Interference Diffraction Polarization interference diffraction polarization light as

Interference, Diffraction & Polarization interference, diffraction & polarization

light as waves Ø so far, light has been treated as if it travels in straight lines Ø ray diagrams Ø refraction, reflection Ø To describe many optical phenomena, we have to treat light as waves. Ø Just like waves in water, or sound waves, light waves can interact and form interference patterns. Remember c = f 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 interference, diffraction & polarization 3

demo: interference Interference III + + = constructive interference waves in phase = destructive interference waves ½ out of phase interference, diffraction & polarization 4

Interference in spherical waves maximum of wave r =r 1 2 minimum of wave r 1 r 2 positive constructive interference negative constructive interference destructive interference if r 2 -r 1 = n then constructive interference occurs if r 2 -r 1 = (n+½) then destructive interference occurs interference, diffraction & polarization 5

interference, diffraction & polarization 6

light as waves it works the same for light waves, sound waves, and *small* water waves 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 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 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: d sin = m , 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 (d sin = [m+1/2] , m=0, 1, 2, 3…) demo interference, diffraction & polarization 10

distance between bright spots tan =y/L L if is small, then sin tan so: d sin = m , m=0, 1, 2, 3… converts to dy/L = m difference between maximum m and maximum m+1: ym+1 -ym= (m+1) L/d-m L/d= L/d demo ym=m L/d interference, diffraction & polarization 11

question Ø 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 distance is equal Ø a) constructive interference so 1/2 difference: Ø b) destructive interference destructive int. +1/2 interference, diffraction & polarization 12

question Ø 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 d sin = m with m=3 =sin-1(m /d)=sin-1(3 x 600 x 10 -9/0. 01)=0. 01030 b) Ym = m L/d m=0: y 0 =0 m=3: y 3 = 3 x 600 x 10 -9 x 5/0. 01 = 9 x 10 -4 m = 0. 9 mm interference, diffraction & polarization 13

other ways of causing interference Ø remember equivalent to: n 1>n 2 n 1<n 2 1 2 interference, diffraction & polarization 1 2 14

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 1/2. This will not happen if n 1>n 2 1 2 n 1<n 2 1/2 phase change no phase change In a medium with index of refraction n, the wavelength changes (relative to vacuum) to /n interference, diffraction & polarization 15

thin film interference 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? interference, diffraction & polarization 16

thin film analysis 1 n=1. 5 2 1. 2. 3. 4. top surface? bottom surface? conditions? d? 1. top surface: n 1<n 2 so phase inversion 1/2 n=1 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 1/2 4. if phase inversion would not take place at any of the surfaces: constructive: 2 d=m (difference in path length=integer number of wavelengths) due to phase inversion at top surface: 2 d=(m+1/2) since the ray travels through film: 2 d=(m+1/2) film =(m+1/2) /nfilm destructive: 2 d=m film =m /nfilm interference, diffraction & polarization 17

Note The interference is different for light of different wavelengths interference, diffraction & polarization 18

question Phase inversion will occur at a) top surface b) bottom surface c) top and bottom surface d) neither surface na=1 nb=1. 5 n 1<n 2 in both cases nc=2 constructive interference will occur if: a) b) c) d) 2 d=(m+1/2) /nb 2 d=m /nb 2 d=(m+1/2) /nc 2 d=m /nc note: if destructive 2 d=(m+1/2) /nb this is used e. g. on sunglasses to reduce reflections interference, diffraction & polarization 19

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 2 t=(m+1/2) constructive interference if 2 t=m destructive interference, diffraction & polarization 20

question Given h=1 x 10 -5 m 30 bright fringes are seen, with a dark fringe at the left and the right. What is the wavelength of the light? 2 t=m destructive interference. m goes from 0 (left) to 30 (right). =2 t/m=2 h/m=2 x 1 x 10 -5/30=6. 67 x 10 -7 m=667 nm interference, diffraction & polarization 21

demo newton’s rings spacing not equal interference, diffraction & polarization 22

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 interference, diffraction & polarization 23

question 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 interference, diffraction & polarization 24

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 interference, diffraction & polarization 25

interference, diffraction & polarization 26

interference pattern from a single slit pick two points, 1 and 2, one in the top half of the slit, one in the bottom half of the slit. Light from these two points interferes destructively if: x=(a/2)sin = /2 so sin = /a we could also have divided up the slit into 4 pieces: x=(a/4)sin = /2 so sin =2 /a 6 pieces: x=(a/6)sin = /2 so sin =3 /a Minima occur if sin = m /a m=1, 2, 3… In between the minima, are maxima: sin = (m+1/2) /a m=1, 2, 3… AND sin =0 or =0 interference, diffraction & polarization 27

slit width a a if >a sin = /a > 1 Not possible, so no patterns <a : interference pattern is seen interference, diffraction & polarization if <<a sin =m /a is very small diffraction hardly seen 28

the diffraction pattern The intensity is not uniform: I=I 0 sin 2( )/ 2 = a(sin )/ a a a interference, diffraction & polarization 29

question light with a wavelength of 500 nm is used to illuminate a slit of 5 m. At which angle is the 5 th minimum in the diffraction pattern seen? sin = m /a = sin-1(5 x 500 x 10 -9/(5 x 10 -6))=300 interference, diffraction & polarization 30

diffraction from a single hair instead of an slit, we can also use an inverse image, for example a hair! demo interference, diffraction & polarization 31

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 asin =m , m=1, 2, 3… maxima asin =(m+1/2) , m=1, 2, 3… and =0 a: width of individual slit due to 2 -slit interference maxima dsin =m , m=0, 1, 2, 3… minima dsin =(m+1/2) , m=0, 1, 2, 3… d: distance between two slits interference, diffraction & polarization 32

double-slit experiment a d if >d, each slit acts as a single source of light and we get a more or less prefect double-slit interference spectrum if <d the interference spectrum is folded with the diffraction pattern. interference, diffraction & polarization 33

question 7 th 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: dsin =7 so sin =7 /d 1 st diffraction minimum: asin =1 so sin = /a sin must be equal for both, so /a=7 /d and a=d/7=1/7 mm interference, diffraction & polarization 34

diffraction grating d consider a grating with many slits, each separated by a distance d. Assume that for each slit >d. We saw that for 2 slits maxima appear if: d sin = m , 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 glass 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 interference, diffraction & polarization 35

separating colors d sin = m , m=0, 1, 2, 3… for maxima (same as for double slit) so = sin-1(m /d) depends on , the wavelength. cd’s can act as a diffraction grating (DVD’s work even better because their tracks are more closely spaced. ) interference, diffraction & polarization 36

question Ø 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? Ø a) the more slits, the larger the separation between maxima. Ø b) the more slits, the narrower each of the bright spots and thus easier to see Ø c) the more slits, the more light reaches each maximum and the maxima are brighter Ø d) there is no advantage interference, diffraction & polarization 37

question An diffraction grating has 5000 lines per cm. The angle between the central maximum and the fourth order maximum is 47. 20. What is the wavelength of the light? d sin = m , m = 0, 1, 2, 3… d = 1/5000 = 2 x 10 -4 cm = 2 x 10 -6 m m = 4, sin(47. 2)=0. 734 so = d sin /m = 2 x 10 -6 x 0. 734/4 = 3. 67 x 10 -7 m = 367 nm interference, diffraction & polarization 38

polarization Ø 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 Efield) could point in any direction E-vectors could point anywhere: unpolarized propagation into screen interference, diffraction & polarization 39

polarized light Ø light can be linearly polarized, which means that the E-field only oscillates in one direction (and the B-field perpendicular to that) Ø The intensity of light is proportional to the square of amplitude of the E-field. I~Emax 2 interference, diffraction & polarization 40

How to polarize? Ø absorption Ø reflection Ø scattering interference, diffraction & polarization 41

polarization by absorption Ø certain material (such as polaroid used for sunglasses) only transmit light along a certain ‘transmission’ axis. Ø because only a fraction of the light is transmitted after passing through a polarizer the intensity is reduced. Ø If unpolarized light passes through a polarizer, the intensity is reduced by a factor of 2 interference, diffraction & polarization 42

polarizers and intensity polarization axis direction of E-vector If E-field is parallel to polarization axis, all light passes For unpolarized light, on average, the E-field has an angle of 450 with the polarizer. I=I 0 cos 2(45)=I 0/2 If E-field makes an angle pol. axis only the component parallel to the pol. axis passes: E 0 cos So I=I 0 cos 2 interference, diffraction & polarization 43

question Ø 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 300 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 230=0. 75 I 1=0. 375 I 0 interference, diffraction & polarization 44

polarization by reflection n 1 n 2 Ø If unpolarized light is reflected, than the reflected light is partially polarized. Ø if the angle between the reflected ray and the refracted ray is exactly 900 the reflected light is completely polarized Ø the above condition is met if for the angle of incidence the equation tan =n 2/n 1 Ø the angle =tan-1(n 2/n 1) is called the Brewster angle Ø the polarization of the reflected light is (mostly) parallel to the surface of reflection interference, diffraction & polarization 45

question direction of polarization of reflected light vertical horizontal Ø 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. Ø a) horizontally Ø b) vertically interference, diffraction & polarization 46

sunglasses wearing sunglasses will help reducing glare (reflection) from flat surfaces (highway/water) without interference, diffraction & polarization with sunglasses 47

polarization by scattering Ø 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 interference, diffraction & polarization 48