a and g power point

8/7/2019 A and G Power Point

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A.1.1 Describe the basic structure of the

human eye. The human eye is not simply a ball of mush and magic thatallows us to see. On the contrary, it is actually a very complexstructure. The eye actually consists of three layered structures.The first layer is made up of the cornea (which refracts lightwith the lens and serves as the overall focus) and the sclera(the white of the eye). The middle layer of the eye is made up

of the choroid, the ciliary body (aqueous humourwaterysubstance between the lens and cornea), and the iris (thecolored part of the eye which controls the size of the pupilsand how much light can enter). The third layer is the retinawhich at its core is made up of a thin layer of tissue thatreceive chemical and electrical impulses that are then sentto the visual parts of the brain through the optic nerve. Soyeah, the eye is a pretty cool human feature!


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A.1.2 State and explain the process ofdepth of vision and accommodation.

Having two eyes enables us to obtain two images of

the same thing and then the brain is then able tojudge distance and motion. The size of any imageformed on the retina of the eye depends on theangle subtended by the object at the eye. The closerthe object is the greater the angle and therefore theangular magnification. The near point and far point

are important terms one must know to be able to fullyunderstand the eye and sight. The near point is theclosest position that an object can be seen in focusby the naked eye. The far point is the farthest positionthat an object can be seen in focus by the nakedeye.


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A.1.3 State that the retina contains rods andcones, and describe the variation in densityacross the surface of the retina.

The retina contains rods and cones whichare photoreceptors with complementaryproperties. Rods: fast response rates,sensitive to low light levels, insensitive tocolor. Cones: slow response rates, insensitiveto low light levels, sensitive to particularwavelengths of light and allow us to havecolor vision. There are far more rods thancones in the retina; 120 million compared to6.5 million.


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A.1.4 Describe the function of the rods andof the cones in photopic and scotopicvision.

Rods are utilized during scotopic vision,which is vision at low light levels. Howeverthey do not mediate color and thereforeare considered to have low spatial

resolution. Rods are great photoreceptorsbecause of their high sensitivity.

Cones are used during scotopic vision,which is vision at high light levels. Cones are

able to pick up color.


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A.1.5 Describe color mixing of light by

addition and subtraction.

All visible colors can be seen by mixingthe three primary colors of light by

addition or subtraction. Addition of lightoccurs when light is added to a dark

background. Subtraction of light occurs

when pigments are used to selectively

block out white light.


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A.1.6 Discuss the effect of light and dark,

and color, on the perception of objects.

Certain connotations come along with

light and dark and different colors. Theseconnotations and actually discrepancies

create varying ways people perceive

object based on the way they physicallysee them.


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A.2.1 Describe the nature ofstanding (stationary) waves.

A standing wave is a wave thatremains in a constant position.This phenomenon can occurbecause the medium is movingin the opposite direction to thewave, or it can arise in astationary medium as a result of

interference between twowaves traveling in oppositedirections.


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A.2.2 Explain the formation of one-dimensional standing waves.

One-dimensional standing waves areformed when two waves of the same speedand wavelength and equal or almost equalamplitudes, travelling in opposite directionsmeet.

A standing wave has nodes, points at whichthe displacement is always zero, andantinodes, points at which displacement isat maximum.


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A.2.4 Compare standing waves and

travelling waves.

Property Travelling Wave Standing Wave

Energy

Propagation

Propagated Not Propagated

Amplitude SingleAmplitude

VariableAmplitude

Phase

Difference

All phase

differencesbetween 0 and2

Only 0, 2, and phase difference


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A.2.5 Solve problems involving standingwaves.


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A.3.2 Explain the Doppler effect byreference to wavefront diagrams formoving-detector and moving-source

situations. If the source and observer are moving

towards each other the frequency will goup since the wavefronts will be closer to

one another (the relative speed of soundcompared to the observer is decreasing).

If they are moving away from each otherthe opposite applies.


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A.3.3 Apply the Doppler effect equations

for sound.

Object moving towards

source.

Object moving away

from source.


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A.3.5 Solveproblems on theDoppler effect forelectromagneticwaves using the

approximationf=f*(v/c).


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A.3.6 Outline anexample in which theDoppler effect is usedto measure speed.

In blood-flowmeasurements a socalled Dopplervelocimeter can sendout light beam bursts,and measure theDoppler shift inwavelengths ofreflections fromparticles moving withthe flow.


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A.4.1 Sketch the variation with angle of

diffraction of the relative intensity of light

diffracted at a single slit.


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A.4.3 Solve problems involving single-slitdiffraction.


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A.5.1 Sketch the variation with angle of

diffraction of the relative intensity of lightemitted by two point sources that has

been diffracted at a single slit.


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A.5.4 Solve problems involving resolution.


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A.6.1 Describe what is meant by

polarized light.

Polarized light is light whose waves onlyoscillate along one plane.


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A.6.2 Describe polarization by reflection.Reflected light is partially polarized such that

oscillations are (to a degree) parallel to the

surface of reflection, depending upon its

nature. Light reflected off non-metallicsubstances is polarized such that a large

quantity of the wave oscillations occur

parallel to the water-surface.


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A.6.3 State and apply Brewsters law. Brewsters angle is an angle of incidence

at which light with a particularpolarization is perfectly transmitted

through a surface, with no reflection.


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A.6.4 Explain the termspolarizer and analyzer. Polarizer: a sheet of

material with molecularstructure that only allows

a specific orientation ofthe electric field to gothrough.

Analyzer: the secondpolarizer when two is

placed in a row,perpendicular to eachother (used for thepurpose of determiningwhether light is polarized).


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A.6.5 Calculate theintensity of atransmitted beam ofpolarized light usingMalus law.

Malus law states thatwhen a perfectpolarizer is placed in apolarized beam oflight, the intensity, I, ofthe light that passesthrough is given by:

(Where = the anglebetween polarizationdirection andpolarizer.)


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A.6.6 Describe what is meant by anoptically active substance.

Optically active substances are

substances that rotate the plane of

polarization.


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A.6.7 Describe theuse of polarization inthe determination ofthe concentration of

certain solutions.

The optical activity of a

substance depends upon itsmolecular study. Therefore,

measures of the polarization-properties of a solution (e.g.

sugar solution) can help to revealthe molecular structure of the

solute.


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A.6.8 Outlinequalitatively howpolarization may beused in stress analysis.

For certain materials,

the degree to which thesubstance becomesoptically active isproportional to thestress. Therefore

examination of thepolarization patternoccurring through thismaterial will giveinformation as to itslevel/state of stress.


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A.6.9 Outline qualitatively the actionof liquid-crystal displays (LCDs).

A pixel of an LCD is a cell of alignedliquid crystal between twotransparent plates, between a

crossed polarizer and analyzer (that is,their axes of transmission areperpendicular to each other). Whenthere is no electric field applied, theplane of polarization twists as the lightpasses through the crystals (they arearranged as a helix); when it emerges,it is polarized parallel to the

transmission axes of the analyzer, andso exits the cell appearing light. Whenan electric field is applied, the liquidcrystals are aligned with each othersuch that when the light emerges, it isstill polarized perpendicular to theanalyzer, and thus is not transmitted.This creates a dark spot.


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A.6.10 Solve problems involving the

polarization of light.


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G.1.1 Outline the nature of electromagnetic(EM) waves.

Light is an EM wave. The speed of light isindependent of the velocity of the sourceof the light.

Electromagnetic waves consists ofoscillating magnetic and electric fields thatare at 90 degrees to each other and inphase. This can be visualized as twotransverse waves perpendicular to eachother, propagating in the same direction.


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G.1.5 Distinguish between transmission,absorption and scattering of radiation.

The wave is transmitted if it goes straight

through The wave is absorbed if the substances in

the medium (molecules in air) absorb itsenergy. This energy can be re-emitted.

The wave is scattered if it bounces ofparticles in the medium into randomdirections.

Waves can actually be partiallytransmitted/absorbed/scattered.


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G.1.6 Discuss examples of the transmission, absorption andscattering of EM radiation.

During the day: blue light is scattered by small dustparticles in the atmosphere. If there was no atmosphere,the sky would be black.

During sunsets: the light must travel through moreatmosphere. Blue light is scattered more than red light

(because it has a shorter wavelength) so the blue light isscattered too much before it reaches you, therefore weonly see the red light.

Other examples include; UV light being absorbed by theozone layer and infra-red radiation being absorbed bygreenhouse gases (causing global warming).


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G.1.7 Explain the termsmonochromatic and coherent.

Monochromatic (mono=one,

chromatic=color) light is of a smallrange of wavelengths, forexample from 500-501 nm. Lightbulbs are NOT monochromatic aswhite light is a mixture of manydifferent frequencies.

Coherent light means the wavesare linked together, that is all thephotons emitted are in phase withone another.


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G.1.8 Identify laser light as a source of

coherent light.

DUH?!


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G.1.9 Outline the mechanism for the production oflaser light.

"Light Amplification by Stimulated Emission of

Radiation". Stimulated emission, refers to a process where

'electrons that are in an excited state' emit photonsand drop to a lower state. This is caused by anotherphoton passing by the excited electron. By nature,both photons are of the same wavelength and are in

phase. Lasers work by pumping lots of energy into the laser

such that over 50% of the electrons in a medium arein an excited state. It then has two mirrors, one ofwhich lets 1% of photons through.


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G.1.10 Outline anapplication of theuse of a laser.

Lasers are used toread informationon CDs by sensingpits and bumps

that can beinterpreted andchanged intosound.


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G.2.2 Define the power of a convex lens

and the dioptre.

Power is the reciprocal of the focal

length of the lens, measured in dioptres.


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G.2.6 Apply theconvention "real ispositive, virtual isnegative" to the thin

lens formula. (1/S1) + (1/S2) = (1/f)

If S1 > f then it will mostlikely be a real image.

If S1 < f then it will mostlikely be a virtualimage.


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G.2.7 Solve problems for a single convex

lens using the thin lens formula.


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G.2.9 Define angular magnification.

Angular magnification: ratio of the angle

subtended by the image at the eye withthe instrument to the angle subtended

by the object at the eye without theinstrument.


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G.2.10 Derive an expression for theangular magnification of a simplemagnifying glass formed at the near

point and at infinity.


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G.2.11 Construct a ray diagram for a

compound microscope with final imageformed close to the near point of the

eye (normal adjustment).


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G.2.12 Construct a ray diagram for an

astronomical telescope with the final

image at infinity (normal adjustment).


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G.2.13 State the equation relating

angular magnification to the focal

lengths to lenses in an astronomicaltelescope in normal adjustment.


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G.2.14 Solve problems involving thecompound and the astronomical

telescope.


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G.2.15 Explain the meaningof spherical aberration and

of chromatic aberration asproduced by a single lens.

Spherical aberration: rays far

from the lens axis do notfocus at the focal point.

Chromatic aberration: lightof different wavelengths hasdifferent indices of refraction

and focuses at differentpoints.


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G.2.16 Describe how spherical

aberration in a lens may be reduced.

The solution to reduce spherical

aberration in a lens is to use acompound-lens systems (combination ofsimple lenses) or to use only central part

of lens.


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G.2.17 Describe how chromatic

aberration in a lens may be reduced.

The solution to reduce chromatic

aberration in a lens is to use anachromatic doublet, made of lenses oftwo different materials.


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G.3.1 State the conditions necessary toG.3.1 State the conditions necessary toobserve interference between twoobserve interference between twosources.sources.

Conditions necessary:Conditions necessary: Same frequency, wavelength, & phaseSame frequency, wavelength, & phase

Standing wave pattern (nodes and antinodes)Standing wave pattern (nodes and antinodes)

CoherentCoherent----the waves maintain a constantthe waves maintain a constantphase angle between them.phase angle between them.

MonochromaticMonochromatic----the waves have thethe waves have thesame wavelengthsame wavelength


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G.3.2 Explain, by means of the principle of superposition, theinterference pattern produced by waves from two coherentpoint sources.

The superposition principle states that, for all

linear systems, the net response at a given placeand time caused by two or more stimuli is the sumof the responses which would have been causedby each stimulus individually.

This is important to understand the concept ofsuperposition to determine if the sources weremonochromatic or coherent based on theirinterference pattern and where the constructiveand destructive interference occurs.


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G.3.4 Solve problems involving two-

source interference.


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G.4.1 Describe the effect on the double-slit intensity distribution of increasing the

number of slits.

As the number of slit increase

the bright fringes becomemore concentrated and

areas of separation betweenbright fringes become evenmore dark. The total amountof light stays the same butwith higher peaks and darkerdark fringes.


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G.4.2 Derive the diffraction grating formula for normalincidence.

Parallel rays of a monochromatic light of wavelength 1

are incident on a diffraction grating in which the slitseparation is d. If the grating has N lines per meter, the

grating spacing is given by: d = (1/N) Light from A must be in phase with light from B, and this

can only happen when the path difference is a wholenumber of complete wavelengths (even number of

half-wavelengths).

So: AC = nl, where n = 0, 1, 2, 3...

AC = d sin q. and so d sin q = nl


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G.4.3 Outline the use of a diffraction gratingto measure wavelengths.

A diffraction grating can be used to split light into differentwavelengths with a high degree of accuracy, much more sothan glass prisms. A diffraction grating usually consists of a

piece of glass with very closely spaced lines ruled on it. The diffraction grating has the advantage over the double slit

method of measuring wavelength in that:

-the maxima are more sharply defined;

-the beam passes through more slits than two, so the intensityis brighter;

-the angles are larger so that they can be measured withgreater precision.


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G.5.1 Outline the experimental arrangement for theproduction of X-rays.

X-rays are produced by accelerating electrons with a

high voltage and allowing them to collide with ametal target. X-rays are produced when the electronsare suddenly decelerated upon collision with themetal target. If the bombarding electrons haveenough energy, they can knock an electron out of aninner shell of the target metal atoms. Then electrons

from higher states drop down to fill the vacancy,emitting x-ray photons with precise energiesdetermined by the electron energy levels.


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G.5.3 Explain the origins of the features of a characteristic X-rayspectrum.

Continuous features: As the incoming electrons collide withthe target atoms, they are decelerated. This decelerationmeans that x-rays are emitted. The energy of the x-rayphoton depends on the energy lost in the collisions. Themaximum amount of energy that can be lost is al the initial

kinetic energy of the electrons. The maximum energyavailable means that there is a maximum frequency of x-raysproduced. This corresponds to a minimum wavelength limitshown on the graph.

Characteristic features: In some circumstances, the collisionsbetween the incoming electrons and the target atoms cancause electrons from the inner orbital of the target atom tothe promoted up to a higher energy levels. When theseelectrons fall back down they emit x-rays of a particularfrequency which is fixed by the energy levels available.


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G.5.5 Explain how X-ray diffraction arisesfrom the scattering of X-rays in a crystal.

Crystals have spacing between their

layers that is ideal for diffracting X-rays,as can be seen in the diagram below.


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G.5.7 Outline how cubic crystals may beused to measure the wavelength of X-rays.

When X-rays are scattered from a crystallattice, peaks of scattered intensity areobserved which correspond to thefollowing conditions:

- The angle of incidence = angle ofscattering.- The path length difference is equal toan integer number of wavelengths.


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G.5.8 Outline how X-rays may be used to determine thestructure of crystals.

When X-rays are incident on a regular structure the majorityof X-rays will pass through the material. At particular angles

from the straight through direction, high intensity X-ray signalsare recorded.These angles correspond to points of constructiveinterference of the X-rays scattered from different planes ofatoms in the crystal.

A regular crystal structure contains many different

lattice planes that can cause the interference sotypically a crystal structure will give rise to manydifferent constructive interference positions. Apowdered sample of crystal will contain everyorientation of these planes so the resulting X-raydiffraction picture will contain circles rather than points.


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G.5.9 Solve problems involving the Bragg

equation.


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G.6.1 Explain the production of

interference fringes by a thin air wedge.

There is a path difference betweenthe rays of light reflecting from the topand from the bottom surfaces of the

wedge. This results in parallel lines ofequally spaced constructive and

destructive interference fringes.


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G.6.2 Explain howwedge fringes can beused to measure very

small separations. Since the eye has a

small aperture thesefringes, unless viewed at

near to normalincidence (=0), willonly be observed if thefilm is very thin.


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G.6.7 State the conditions for constructive

and destructive interference.

Constructive interference occurs when

the waves are in phase.

Destructive interference occurs whenthe waves are out of phase.


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G.6.8 Explain the formation of colored fringeswhen white light is reflected from thin films,such as oil and soap films.

If white light is used then the situationbecomes more complex. Provided thethickness of the film is small, then one ortwo colors may reinforce along adirection in which others cancel. The

appearance of the film will be brightcolors, such as can be seen when lookingat an oil film on the surface of water orsoap bubbles.


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G.6.9 Describe the difference betweenfringes formed by a parallel film and awedge film.

For a parallel film, the fringes are of equalinclination, that is they form arcs of acircle whose center is located at the endof a perpendicular drawn from the eye tothe surface of the film.

For a wedge film, the fringes are paralleland of equal thickness.


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G.6.10 Describe applications of parallel thin films.

The design of non-reflecting radar coatings for military aircraft.If the thickness of the extra contain is designed so that radarsignals destructively interfere when they reflect from bothsurfaces, then no signal will be reflected and an aircraft could

go undetected. Measurements of thickness of oil slicks caused by spillage.

Measurements of the wavelengths of electromagnetic signalsthat give constructive and destructive interference (at knownangles) allow the thickness of the oil to be calculated.

Design of non-reflecting surfaces for lenses (blooming), solar

panels, and solar cells, A strong reflection at any of thesesurfaces could reduce the amount of energy being usefullytransmitted. A think surface film can be added so thatdestructive interference takes place for a typical wavelengthand thus maximum transmittance takes place at hiswavelength.


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G.6.11 Solve problems involving parallel

films.

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