35
diffraction grating d consider a grating with many slits, each separated by a distance d. Assume that for each slit λ>d. We saw that for 2 slits maxima appear if: dsinθ=mλ, m=0,1,2,3 This condition is not changed for in the case of n slits. diffraction gratings can be made by scratching lines on glas and are often used to analyze light instead of giving d, one usually gives the number of slits per unit distance: e.g. 300 lines/mm d=1/(300 lines/mm)=0.0033 mm PHY232 - Remco Zegers - interference, diffraction & polarization 36
separating colors dsinθ=mλ, m=0,1,2,3 for maxima (same as for double slit) so θ=sin -1 (mλ/d) depends on λ, the wavelength. cd s can act as a diffraction grating PHY232 - Remco Zegers - interference, diffraction & polarization 37
quiz question (extra credit) why is it not possible to produce an interference pattern in a double-slit experiment if the separation of the slits is less than the wavelength of the light used? a) the very narrow slits required would generate different wavelength, thereby washing out the interference pattern b) the two slits would not emit coherent light c) the fringes would be too close together d) in no direction could a path difference as large as one wavelength be obtained PHY232 - Remco Zegers - interference, diffraction & polarization 38
question If the interference conditions are the same when using a double slit or a diffraction grating with thousands of slits, what is the advantage of using the grating to analyze light? a) the more slits, the larger the separation between maxima. b) the more slits, the narrower each of the bright spots and thus easier to see c) the more slits, the more light reaches each maximum and the maxima are brighter d) there is no advantage PHY232 - Remco Zegers - interference, diffraction & polarization 39
question An diffraction grating has 5000 lines per cm. The angle between the central maximum and the fourth order maximum is 47.2 0. What is the wavelength of the light? dsinθ=mλ, m=0,1,2,3 d=1/5000=2x10-4 cm=2x10-6 m m=4, sin(47.2)=0.734 so λ= dsinθ/m=2x10-6 x0.734/4=3.67x10-7 m=367 nm PHY232 - Remco Zegers - interference, diffraction & polarization 40
lon-capa do question 6 from lon-capa 9 PHY232 - Remco Zegers - interference, diffraction & polarization 41
polarization We saw that light is really an electromagnetic wave with electric and magnetic field vectors oscillating perpendicular to each other. In general, light is unpolarized, which means that the E-field vector (and thus the B-field vector as long as it is perpendicular to the E- field) could point in any direction E-vectors could point anywhere: unpolarized propagation into screen PHY232 - Remco Zegers - interference, diffraction & polarization 42
polarized light light can be linearly polarized, which means that the E- field only oscillated in one direction (and the B-field perpendicular to that) The intensity of light is proportional to the square of amplitude of the E-field. I~E 2 max PHY232 - Remco Zegers - interference, diffraction & polarization 43
How to polarize? absorption reflection scattering PHY232 - Remco Zegers - interference, diffraction & polarization 44
polarization by absorption certain material (such as polaroid used for sunglasses) only transmit light along a certain transmission axis. because only a fraction of the light is transmitted after passing through a polarizer the intensity is reduced. If unpolarized light passes through a polarizer, the intensity is reduced by a factor of 2 PHY232 - Remco Zegers - interference, diffraction & polarization 45
polarization axis polarizers and intensity direction of E-vector θ For unpolarized light, on average, the E-field has an angle of 45 0 with the polarizer. I=I 0 cos 2 θ=i 0 cos 2 (45)=I 0 /2 If E-field is parallel to polarization axis, all light passes If E-field makes an angle θ pol. axis only the component parallel to the pol. axis passes: E 0 cosθ So I=I 0 cos 2 θ PHY232 - Remco Zegers - interference, diffraction & polarization 46
question unpolarized light with intensity I 0 passes through a linear polarizer. It then passes through a second polarizer (the second polarizer is usually called the analyzer) whose transmission axis makes and angle of 30 0 with the transmission axis of the first polarized. What is the intensity of the light after the second polarizer, in terms of the intensity of the initial light? After passing through the first polarizer, I 1 =I 0 /2. After passing through the second polarizer, I 2 =I 1 cos 2 30=0.75I 1 =0.375I 0 PHY232 - Remco Zegers - interference, diffraction & polarization 47
n 1 n 2 polarization by reflection If unpolarized light is reflected, than the reflected light is partially polarized. if the angle between the reflected ray and the refracted ray is exactly 90 0 the reflected light is completely polarized the above condition is met if for the angle of incidence the equation tanθ=n 2 /n 1 the angle θ=tan -1 (n 2 /n 1 ) is called the Brewster angle the polarization of the reflected light is (mostly) parallel to the surface of reflection PHY232 - Remco Zegers - interference, diffraction & polarization 48
question direction of polarization of reflected light vertical horizontal Because of reflection from sunlight of the glass window, the curtain behind the glass is hard to see. If I would wear polaroid sunglasses that allow polarized light through, I would be able to see the curtain much better. a) horizontally b) vertically PHY232 - Remco Zegers - interference, diffraction & polarization 49
sunglasses wearing sunglasses will help reducing glare (reflection) from flat surfaces (highway/water) without with sunglasses PHY232 - Remco Zegers - interference, diffraction & polarization 50
polarization by scattering certain molecules tend to polarize light when struck by it since the electrons in the molecules act as little antennas that can only oscillate in a certain direction PHY232 - Remco Zegers - interference, diffraction & polarization 51
lon-capa do problem 9 from set 9 PHY232 - Remco Zegers - interference, diffraction & polarization 52