Option E - Electromagnetic Waves

Question 1

Outline the nature of electromagnetic (EM) waves.

Answer 1

• oscillating electric charge production varying electric and magnetic fields
• transverse
• all same speed in vacuum
• interacting electric and magnetic fields are at right angles to one another and are in the direction of propagation of energy

Question 2

Describe the different regions of the electromagnetic spectrum.

Answer 2

• 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.

Question 3

Describe what is meant by the dispersion of EM waves.

Answer 3

• splitting of white light into its component colours
• a prism causes the dispersion of light because the refractive indexes are slightly different for each of the different colours
• red light is bent the least
• blue light is bent the most

Question 4

Describe the dispersion of EM waves in terms of the dependence of refractive index on wavelength. State the approx. wavelengths of blue and red light.

Answer 4

• dispersion is dependent on refractive index
• influenced by wavelength of light
• red light is bent the least 700nm
• blue light is bent the most 400nm

Question 5

Describe transmission of radiation.

Answer 5

• transfer of radiation from one medium to another
• when enough energy is supplied to the source, causes charge to accelerate
• energy transmitted in straight lines

Question 6

Describe absorption of radiation.

Answer 6

• absorption of radiation by material
• absorbing energy
• causing temperature to increase and decrease energy transmitted

Question 7

Describe scattering of radiation.

Answer 7

•deflection of radiation from its intended path due to collisions with particles within material

Question 8

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

Answer 8

•medium may re-emit some of its absorbed energy
EFFECT OF EARTHS ATMOSPHERE ON INCIDENTED EM RADIATION:
•blue light is scattered in all directions as a result of interaction with small dust particles in atmosphere
→shorter wavelength of light (blue) absorbed and reradiated more readily than longer wavelengths
•grazing incidence to the atmosphere at sunset and sunrise means that light from sun travels a greater length through the atmosphere and sunsets and sunrises appear red
•harmful UV radiation is absorbed by ozone layer in atmosphere
•increasing the carbon dioxide content of atmosphere will increase the absorption of infra red and result in global warming

Question 9

Explain monochromatic.

Answer 9

•a wave of a single wavelength and frequency

Question 10

Explain coherent.

Answer 10

• wave of one frequency and one wavelength with a constant phase relationship
• same direction and constant phase difference

Question 11

Describe laser light.

Answer 11

• laser light source of coherent light
• stands for Light Amplification by Stimulated Emission of Radiation
• light photons are produced when an atomic electron falls from a higher energy level to a lower energy level
• monochromatic and coherent parallel beams

Question 12

Outline the mechanism for the production of laser light.

Answer 12

STEPS:

1. electrons pumped to higher energy level (metastable state→excited state where electrons stay for longer times than normal excited states) by a flash of light (population inversion)
2. excited atoms start to de-excite giving out photons of light in all directions
3. photons travel past atoms that are still excited causing them to dexcite (stimulated emission)
4. results in amplification of light
5. production of laser light involves repeated reflections between carefully aligned mirrors

Question 13

Describe population inversion.

Answer 13

•more excited atoms than non excited

Question 14

Outline an application of the use of a laser.

Answer 14

• 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.

Question 15

Define principle axis.

Answer 15

•line going directly through the middle of the lens

Question 16

Define focal point.

Answer 16

• also known as principle focus
• point on principle axis to which rays that were parallel to the principle axis are brought to focus after passing through the lens
• convergence point of rays

Question 17

Define focal length.

Answer 17

•distance between the centre of the lens and the focal point

Question 18

Define linear magnification.

Answer 18

m=image size/object size=hi/h₀=image distance/object distance

Question 19

Define the power of a convex lens and the dioptre.

Answer 19

P=1/f

P is measured in m⁻¹ or dioptres (dpt)

Question 20

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

Answer 20

IMPORTANT RAYS:
•ray that was travelling parallel to the principle axis will be refracted towards the focal point on the other side of the lens
•ray that travelled through the focal point will be refracted parallel to the principle axis
•ray that goes through the centre of the lens will be undeviated

Question 21

Distinguish between a real image and a virtual image.

Answer 21

VIRTUAL:
•images created when rays of light seem to come from a single point when they actually do not pass through that point (when doing calculations use -ve v)
•produced by concave lenses, convex lenses object placed between focus and lens, mirror
REAL:
•light rays do actually pass through a single point
•produced by convex lens as long as it is behind focus
•Image will be:
→upside down
→diminished

Question 22

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

Answer 22

•+v for real image
•-v for virtual image
for virtual image: m=-v/u

Question 23

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

Answer 23

1/u+1/v=1/f (given)
\+f for converging lenses
-f for diverging lenses
u=object distance 
v=image distance

Question 24

Describe the image formed by reflection in a plane mirror

Answer 24

• same distance behind mirror as object is infront
• upright
• same size
• laterally inverted
• virtual

Question 25

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

Answer 25

FAR POINT: •distance between the eye and the furthest object that can be brought into focus •infinity for normal vision NEAR POINT: •nearest object that can be brought into clear focus •convention is taken to be 25cm for normal vision

Question 26

Define angular magnification.

Answer 26

•ratio between the angle that an object subtends normally and the angle that is image subtends as a result of the optical instrument Angular magnification, M=θi/θ₀ (given)

Question 27

Derive an expression for the angular magnification of a simple magnifying glass for an image formed at the near point and at infinity.

Answer 27

``` IMAGE FORMED AT NEAR POINT: •virtual image is located at near point D=-v M=θi/θ₀=(hi/D)/(h₀/D)=hi/h₀=D/d M=D/f +1 D=image distance d=object distance IMAGE FORMED AT INFINITY: •object placed at focal point u=f •for small angles sinθ=θ=tanθ θi=h/f M=θi/θ₀=(h/f)/(h/D)=D/f ```

Question 28

Construct a ray diagram for a compound microscope with final image formed close to the near point of the eye (normal adjustment).

Answer 28

•consists of two lenses →objective lens: forms a real magnified image of the object being viewed →eyepiece: acts as a magnifying lens, form a virtual magnified image M=linear magnification of eyepiece x linear magnification by objective lens SEE NOTES FOR DIAGRAM

Question 29

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

Answer 29

•consists of two lenses →objective lens: forms real but diminished image of distant object →eyepiece: rays from real image travel into this lens and form a virtual magnified image M=θi/θ₀=(h₁/fe)/(h₁/f₀)=f₀/fe (given) Length of telescope=f₀+fe SEE NOTES FOR DIAGRAM

Question 30

State the equation relating angular magnification to the focal lengths of the lenses in an astronomical telescope in normal adjustment.

Answer 30

M=θi/θ₀=(h₁/fe)/(h₁/f₀)=f₀/fe (given)

Question 31

Solve problems involving the compound microscope and the astronomical telescope.

Answer 31

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

Question 32

Explain the meaning of spherical aberration as produced by a single lens.

Answer 32

* rays striking outer regions of a spherical lens will be brought to a slightly different focus point from those striking the inner regions of the same lens * point object will focus into a small circle of light, rather than a perfect point * rays incident on edge of lens are refracted more than those in centre

Question 33

Explain the meaning of chromatic aberration as produced by a single lens.

Answer 33

* rays of different colours (wavelength and frequency) will be brought to slightly different focus points by the same lens * point object will produce a blurred image of different colours

Question 34

Describe how spherical aberration in a lens may be reduced.

Answer 34

* shape of lens could be altered to slightly non spherical shape * decrease the aperture, this is stopping down the aperture

Question 35

Describe how chromatic aberration in a lens may be reduced.

Answer 35

* using two different materials to make up a compound lens, called an achromatic doublet * the two types of glass produce equal but opposite dispersion

Question 36

State the conditions necessary to observe interference between two sources.

Answer 36

•principle of super position for two coherent sources having roughly the same amplitude •must be coherent: →same frequency →same wavelength →constant phase relationship between the sources

Question 37

Explain, by means of the principle of superposition, the interference pattern produced by waves from two coherent point sources.

Answer 37

* regions where waves are in phase (nλ) → constructive interference * regions where waves are out of phase ((n+½)λ) → destructive interference

Question 38

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

Answer 38

YOUNGS DOUBLE SPLIT EXPERIMENT: •light from the twin slits (the sources) interferes and patterns of light and dark regions, called fringes, can be seen on the screen •monochromatic light passes through first slit and spreads out due to diffraction so it can pass through two more slits •if laser light used, it is already coherent so first slit not necessary •slit width small compared to slit separation SEE NOTES FOR INTENSITY DISTRIBUTION AND DIAGRAM

Question 39

Solve problems involving two-source interference.

Answer 39

``` INTERFERENCE: sinθ=nλ/d DOUBLE SLIT- BRIGHT FRINGES: x/D=nλ/d DOUBLE SLIT- DARK FRINGES: x/D=((n+½)λ)/d FRING SEPARATION: S=λD/d ```

Question 40

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

Answer 40

•series of parallel slits (at a regular separation) called a diffraction grating •additional slits at same separation will not effect the condition for constructive interference •angle at which the light from the slits add constructively will be unaffected by the number of slits ADDITION OF FURTHER SLITS AT SAME SLIT SEPARATION: •principal maxima maintain same separation •principal maxima become much sharper (fringes sharper) •increase intensity •width of principle maxima narrower on diagram

Question 41

Derive the diffraction grating formula for normal incidence.

Answer 41

FOR CONSTRUCTIVE INTERFERENCE: Path difference=nλ nλ=dsinθ

Question 42

Outline the use of a diffraction grating to measure wavelengths.

Answer 42

* if white light is incident on a diffraction grating, the angle at which constructive interference takes place depends on wavelength * different wavelengths can thus be observed at different angles * accurate measurement of angle provides accurate measurement of the exact wavelength (and thus frequency) of the colour of light

Question 43

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

Answer 43

COOLIDGE XRAY TUBE: •heated cathode releases electrons of high energy •electrons are accelerated by travelling through a potential difference, V •fast moving electrons collide with metal target •collision results in Xrays •heat generated in target kept cool by being rotated in oil •X rays hit walls of chamber •adjustable diaphragm is used to define the beam of X rays •increasing the heater current to the cathode increases temperature thus no of electrons emitted thus the intensity of the x rays •hardness of an x ray beam measures its penetration power. Higher frequency (lower wavelength) x rays are harder SEE NOTES FOR DIAGRAM

Question 44

Draw and annotate a typical X-ray spectrum.

Answer 44

CONTINUOUS FEATURES: •wide range of x ray frequencies produced corresponding to range of λ MINIMUM WAVELENGTH: •energy of X ray photon depends on energy lost in collisions •maximum amount of energy that can eb lost is all the initial KE of electron •max energy available meas there is a max frequency of x ray produced corresponding t min wavelength limit CHARACTERISTIC FEATURES: •in some circumstances, collisions between incoming electrons and target atoms can cause electrons from inner orbital of target atom to be promoted to higher energy levels •when these electrons fall back down they emit X rays of particular frequency which is fixed by energy levels available PEAKS: electron falls from L→K (highest intensity, longer wavelength), M→K and N→K

Question 45

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

Answer 45

* when x rays are incident on regular structure (crystal) the majority of x rays will pass through material * at particular angles, high intensity x ray recorded which corresponds to points of constructive interference of x rays scatted from different planes of atoms in crystal * crystals have many different lattice planes causing many different constructive interference positions * powdered sample of crystals contains every orientation of these planes resulting in circular x ray diffraction picture

Question 46

Derive the Bragg scattering equation.

Answer 46

•relates path length difference between two lattice planes to wavelength of x rays constructive interference path difference=2dsinθ=nλ SEE NOTES

Question 47

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

Answer 47

•using Brag equation possible to →measure lattice plane distances using x rays of known wavelength →use cubic crystals of known lattice plane distance to measure wavelength of x rays

Question 48

Explain the production of interference fringes by a thin air wedge.

Answer 48

* two glass plates are at small angle to one another * gap between two plates called air wedge * there is a path difference the rays of light reflecting from the top and from the bottom of the surfaces of air wedge * results in parallel lines of equally spaced constructive and destructive interference fringes

Question 49

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

Answer 49

•interference pattern produced can be used to measure its thickness •path difference=2t+λ/2=mλ [or 2t=(m+½)λ] If wedge has refractive index n Constructive interference occurs when 2nt=(m+½)λ Destructive interference occurs when 2nt=mλ

Question 50

Describe applications of measuring thickness.

Answer 50

* if separation of fringes is measured and wavelength of light is known, angle and thus the thickness of wedge at any point can be calculated * measuring thickness of film of moisture (tear film) on surface of eye

Question 51

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

Answer 51

* region that is designed to be completely smooth is sometimes known as an optic flat * any deviation from complete smoothness will affect the fringes formed as a result of reflection from the surface

Question 52

State the condition for light to undergo either a phase change of π,or no phase change, on reflection from an interface.

Answer 52

* a phase change is the inversion of the wave that can take place at a reflection interference * it depends on the two media involved * when light is reflected back from an optically denser medium (higher refractive index) there is a phase change of π * when light is reflected back from an optically less dense medium (lower refractive index) there is no phase change * there is no phase change for transmitted waves

Question 53

Describe how a source of light gives rise to an interference pattern when the light is reflected at both surfaces of a parallel film.

Answer 53

•there are two possible paths →reflect back of first boundary →go through first boundary and reflect off boundary 2 •light reflected at second boundary travels extra distance →if equal to whole number of wavelengths nλf, results in constructive interference producing a bright fringe →if extra distance ½nλf then destructive interference occurs and dark band produced λf is the wavelength of the film: λf=λ/n

Question 54

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

Answer 54

•for one angle of viewing, one colour interferes destructively/another interferes constructively; •colour seen is determined by colour that interferes constructively (one or two colours may reinforce along a direction which others cancel) •at different viewing angle, different colour interferes destructively/constructively •appearance of film will be bright colours, such as can be seen when looking at →an oil film on surface of water →soap bubbles.

Question 55

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

Answer 55

* fringes formed by wedges due to changing size of wedge | * for parallel film, viewing angle affects fringes formed

Question 56

Describe applications of parallel thin films.

Answer 56

* design of non-reflecting radar coatings for military aircraft * measurement of thickness of oil slicks caused by spillage * design of non-reflecting surfaces for lenses (blooming), solar panels and solar cells.