Diffraction single slit How can we explain the

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Diffraction: single slit How can we explain the pattern from light going through a

Diffraction: single slit How can we explain the pattern from light going through a single slit? screen w x L

Diffraction: single slit If we break up the single slit into a top half

Diffraction: single slit If we break up the single slit into a top half and a bottom half, then we can consider the interference between the two halves. screen w x L

Diffraction: single slit The path difference between the top half and the bottom half

Diffraction: single slit The path difference between the top half and the bottom half must be l/2 to get a minimum. screen x w L

Diffraction: single slit This is just like the double slit case, except the distance

Diffraction: single slit This is just like the double slit case, except the distance between the “slits” is w/2, and this is the case for minimum: (w/2) sin(q) = l/2 screen x w L

Diffraction: single slit In fact, we can break the beam up into 2 n

Diffraction: single slit In fact, we can break the beam up into 2 n pieces since pieces will cancel in pairs. This leads to: (w/2 n) sin(qn) = l/2 , or w sin( n) = n for MINIMUM. screen x w L

Diffraction: single slit REVIEW: • For double (and multiple) slits: n = d sin(

Diffraction: single slit REVIEW: • For double (and multiple) slits: n = d sin( n) for MAXIMUM (for ALL n) • For single slit: n = w sin( n) for MINIMUM (for all n EXCEPT 0)

Diffraction: single slit NOTES: • For double slit, bright spots are equally separated. •

Diffraction: single slit NOTES: • For double slit, bright spots are equally separated. • For single slit, central bright spot is larger because n=0 is NOT a dark spot. • To have an appreciable , d and w must be about the same size as & a little larger than . • Recall that for small angles, sin(q) = tan(q) = x/L

Diffraction: single slit -2 -1 0 -2 -1 1 2 REVIEW: • For double

Diffraction: single slit -2 -1 0 -2 -1 1 2 REVIEW: • For double (and multiple) slits: n = d sin( n) for MAXIMUM (for ALL n) 0 1 2 • For single slit: n = w sin( n) for MINIMUM (for all n EXCEPT 0)

Diffraction: circular opening If instead of a single SLIT, we have a CIRCULAR opening,

Diffraction: circular opening If instead of a single SLIT, we have a CIRCULAR opening, the change in geometry makes: • the single slit pattern into a series of rings; and • the formula to be: 1. 22 n = D sin( n) • Computer Homework CH 5 -4 shows the rings in several diagrams and the use of this equation.

Diffraction: circular opening Since the light seems to act like a wave and spreads

Diffraction: circular opening Since the light seems to act like a wave and spreads out behind a circular opening, and since the eye (and a camera and a telescope and a microscope, etc. ) has a circular opening, the light from two closely spaced objects will tend to overlap. This will hamper our ability to resolve the light (that is, it will hamper our ability to see clearly).

Limits for Resolving Ability The dots represent either cone cells or diffraction spots. Can

Limits for Resolving Ability The dots represent either cone cells or diffraction spots. Can you tell what letter this is?

e Limits for Resolving Ability The dots represent either cone cells or diffraction spots.

e Limits for Resolving Ability The dots represent either cone cells or diffraction spots. You can see from the filled green dots that the letter is an e.

Limits for Resolving Ability Can you tell what letter this is?

Limits for Resolving Ability Can you tell what letter this is?

Limits for Resolving Ability e You can’t tell from the green dots that the

Limits for Resolving Ability e You can’t tell from the green dots that the letter is an e.

Diffraction: circular opening How close can two points of light be to still be

Diffraction: circular opening How close can two points of light be to still be resolved as two distinct light points instead of one? One standard, called the Rayleigh Criterion, is that the lights can just be resolved when the angle of separation is the same as the angle of the first dark ring of the diffraction pattern of one of the points: limit = 1 from 1. 22 * = D sin( 1).

Rayleigh Criterion: a picture The lens will focus the light to a fuzzy DOT

Rayleigh Criterion: a picture The lens will focus the light to a fuzzy DOT rather than a true point. D lens

Rayleigh Criterion: a picture The Rayleigh minimum angle, limit = sin-1(1. 22 /D) and

Rayleigh Criterion: a picture The Rayleigh minimum angle, limit = sin-1(1. 22 /D) and tan( limit) = x’ / s’. D lens x’ s’

Rayleigh Criterion: a picture If a second point of light makes an angle of

Rayleigh Criterion: a picture If a second point of light makes an angle of qlimit with the first point, then it can just be resolved. D lens x x’ s s’

Rayleigh Criterion: a picture In this case: qlimit = sin-1(1. 22 l/D) and for

Rayleigh Criterion: a picture In this case: qlimit = sin-1(1. 22 l/D) and for small angles qlimit(in radians) tan(qlimit) = x’/s’ = x/s. D lens x x’ s s’

Rayleigh Criterion: an example • • Consider the (ideal) resolving ability of the eye

Rayleigh Criterion: an example • • Consider the (ideal) resolving ability of the eye Estimate D, the diameter of the pupil Use l = 550 nm (middle of visible spectrum) Now calculate the minimum angle the eye can resolve. • Now calculate how far apart two points of light can be if they are 5 meters away.

Rayleigh Criterion: an example • with D = 5 mm and l = 550

Rayleigh Criterion: an example • with D = 5 mm and l = 550 nm, qlimit = sin-1 (1. 22 x 5. 5 x 10 -7 m/. 005 m) = 1. 34 x 10 -4 radians = 7. 7 x 10 -3 degrees =. 46 arc minutes so x/s = tan(qlimit), and so x = 5 m * tan(7. 7 x 10 -3 degrees) =. 67 mm

Rayleigh Criterion: an example • Estimate how far it is from the lens of

Rayleigh Criterion: an example • Estimate how far it is from the lens of the eye to the retinal cells on the back of the eye. • With your same D and l (and so same qlimit), now calculate how far the centers of the two dots of light on the retina are. • How does this distance compare to the distance between retinal cells (approx. diameter of the cells)?

Rayleigh Criterion: an example • s’ = 2 cm (estimation of distance from lens

Rayleigh Criterion: an example • s’ = 2 cm (estimation of distance from lens to retinal cells) • from previous part, qlimit = 7. 7 x 10 -3 degrees • so x’ = 2 cm * tan(7. 7 x 10 -3 degrees) = 2. 7 mm.

Limits on Resolution: • Imperfections in the eye (correctable with glasses) • Rayleigh Criterion

Limits on Resolution: • Imperfections in the eye (correctable with glasses) • Rayleigh Criterion due to wavelength of visible light • Graininess of retinal cells

Limits on Resolution: further examples • hawk eyes and owl eyes eagle eyes have

Limits on Resolution: further examples • hawk eyes and owl eyes eagle eyes have 450, 000 cone cells per mm 2 compared to humans who have 200, 000 cone cells per mm 2. • cameras: – lenses (focal lengths, diameters) – films (speed and graininess) – shutter speeds and f-stops • Amt of light a D 2 t • f-stop = f/D – f-stops & resolution: resolution depends on D

Limits on Resolution: another example: Microscope 1. 22 n = D sin( n) where

Limits on Resolution: another example: Microscope 1. 22 n = D sin( n) where q 1 = qlimit hlimit/s 1. 22λ /D; 1/s + 1/s’ = 1/f and M = s’/s, so 1/s’ = M/s, so 1/s + 1/s’ = 1/s + M/s = (M+1)/s = 1/f, or f = s/(M+1), so s = f(M+1); [(ngl-1)/1]*[(1/R 1)+(1/R 2)] = 1/f, and if R 1=R 2 and ngl=2, then 1/f = 2/R, so D < 2 R, so f D/4; finally hlimit/s 1. 22λ /D or hlimit λs/D = λ(M+1)f/D = λ(M+1)(D/4)/D = λ(M+1)/4 microscopes: smallest angle, qlimit, is limited by λ, so smallest size on object is also limited by λ. Net result is: smallest size resolvable » l =. 5 mm – can easily see. 5 mm, so M-max = 1, 000 – can reduce l by immersing in oil, use blue light

Limits on Resolution: another example: the Telescope Light from far away is almost parallel.

Limits on Resolution: another example: the Telescope Light from far away is almost parallel. Remember that light leaving the telescope will also be almost parallel but the eye will focus that light on the retinal cells to form a point of light. eyepiece objective lens eye Light from star far away fobj feye Starlight focused on retina

Limits on Resolution: the Telescope The telescope collects and concentrates light which the eye

Limits on Resolution: the Telescope The telescope collects and concentrates light which the eye then focuses on the retinal cells. eyepiece objective lens eye fobj feye

Limits on Resolution: the Telescope Light coming in at an angle, in , is

Limits on Resolution: the Telescope Light coming in at an angle, in , is magnified up to out. Note that the object is not magnified, rather the incoming angle is magnified! eyepiece objective lens eye x fobj feye

Limits on Resolution: the Telescope in = x/fo, out = x/fe; M = out/

Limits on Resolution: the Telescope in = x/fo, out = x/fe; M = out/ in = fo/fe eyepiece objective lens eye x fobj feye

Limits on Resolution: the Telescope • telescopes – magnification: M = qout/qin = fo

Limits on Resolution: the Telescope • telescopes – magnification: M = qout/qin = fo /fe – light gathering: Amt a D 2 – resolution: 1. 22 l = D sin(qlimit) so qin = qlimit and qout = 5 arc minutes so qlimit a 1/D implies Museful = 60 * D where D is in inches – surface must be smooth on order of l

Limits on Resolution: Telescope • Mmax useful = qout/qin = qeye/qlimit = 5 arc

Limits on Resolution: Telescope • Mmax useful = qout/qin = qeye/qlimit = 5 arc min / (1. 22 * l / D) radians = (5/60)*(p/180) / (1. 22 * 5. 5 x 10 -7 m / D) = (2167 / m) * D * (1 m / 100 cm) * (2. 54 cm / 1 in) = (55 / in) * D

Limits on Resolution: Telescope Example What diameter telescope would you need to read letters

Limits on Resolution: Telescope Example What diameter telescope would you need to read letters the size of license plate numbers from a spy satellite?

Limits on Resolution: Telescope Example • need to resolve an “x” size of about

Limits on Resolution: Telescope Example • need to resolve an “x” size of about 1 cm • “s” is on order of 100 miles or 150 km • qlimit then must be (in radians) = 1 cm / 150 km = 7 x 10 -8 • qlimit = 1. 22 x 5. 5 x 10 -7 m / D = 7 x 10 -8 so D = 10 m (Hubble has a 2. 4 m diameter)

Limits on Resolution: Telescope & Satellite Dish • other types of light – x-ray

Limits on Resolution: Telescope & Satellite Dish • other types of light – x-ray diffraction (use atoms as slits) – IR – radio & microwave • surface must be smooth on order of l

Wave Interference with Sound: Noise Cancellation Headsets How do noise cancelling headsets work?

Wave Interference with Sound: Noise Cancellation Headsets How do noise cancelling headsets work?

Noise Cancellation How do noise cancelling headphones work? You can try to do this

Noise Cancellation How do noise cancelling headphones work? You can try to do this passively by absorbing (some of) the outside noise in ear covers. Headphones will do some of this. In general, waves penetrate a material but decay exponentially as they penetrate. The amount of decay depends on the material, but also on the depth of penetration in relation to the wavelength. That means high frequency waves (with short wavelengths) are absorbed better than low frequency waves (with long wavelengths). In water, the higher speed of sound means that the same frequencies in air will have a longer wavelength in water than in air and so will be able to go further in water. With the longer wavelengths, though, there is more diffraction and so locating something by sound will be less accurate.

Noise Cancellation You can also use destructive interference to do this actively by using

Noise Cancellation You can also use destructive interference to do this actively by using a microphone to pick up the outside noise, an electrical circuit to analyze and delay it, and then a transmitter (speaker) to transmit that noise that is delayed by half a period, so you get cancellation of the outside noise. This combination is actually called a headset, not a headphone. Since high frequency sound is reduced by absorption, we mainly have to worry about low frequency sound. The low frequency sound has a longer period, and so provides more time for the microphone, delaying circuit, and speaker to respond.

Noise Cancellation It sounds strange that you get less sound by emitting sound, but

Noise Cancellation It sounds strange that you get less sound by emitting sound, but that can happen with wave interference. The extra energy supplied by the transmitter is converted into heat in this cancellation process, but the extra sound energy is very small. Note: the wave cancellation headsets require a battery. Headphones do not since they have the signal with its energy supplied by the sound source (which does have a battery).

Polarization • Experiment with polarizers • Particle Prediction? • Wave Prediction? – Electric Field

Polarization • Experiment with polarizers • Particle Prediction? • Wave Prediction? – Electric Field is a vector: 3 directions • Parallel to ray (1 longitudinal) (Maxwell’s Equations forbid longitudinal) • Two Perpendicular to ray (2 transverse)

Polarization: Wave Theory #1 Polarization by absorption (Light is coming out toward you) unpolarized

Polarization: Wave Theory #1 Polarization by absorption (Light is coming out toward you) unpolarized light polarizer only lets vertical component through no light gets through polarizer only lets horizontal component through

Polarization: Wave Theory Three polarizers in series: Sailboat analogy: North wind sail force on

Polarization: Wave Theory Three polarizers in series: Sailboat analogy: North wind sail force on sail boat goes along direction of keel

Polarization: Wave Theory #2 Polarization by reflection – Brewster Angle: when refracted + reflected

Polarization: Wave Theory #2 Polarization by reflection – Brewster Angle: when refracted + reflected = 90 o – Sunglasses and reflected glare incident ray vertical horizontal reflected ray no problem with horizontal almost no vertical since vertical is essentially longitudinal now surface refracted ray vertical can be transmitted

Polarization: Wave Theory #3 Polarization by double refraction – due to different bondings in

Polarization: Wave Theory #3 Polarization by double refraction – due to different bondings in different directions in the crystal, we have different indices of refraction in different polarization directions. incident ray vertical horizontal surface refracted rays One polarization is bent differently than the other

Polarization: Wave Theory #4 Polarization by scattering (more about this in section three on

Polarization: Wave Theory #4 Polarization by scattering (more about this in section three on Compton Scattering)