Waves in general sine waves are nice other

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Waves (in general) • sine waves are nice • other types of waves (such

Waves (in general) • sine waves are nice • other types of waves (such as square waves, sawtooth waves, etc. ) can be formed by a superposition of sine waves - this is called Fourier Series. This means that sine waves can be considered as fundamental.

Fourier Series Example

Fourier Series Example

Waves (in general) • E = Eo sin( ) where is a phase angle

Waves (in general) • E = Eo sin( ) where is a phase angle which describes the location along the wave = 90 degrees is the crest = 270 degrees is the trough

Waves (in general) E = Eo sin( ) where is a phase angle in

Waves (in general) E = Eo sin( ) where is a phase angle in a moving wave, and changes with both – time (goes 2 radians in time T) and – distance (goes 2 radians in distance ) so = (2 / )*x +/- (2 /T)*t where 2 /T = and 2 / = k. If we follow the crest, = /2, then the above equation for theta becomes (2 / )*x = +/- (2 /T)*t , or with = /2 xcrest = ( /2)/ (2 / ) +/- (2 /T)*t/ (2 / ) , so that phase speed: v = dxcrest /dt = /T = f = /k

Waves (in general) • Solution of the wave equation demands a function that depends

Waves (in general) • Solution of the wave equation demands a function that depends on x±vt, f(x±vt). • Sine waves are one such function: E = Eosin(k{x±vt}) = Eosin(kx ± t) where kv= , or v = /k = fλ. • For waves in general, we can break any repeating wave pattern into component sine waves; this is called spectral analysis.

Light and Shadows • Consider what we would expect from particle theory: sharp shadows

Light and Shadows • Consider what we would expect from particle theory: sharp shadows dark light dark

Light and Shadows • Consider what we would expect from wave theory: shadows NOT

Light and Shadows • Consider what we would expect from wave theory: shadows NOT sharp crest dark dim light dim dark

Light and Shadows • What DOES happen? look at a very bright laser beam

Light and Shadows • What DOES happen? look at a very bright laser beam going through a vertical slit. (A laser has one frequency unlike white light. )

Double Slit Experiment We will consider the single slit situation but only after we

Double Slit Experiment We will consider the single slit situation but only after we consider another: the DOUBLE SLIT experiment. Below the blue indicates crests of the light coming from the top slit, and the red the crests from the bottom slit. The light itself is the same color.

Double Slit Experiment • Note that along the green lines are places where crests

Double Slit Experiment • Note that along the green lines are places where crests meets crests and troughs meet troughs. crest on crest followed by trough on trough

Double Slit Experiment • Note that along the dotted lines are places where crests

Double Slit Experiment • Note that along the dotted lines are places where crests meets troughs and troughs meet crests. crest on trough followed by trough on crest followed by trough on trough

Double Slit Experiment Further explanations are in the Introduction to the Computer Homework Assignment

Double Slit Experiment Further explanations are in the Introduction to the Computer Homework Assignment on Young’s Double Slit, Vol 5, #3. crest on trough followed by trough on crest followed by trough on trough

Double Slit Experiment Our question now is: How is the pattern of bright and

Double Slit Experiment Our question now is: How is the pattern of bright and dark areas related to the parameters of the situation: , d, x and L? bright x d bright L SCREEN

Young’s Double Slit Formula λ/d = sin( ) ≈ tan( ) = x/L bright

Young’s Double Slit Formula λ/d = sin( ) ≈ tan( ) = x/L bright q + q = 180 o and q » 90 o x d bright λ L The two (black) lines from the two slits to the first bright spot are almost parallel, so the three angles are almost 90 degrees, so the two ’s are almost equal.

Double slit: an example n = d sin( ) d x / L •

Double slit: an example n = d sin( ) d x / L • d = 0. 15 mm = 1. 5 x 10 -4 m • x = ? ? ? measured in class • L = ? ? ? measured in class • n = 1 (if x measured between adjacent bright spots) • d x / L = (you do the calculation) Notice that with reasonably measured values of d, x, and L, we can determine λ’s down to less than 1 micron!

Interference: Diffraction Grating The same Young’s formula n = d sin( ) d x

Interference: Diffraction Grating The same Young’s formula n = d sin( ) d x / L works for multiple slits as it did for 2 slits. lens bright s 1 d s 2 s 3 s 4 s 5 λ s 2 = s 1 + s 3 = s 2 + = s 1 + 2 s 4 = s 3 + = s 1 + 3 s 5 = s 4 + = s 1 + 4 bright

Interference: Diffraction Grating With multiple slits, get MORE LIGHT and get sharper bright spots

Interference: Diffraction Grating With multiple slits, get MORE LIGHT and get sharper bright spots – see the next slide. lens bright s 1 d s 2 s 3 s 4 s 5 λ s 2 = s 1 + s 3 = s 2 + = s 1 + 2 s 4 = s 3 + = s 1 + 3 s 5 = s 4 + = s 1 + 4 bright

Interference: Diffraction Grating With two slits, we only get complete cancellation when s =

Interference: Diffraction Grating With two slits, we only get complete cancellation when s = 0. 5 . With 5 slits, we get cancellation when s = 0. 8. This makes the bright spots much sharper when s = 1. 0. lens s 1 d s 2 s 3 s 4 s 5 λ bright dark s 2 = s 1 +. 8 bright s 3 = s 2 +. 8 = s 1 + 1. 6 s 4 = s 3 +. 8 = s 1 + 2. 4 s 5 = s 4 +. 8 = s 1 + 3. 2

Colors and Wavelengths for visible light, using the diffraction grating, we come up with

Colors and Wavelengths for visible light, using the diffraction grating, we come up with the following ranges (these are approximate and vary slightly with individuals): violet 400 - 450 nm blue 450 - 500 nm green 500 - 550 nm yellow 550 - 600 nm orange 600 - 650 nm red 650 - 700 nm

Colors and wavelengths The wavelengths specified for the colors on the previous slide are

Colors and wavelengths The wavelengths specified for the colors on the previous slide are only approximate, and they do vary slightly with each person. On any test, I will give you one color leeway, e. g. , if the wavelength falls in the green range (previous slide), but you answer either blue or yellow, I will count your answer correct. Is the way we see different wavelengths (frequencies) of light similar to the way we hear different frequencies of sound?

Sound and frequencies For sound, our ears function like a Fourier analyzer – we

Sound and frequencies For sound, our ears function like a Fourier analyzer – we can actually hear each frequency. If we mix frequencies, we hear a chord. We can distinguish the base from the treble; we can distinguish the guitar from the piano.

Colors and frequencies For light, our eyes do NOT act this way. If we

Colors and frequencies For light, our eyes do NOT act this way. If we mix frequencies, we see only one color which is different from either of the “pure” colors (frequencies). We have two different types of retinal cells: rods and cones. The rod cells are not that closely packed so do not give very good resolution, are good for dim light, and only send a black and white signal. There are three different types of cone cells which are tightly packed for good resolution but need bright light, and which act as color light receptors. The next slide shows only a rough picture. You should look at more advanced texts to get a more accurate picture, or clink on the web info below: https: //www. ecse. rpi. edu/~schubert/Light-Emitting-Diodes-dot-org/Sample-Chapter. pdf

Cone cells Cone cell sensitivity to different wavelengths 400 450 500 550 600 650

Cone cells Cone cell sensitivity to different wavelengths 400 450 500 550 600 650 700 (in nm) If only the “blue” cone is activated, the color is violet. If both the “blue” and “green” cones are activated, and the “blue” gives a stronger signal, the color is blue. If both the “blue” and “green” cones are activated, and the “green” gives a stronger signal, the color is green.

Cone cells Cone cell sensitivity to different wavelengths 400 450 500 550 600 650

Cone cells Cone cell sensitivity to different wavelengths 400 450 500 550 600 650 700 (in nm) If both the “green” and “red” cones are activated, and the “green” gives a stronger signal, the color is yellow. If both the “green” and “red” cones are activated, and the “red” gives a stronger signal, the color is orange. If only the “red” cone is activated, the color is red. If we shine both red and green light at a point, the eye does not see both red and green like the ear would hear a chord from two notes, instead it sees yellow – even though there is no yellow color.

Colors: frequencies & wavelengths (in vacuum) • • AM radio 1 MHz (106 Hz)

Colors: frequencies & wavelengths (in vacuum) • • AM radio 1 MHz (106 Hz) 100’s of m FM radio 100 MHz (108 Hz) m’s microwave 10 GHz (1010 Hz) cm - mm Infrared (IR) 1012 - 4 x 1014 Hz mm - 700 nm visible 4 x 1014 - 7. 5 x 1014 Hz 700 nm – 400 nm Ultraviolet (UV) 7. 5 x 1014 – 1017 Hz 400 nm - 1 nm x-ray & ray > 1017 Hz < 1 nm

Different Types of Light If all of the different types of light are all

Different Types of Light If all of the different types of light are all actually E&M waves, what separates the types of light? What separates the types of light are the ways of making the light: Radio uses resonant LRC circuits; Microwaves use klystrons; IR, Visible, and UV use heat (but the sensitivity of the eye distinguishes IR, visible, and UV); and X-rays and Gamma rays use x-ray machines and nuclear processes.

Diffraction Grating: demonstrations • look at the white light source (incandescent light due to

Diffraction Grating: demonstrations • look at the white light source (incandescent light due to hot filament) • look at each of the excited gas sources (one is Helium, one is Mercury)

Interference: Thin Films Before, we had several different parts of a wide beam interfering

Interference: Thin Films Before, we had several different parts of a wide beam interfering with one another. Can we find other ways of having parts of a beam interfere with other parts?

Interference: Thin Films We can also use reflection and refraction to get different parts

Interference: Thin Films We can also use reflection and refraction to get different parts of a beam to interfere with one another by using a thin film. air film water reflected red part interferes with refracted/reflected/refracted blue.

Interference: Thin Films Blue travels an extra distance of 2 t in the film.

Interference: Thin Films Blue travels an extra distance of 2 t in the film. [Actually, the distance travelled in the film is 2 t/cos(qf) rather than 2 t, but we’ll consider qf close to 0 o – near normal incidence. ] air film water reflected red part interferes with refracted/reflected/refracted blue. t

Interference: Thin Films Also, blue undergoes two refractions (into and out of the film)

Interference: Thin Films Also, blue undergoes two refractions (into and out of the film) and reflects off of a different surface (off of the film/water surface rather than off the air/film surface). air film water reflected red part interferes with refracted/reflected/refracted blue. t

Interference: Thin Films When a wave encounters a new medium: – the phase of

Interference: Thin Films When a wave encounters a new medium: – the phase of the refracted wave is NOT affected. – the phase of the reflected wave MAY BE affected.

Interference: Thin Films When a wave on a string encounters a fixed end, the

Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

Interference: Thin Films When a wave on a string encounters a fixed end, the

Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

Interference: Thin Films When a wave on a string encounters a free end, the

Interference: Thin Films When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

Interference: Thin Films When a wave on a string encounters a free end, the

Interference: Thin Films When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

Interference: Thin Films • When light is incident on a SLOWER medium (one of

Interference: Thin Films • When light is incident on a SLOWER medium (one of index of refraction higher than the one it is in), the reflected wave is 180 degrees out of phase with the incident wave. • When light is incident on a FASTER medium, the reflected wave does NOT undergo a 180 degree phase shift.

Interference: Thin Films If na < nf < nw, BOTH red and blue reflected

Interference: Thin Films If na < nf < nw, BOTH red and blue reflected rays will be going from fast to slow, and no difference in phase will be due to reflection. air film water reflected red part interferes with refracted/reflected/refracted blue. t

Interference: Thin Films If na < nf > nw, there WILL be a 180

Interference: Thin Films If na < nf > nw, there WILL be a 180 degree phase difference ( /2)due to reflection. air film water reflected red part interferes with refracted/reflected/refracted blue. t

Interference: Thin Films There will ALWAYS be a phase difference due to the extra

Interference: Thin Films There will ALWAYS be a phase difference due to the extra distance of 2 t/. air film water reflected red part interferes with refracted/reflected/refracted blue. t

Interference: Thin Films When t= /2 the phase difference due to path is 360

Interference: Thin Films When t= /2 the phase difference due to path is 360 degrees (equivalent to no difference). air film water reflected red part interferes with refracted/reflected/refracted blue. t

Interference: Thin Films When t= /4 the phase difference due to path is 180

Interference: Thin Films When t= /4 the phase difference due to path is 180 degrees. air film water reflected red part interferes with refracted/reflected/refracted blue. t

Interference: Thin Films Recall that the light is in the FILM, so the wavelength

Interference: Thin Films Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a/nf. air film water reflected red part interferes with refracted/reflected/refracted blue. t

Interference: Thin Films • reflection: no difference if 180 degree difference if • distance:

Interference: Thin Films • reflection: no difference if 180 degree difference if • distance: no difference if 180 degree difference if n f < n w; n f > n w. t = f/2 = a/2 nf t = f/4 = a/4 nf • Total phase difference is sum of the above two effects.

Interference: Thin Films Total phase difference is sum of the two effects of distance

Interference: Thin Films Total phase difference is sum of the two effects of distance and reflection • For minimum reflection, need total to be 180 degrees or 180 o + 360 o. – anti-reflective coating on lens • For maximum reflection, need total to be 0 degrees or 360 o. – colors on oil slick

Thin Films: an example An oil slick preferentially reflects green light. The index of

Thin Films: an example An oil slick preferentially reflects green light. The index of refraction of the oil is 1. 65, that of water is 1. 33, and of course that of air is 1. 00. What is the thickness of the oil slick?

Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase

Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).

Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase

Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees). Since we have nf > nw, we have 180 degrees due to reflection.

Thin Films: an example • Since we have preferentially reflected green light, the TOTAL

Thin Films: an example • Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees). • Since we have nf > nw, we have 180 degrees due to reflection. Therefore, we need 180 degrees due to extra distance, so need t = a/4 nf where a = 500 nm, nf = 1. 65, and so t = 500 nm / 4(1. 65) = 76 nm.

Michelson Interferometer Split a beam with a Half Mirror, then use mirrors to recombine

Michelson Interferometer Split a beam with a Half Mirror, then use mirrors to recombine the two beams. Note that both beams will be refracted once and reflected twice, so no phase difference due to refraction. Mirror Half Mirror Light source Mirror Screen

Michelson Interferometer If the red beam goes the same length as the blue beam,

Michelson Interferometer If the red beam goes the same length as the blue beam, then the two beams will constructively interfere and a bright spot will appear on screen. Mirror Half Mirror Light source Mirror Screen

Michelson Interferometer If the blue beam goes a little extra distance, s, the screen

Michelson Interferometer If the blue beam goes a little extra distance, s, the screen will show a different interference pattern. Mirror Light source Half Mirror Screen s

Michelson Interferometer If s = /4, then the interference pattern changes from bright to

Michelson Interferometer If s = /4, then the interference pattern changes from bright to dark. Mirror Light source Half Mirror Screen s

Michelson Interferometer If s = /2, then the interference pattern changes from bright to

Michelson Interferometer If s = /2, then the interference pattern changes from bright to dark back to bright (a fringe shift). Mirror Light source Half Mirror Screen s

Michelson Interferometer By counting the number of fringe shifts, we can determine how far

Michelson Interferometer By counting the number of fringe shifts, we can determine how far s is! Mirror Light source Half Mirror Screen s

Michelson Interferometer If we use the red laser ( =632 nm), then each fringe

Michelson Interferometer If we use the red laser ( =632 nm), then each fringe shift corresponds to a distance the mirror moves of /2 = 316 nm (about 1/3 of a micron)! Mirror Light source Half Mirror Screen s

Michelson Interferometer We can also use the Michelson interferometer to determine the index of

Michelson Interferometer We can also use the Michelson interferometer to determine the index of refraction of a gas (such as air). Put a cylinder with transparent ends into one of the beams.

Michelson Interferometer • Evacuate the cylinder with a vacuum pump • Slowly allow the

Michelson Interferometer • Evacuate the cylinder with a vacuum pump • Slowly allow the gas to seep back into the cylinder and count the fringes. Mirror Light source Half Mirror cylinder Screen

Michelson Interferometer In vacuum, #v v = 2 L where # indicates the number

Michelson Interferometer In vacuum, #v v = 2 L where # indicates the number of wavelengths. In the air, #a a = 2 L. Since va < c, a < v and #a > #v. 1. Knowing v and L, can calculate #v. (continued on the next slide)

Michelson Interferometer 1. Knowing v and L, can calculate #v. 2. By counting the

Michelson Interferometer 1. Knowing v and L, can calculate #v. 2. By counting the number of fringe shifts, we can determine #. 3. Since # = #a - #v , we can calculate #a. 4. Now knowing L and #a, we can calculate a.

Michelson Interferometer We now know v and a, so: with vf = c and

Michelson Interferometer We now know v and a, so: with vf = c and af = va , we can use na = c/va = vf / af = v / a.

Michelson Interferometer an example If L = 6 cm, and if v = 632

Michelson Interferometer an example If L = 6 cm, and if v = 632 nm, and if 50 fringes are counted when air is let back into the cylinder, then: #v = 2 L/ v = 2 *. 06 m / 632 x 10 -9 m = 189, 873 #a = #v + # = 189, 873 + 50 = 189, 923 a = 2 L/#a = 2 *. 06 m / 189, 923 = 631. 83 nm na = v / a = 632. 00 nm / 631. 83 nm = 1. 00026