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

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. w screen L x

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

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(  ) = /2 w screen L x

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

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. 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  ) = tan(  ) = x/L

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 ) 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 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).

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 rather than a true point. lens D

Rayleigh Criterion: a picture The Rayleigh minimum angle,  limit = sin -1 (1.22 /D) = x’ / s’. lens D s’ x’

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

Rayleigh Criterion: a picture In this case:  limit = sin -1 (1.22 /D) = x’/s’ = x/s. lens D s’ x’ s x

Rayleigh Criterion: an example Consider the (ideal) resolving ability of the eye Estimate D, the diameter of the pupil Use = 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 = 550 nm,  limit = sin -1 (1.22 x 5.5 x m/.005 m) = 7.7 x degrees =.46 arc minutes so x/L = tan(  limit ), and x = 5m * tan(7.7 x degrees) =.67 mm

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   and so same  limit ), 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 L = 2 cm (estimation of distance from lens to retinal cells) from previous part,  limit = 7.7 x degrees so x = 2 cm * tan(7.7 x degrees) = 2.7  m.

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 cameras: –lenses (focal lengths, diameters) –films (speed and graininess) –shutter speeds and f-stops Amt of light  D 2 t f-stop = f/D –f-stops & resolution: resolution depends on D

Limits on Resolution: further examples 1.22 n = D sin(  n ) where  1 =  limit microscopes: smallest size =  =.5  m –can easily see.5 mm, so M-max = 1000 –can reduce by immersing in oil, use blue light

Limits on Resolution: further examples: the telescope Light from far away is almost parallel. objective lens eyepiece f obj f eye

Limits on Resolution: further examples The telescope collects and concentrates light objective lens eyepiece f obj f eye

Limits on Resolution: further examples Light coming in at an angle,  in is magnified. objective lens eyepiece f obj f eye x

Limits on Resolution: further examples  in = x/f o,  out = x/f e ; M =  out /  in = f o /f e objective lens eyepiece f obj f eye x

Limits on Resolution: further examples telescopes –magnification: M =  out /  in = f o /f e –light gathering: Amt  D 2 –resolution: 1.22 = D sin(  limit ) so  in =  limit and  out = 5 arc minutes  so  limit  1/D implies M useful = 60 * D where D is in inches –surface must be smooth on order of

Limits on Resolution: Telescope M max useful =  out /  in =  eye /  limit = 5 arc min / (1.22 * / D) radians = (5/60)*(  /180) / (1.22 * 5.5 x 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 the size of license plate numbers from a spy satellite?

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  limit then must be (in radians) = 1 cm / 150 km = 7 x  limit = 1.22 x 5.5 x m / D = 7 x so D = 10 m (Hubble has a 2.4 m diameter)

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

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

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

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 = 90 o – Sunglasses and reflected glare surface incident ray vertical horizontal refracted ray reflected ray no problem with horizontal almost no vertical since vertical is essentially longitudinal now vertical can be transmitted

Polarization: Wave Theory #3 Polarization by double refraction –different n’s in different directions due to different bonding #4 Polarization by scattering