Option G: Electromagnetic waves

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Outline the nature of electromagnetic (EM) waves.

Students should know that an oscillating electric charge produces varying electric and magnetic fields. Students should know that electromagnetic waves are transverse waves and all have the same speed in a vacuum.

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Oscillating chargeClick to play

Outline the nature of electromagnetic (EM) waves.

Students should know that an oscillating electric charge produces varying electric and magnetic fields. Students should know that electromagnetic wavesare transverse waves and all have the same speed in a vacuum.

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Aim 8 and TOK: Students could considerthe possible health hazards associated withtransmission lines.

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Describe the different regions of the electromagnetic spectrum.

Students should know the order of magnitude of the frequencies and wavelengths of different regions, and should also be able to identify a source for each region.

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Construct ray diagrams to locate the image formed by a convex lens.Students should appreciate that all rays incident on the lens from the object will

be focused, and that the image will be formed even if part of the lens is covered.

3 predictable rays

Describe the different regions of the electromagnetic spectrum.

Students should know the order of magnitude of the frequencies and wavelengths of different regions, and should also be able to identify a source for each region.

Hamper Page 412 Q’s 1-9. Review pack Q1

Describe what is meant by thedispersion of EM waves.

Different wavelengths of light travel at different speeds through glass (and other transparent media). Hence, the refractive index is different for different wavelengths (or colours) of light.

In the diagram above is called the angle of dispersion.

Describe the dispersion of EM waves in terms of the dependence of refractive index

on wavelength.No quantitative discussion is required.

Which colour has the highest refractive index?

Distinguish between transmission,absorption and scattering of radiation.

Light passing through an optical system can be attenuated by absorption and by scattering.

Initial intensity Transmitted intensity

In physics, absorption of electromagnetic radiation is the way by which the energy of a photon is taken up by matter, typically the electrons of an atom. Thus, the electromagnetic energy is transformed to other forms of energy, for example, to heat. The absorption of light during wave propagation is often called attenuation.

Discuss examples of the transmission, absorption and scattering of EM radiation.

Students should study the effect of the Earth’s atmosphere on incident EM radiation. This will lead to simple explanations for the blue colour of the sky, red sunsets or sunrises, the effect of the ozone layers, and the effect of increased CO2 in the atmosphere. This links with 8.5.6.

Blue Sky?

The blue color of the sky is caused by the scattering of sunlight off the molecules of the atmosphere. This scattering, called Rayleigh scattering, is more effective at short wavelengths (the blue end of the visible spectrum). Therefore the light scattered down to the earth at a large angle with respect to the direction of the sun's light is predominantly in the blue end of the spectrum.

Review pack Page 25 G1

Red SunsetSunsets are reddened because for sun positions which are very low or just below the horizon, the light passing at grazing incidence upon the earth must pass through a greater thickness of air than when it is overhead. Just before the sun disappears from view, its actual position is about a diameter below the horizon, the light having been bent by refraction to reach our eyes. Since short wavelengths are more efficiently scattered by Rayleigh scattering, more of them are scattered out of the beam of sunlight before it reaches you. Aerosols and particulate matter contribute to the scattering of blue out of the beam, so brilliant reds are seen when there are many airborne particles, as after volcanic eruptions.

Lasers

The stimulated emission of light is the crucial quantum process necessary for the operation of a laser.

Explain the terms monochromaticand coherent.

Coherence means that the waves have a constant phase relationship.

Monochromatic means “the same wavelength”

Identify laser light as a source ofcoherent light.

Coherence is one of the unique properties of laser light. It arises from the stimulated emission process which provides the amplification. Since a common stimulus triggers the emission events which provide the amplified light, the emitted photons are "in step" and have a definite phase relation to each other.

Outline the mechanism for theproduction of laser light.

Students should be familiar with the term population inversion.

The achievement of a significant population inversion in atomic or molecular energy states is a precondition for laser action. Electrons will normally reside in the lowest available energy state. They can be elevated to excited states by absorption, but no significant collection of electrons can be accumulated by absorption alone since both spontaneous emission and stimulated will bring them back down.

Helium-Neon Laser

One of the excited levels of helium at 20.61 eV is very close to a level in neon at 20.66 eV, so close in fact that upon collision of a helium and a neon atom, the energy can be transferred from the helium to the neon atom.

Outline an application of the use of a laser.

Students should appreciate that lasers have manydifferent applications. These may include:• medical applications• communications• technology (bar-code scanners, laser disks)• industry (surveying, welding and machining metals, drilling tiny holes in metals)• production of CDs• reading and writing CDs, DVDs, etc.

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Tsokos

Page 605 Q’s 4,6,7,8.

Review pack Q 14.

G2 Optical instruments

Principal axis

Focal lengthParallel rays arriving from infinite distance.

Image formation

Define the terms principal axis, focal point, focal length and linear magnification as applied to a

converging (convex) lens.Hyperlink

Define linear magnification.

vu

h0

hi

Construct ray diagrams to locate the image formed by a convex lens.

Students should appreciate that all rays incident on the lens from the object will be focused, and that the image will be formed even if part of the lens iscovered.

Distinguish between a real image and a virtual image.

In optics, a real image is a representation of an object (source) in which the perceived location is actually a point of convergence of the rays of light that make up the image. If a screen is placed in the plane of a real image the image will generally become visible on the screen.

In optics, a virtual image is an image in which the outgoing rays from a point on the object never actually intersect at a point.

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Apply the convention “real is positive, virtual is negative” to the thin lens formula.

u v

Solve problems for a single convexlens using the thin lens formula.

Tsokos Page 621 Q’s 8,9,10.Hamper page 442 Q’s 30 – 35.Review pack Q’s 5-10, G2.

Define the power of a convex lens and the dioptre.

Ray diagram for magnifying glass

Define the terms far point and nearpoint for the unaided eye.

For the normal eye, the far point may be assumed to be at infinity and the near point is conventionally taken as being a point 25 cm from the eye.

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Define angular magnification.

Hamper page 445 Q’s 36 – 39.

θ0

θi

The simple magnifying glassObject at nearpoint

Image at nearpoint, object at unknown distance (u)

Angular magnification is =

Applying the lens formula with v = -ve

But angular magnification is = 25/o

Derive an expression for the angularmagnification of a simple magnifyingglass for an image formed at the nearpoint and at infinity.

At infinity 25/25 goes to 25/∞

The compound microscope and astronomical telescope

Construct a ray diagram for a compound microscope with final image formed close to the near point of the eye

(normal adjustment).

Students should be familiar with the terms objective lens and eyepiece lens.

Image distance at near point of the eye

2.Image formed by objective lens becomes object for the eyepiece lens

Image formed by eyepiece lens

1. Construct ray diagram for objective lens with u between f and 2f

3. Construct ray diagram for eye piece lens with object between lens and f (but close to f).

Construct a ray diagram for an astronomical telescope with the final image at infinity (normal

adjustment).1. Construct an off axis ray diagram for the objective lens wit the object at infinity

2. Image formed by the objective becomes the object for the eyepiece lens

3. Make the distance between the eyepiece lens and the object = f for the eyepiece lens

Rays are parallel

Rays are parallel

State the equation relating angular magnification to the focal lengths of the lenses in an astronomical

telescope in normal adjustment.

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Solve problems involving the compound microscope and the astronomical telescope.

Problems can be solved either by scale ray diagrams or by calculation.

Tsokos Page 622 Q’s 23,26,27.Hamper page 448 Q’s 40 – 42.Review pack Q’s 12,13.

Aberrations

Spherical aberration. A perfect lens (top) focuses all incoming rays to a point on the optic axis. A real lens with spherical surfaces (bottom) suffers from spherical aberration: it focuses rays more tightly if they enter it far from the optic axis than if they enter closer to the axis. It therefore does not produce a perfect focal point.

Explain the meaning of spherical aberration and of chromatic aberration as produced by

a single lens.

A lens will not focus different colours in exactly the same place because the focal length depends on refraction and the index of refraction for blue light (short wavelengths) is larger than that of red light (long wavelengths).

For lenses made with spherical surfaces, rays which are parallel to the optic axis but at different distances from the optic axis fail to converge to the same point. For a single lens, spherical aberration can be minimized by bending the lens into its best form.

The curved surface of most lenses is a small section of a sphere. This is not the ideal shape and the resulting problem is called spherical aberration. After refraction by the lens the rays do not all pass through the same point. This is especially noticeable with the plane/convex lenses used in medium priced optical instruments. The effect is exaggerated in the diagram below.

A method of reducing the spherical aberration of a lens is simply to reduce the aperture using a diaphragm (also called an iris or a stop). Although this reduces the spherical aberration it also reduces the brightness of the image.

Describe how spherical aberration in a lens may be reduced.

For a single lens, spherical aberration can be minimized by

1.Bending the lens into its best form.

2. Using a diaphragm to reduce the aperture.

Describe how chromatic aberration in a lens may be reduced.

These lenses consist of two optical elements, usually of crown and flint glass types, cemented together to form an achromatic doublet. Most often, the crown lens is a biconvex positive lens. Due to the difference in refractive indices, chromatic aberration with respect to two selected wavelengths has been corrected.

This effect can be reduced by having a combination of a convex and a concave lens made of glasses having different refractive indices.

Review pack Q 11.

G3 Two-source interference of waves

State the conditions necessary to observe interference between two sources.

(1) The sources must be coherent (i.e., they must maintain a constant phase relationship with one another).

(2) The sources must be monochromatic (i.e., of a single wavelength). (Or made of distinct wavelengths).

(3) Similar amplitude.

Young's Double Slit Experiment

Explain, by means of the principle of superposition, the interference pattern produced by waves from

two coherent point sources.The effect may be illustrated using water waves and sound waves in addition to EM waves.

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Outline a double-slit experiment for light and draw the intensity distribution of the observed fringe

pattern.This should be restricted to the situation where the slit width is small compared to the slit separation so that diffraction effects of a single slit on the pattern are not considered.

D

s

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Solve problems involving two-source interference.

Tsokos Page 631 Q’s 2-9.Hamper page 425 Q’s 13 – 19.Review pack Q G3

G4 Diffraction grating

Diffraction angle, m()

Zeroth order

First order

Minus first order

Describe the effect on the double-slit intensity distribution of increasing the number of slits.

The image becomes brighter and more distinct.

Derive the diffraction grating formula for normal incidence.

From the enlarged view of the slits it is clear that this path difference (λ )is equal to dsinθ

dsinθ

Other bright fringes will occur for path differences of 2λ, 3 λ etc.

Therefore, for a bright fringe to be seen

nλ = dsinθ

Diffraction grating formula

λ = dsinθ

2λ = 2dsinθ

3λ = 3dsinθ

Outline the use of a diffraction grating to measure wavelengths.

Use of the spectrometer is not included.

Solve problems involving a diffraction grating.

Hamper page 434 Q 25Review pack Q’s 15,16.

Xd

Or diffraction grating

G5 X-rays

Outline the experimental arrangement for the production of X-rays.

A Coolidge tube is sufficient. Students should understand how the intensity (current in the filament) and hardness (tube voltage) of the X-ray beam are controlled. K: filament

A: anodeWin and Wout: water inlet and outlet of the cooling device

Draw and annotate a typical X-rayspectrum.

Students should be able to identify the continuous and characteristic features of the spectrum and the minimum wavelength limit.

Explain the origins of the features of a characteristic X-ray spectrum.

Bremsstrahlung X-Rays

"Bremsstrahlung" means "braking radiation" and is retained from the original German to describe the radiation which is emitted when electrons are decelerated or "braked" when they are fired at a metal target.

Characteristic X-Rays Characteristic x-rays are emitted from heavy elements when their electrons make transitions between the lower atomic energy levels. The characteristic x-rays emission which shown as two sharp peaks in the illustration at left occur when vacancies are produced in the n=1 or K-shell of the atom and electrons drop down from above to fill the gap.

Solve problems involving accelerating potential difference and minimum wavelength.

Hamper page 423 Q’s 10,11,12.

X-ray diffraction

Path difference = 2dsinθ

dsinθdsinθ

Constructive

interference between

these 2 rays when……

Derive the Bragg scattering equation

Explain how X-ray diffraction arises from the scattering of X-rays in a crystal.

This may be illustrated using 3 cm equipment.

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Outline how cubic crystals may be used to measure the wavelength of X-rays.

Students should be aware of the fact that the structure of DNA was discovered by means of X-ray diffraction.

an X-ray diffraction image of sodium salt of DNA

As shown by X-ray crystallography, the hexagonal symmetry of snowflakes.

Outline how X-rays may be used todetermine the structure of crystals.

Solve problems involving the Braggequation.

Hamper page 436 Q’s 26-28.Tsokos page 639 Q’s 7,8,9.Review pack Q G4.

G6 Thin-film interference

Iridescence of peacock feathers is caused bylight reflected from complex layered surface

What happens when a wave is reflected?

From low speed to high speed (high density to low density) i.e no phase change

From high speed to low speed (low density to high density) i.e. π phase change

Air to Glass Glass to Air

Wedge films

Interference from 1 and 2 causes the light and dark fringes.

Fringes arise from Interference between the reflections from the air gap

Explain the production of interferencefringes by a thin air wedge.

Students should be familiar with the terms equal inclination (parallel rays) and equal thickness.

Because there is no phase change from the top of the air wedge, and π phase change at B, constructive interference occurs when

And destructive interference occurs when

Where m is an integer and n is the refractive index of the air (i.e. =1).

AB

The interference occurs from the top and bottom surfaces of the air wedge i.e points A and B.

no phase change

π phase change

Explain how wedge fringes can be used to measure very small separations.

If two flat pieces of glass are held at a very small angle to each other and viewed under a microscope (as shown below), a regular series of bright and dark lines can be observed.

At an external reflection there is a phase change of π rad, so if the path difference, nt (=s), is equal to nλ, destructive interference will occur and a dark fringe will be seen. If the path difference is an odd number of half wavelengths, a bright fringe will be seen.

t

Describe how thin-film interference is used to test optical flats.

Solve problems involving wedgefilms.

Hamper page 432 Q 24.

Parallel films

Applications include measurement of the thickness of the tear film on the eye and oil slicks.

State the condition for light to undergo either a phase change of π (going to a denser medium), or no phase change (going to a less dense medium),

on reflection from an interface.

Constructive

DestructiveNo phase change

phase change of π

t

State the condition for light to undergo either a phase change of π (going to a denser medium), or no phase change (going to a less dense medium),

on reflection from an interface.

phase change of π

Destructive

Constructive

phase change of π t

Describe how a source of light gives rise to an interference pattern when the light is reflected at

both surfaces of a parallel film.

Figure A shows a wave incident on a thin film. Each half wavelength has been numbered, so we can keep track of it. Note that the thickness of the film is exactly half the wavelength of the wave when it is in the film.

Figure B shows the situation two periods later, after two complete wavelengths have encountered the film. Part of the wave is reflected off the top surface of the film; note that this reflected wave is flipped by 180°, so peaks are now troughs and troughs are now peaks. This is because the wave is reflecting off a higher-n medium.

Another part of the wave reflects off the bottom surface of the film. This does not flip the wave, because the reflection is from a lower-n medium.

State the conditions for constructiveand destructive interference.

Destructive or Constructive ? This depends on the phase change at the 2nd boundary. If going from soap film to air, or from oil to water then there is no phase change from the second boundary. If going from an anti reflective coating to glass, then there is a phase change of π

Where n is the refractive index and m is an integer.

Destructive, constructive

Constructive, destructive

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Explain the formation of coloured fringes when white light is reflected from thin films, such as oil

and soap films.

Where n is the refractive index and m is an integer.

Different wavelengths will constructively interfere at different angles, hence the production of colours.

Describe the difference between fringes formed by a parallel film and a wedge film.

A thin film interference pattern will be a series of alternating bright and dark fringes of the same color when it is created by a wedge illuminated by monochromatic light. For example: (1) an air wedge formed between two pieces of glass, or (2) a sliver of glass surrounded by air or water.

As the wedge becomes thinner (θ → 0), the fringes are less numerous as they spread further apart and get wider. Note that the point of contact is a dark fringe. This results not from the thickness of the wedge, but from the net phase inversion between the two reflected rays equals ½ λ. If the plates are placed flat, directly on top of one another, the plates would be entirely dark. If the plates remained parallel, but were slowly raised apart, the pattern would alternate between completely dark and completely bright as the distance between the plates changed by ¼ wavelengths. [2t = ½ λ]. Only when viewed from an angle with non monochromatic light, do we see the colours.

Describe applications of parallel thin films.

Applications should include:• design of non-reflecting radar coatings formilitary aircraft• measurement of thickness of oil slicks caused byspillage• design of non-reflecting surfaces for lenses(blooming), solar panels and solar cells.

Solve problems involving parallelfilms.

These will include problems involving the application of thin films.

Hamper page 431 Q’s 20 – 23.

Questions

Hamper page 450 Q’s 1-8.

Review pack Q’s 17,18,19.