1. Chapter 9 Electromagnetic Waves

2. 9.2 ELECTROMAGNETIC WAVES

3. Introduction Light is considered to consist of oscillating electric and magnetic fields at 90 o to each other and at right angles to the direction of travel. Y Z X E E B B Direction of propagation

4. Production of Electromagnetic Waves Consider a charged particle undergoing simple harmonic motion, the particle is accelerating in one direction then in the other in a regular cycle

5. Production of Electromagnetic Waves The particle is momentarily stationary at the top and bottom of its oscillations, while it is fastest at the point half way between

6. Production of Electromagnetic Waves Because the charged particle is moving, it generates a magnetic field; the magnetic field is changing due to the changing speed of the particle

7. Production of Electromagnetic Waves The constantly changing magnetic field generates a constantly changing electric field, these changing fields radiate out at the speed of light

8. Production of Electromagnetic Waves The electric and magnetic fields continue to create each other and are now self propagating, however one cannot exist without the other

9. Characteristics of Electromagnetic Waves

10. Electromagnetic waves are often generated by accelerating (vibrating) electrons. Electrons which vibrate in a single plane will generate electromagnetic waves with the electric field restricted to a single plane. This is called polarisation

11. Characteristics of Electromagnetic Waves Similarly the magnetic field will also be restricted to a plane at right angles to the electric field. and the magnetic field vectors are confined to a single plane at right angles to the electric field. We define the plane of polarisation to be the plane of the electric field in the polarised electromagnetic wave.

12. Class Problems Conceptual Questions: 1-2

13. 9.3 USE OF ANTENNAE TO TRANSMIT AND RECEIVE SIGNALS

14. Production of Electromagnetic (E-M) Waves by an Antenna When any charged particle (eg. an electron) accelerates, an E-M wave is produced. If a circuit has a vibrating or alternating current in it, the charges are continually accelerating and thus will radiate E-M waves.

15. Production of Electro-Magnetic (E-M) Waves by an Antenna An alternating voltage applied to a length of metal (an antenna) will thus radiate an E-M wave. ~~ E E E E ~ E ++ + -- - ~ ++ + -- - ~ ++ + -- - Two metal rods are connected to an AC generator. The charges on each rod then alternate, creating an alternating Electric field, thus radiating an E-M wave - which travels at the speed of light.

16. Reception of an E-M Wave by an Antenna: When an EM wave hits an antenna, the free electrons in the antenna will be forced to vibrate (by induction). If a simple circuit is connected to the antenna and the circuit is tuned, a narrow band of frequency will cause the electrons to resonate. This signal is the amplified and sent to the appropriate audio visual device. The orientation of the antenna should match the plane of polarization of the EM wave.

17. Class Problems Conceptual Questions: 3-6 Descriptive Questions: 1, 3, 4 Computational Questions: 1-3

18. 9.4 SPEED, FREQUENCY & WAVELENGTH

19. Transverse Waves E-M waves are transverse waves and their properties can be related to water waves However, while water particles do not travel with the waves, the electric and magnetic fields of E-M waves do

20. Frequency The frequency of the wave refers to the frequency of the periodic variations in the electro-magnetic fields (waves per second)

21. Wavelength The wavelength of light refers to the distance between two consecutive points in space where the electric field (or magnetic field) is in phase. Two points are in phase if they have the same displacement in both magnitude and direction

22. Speed In a vacuum any electromagnetic wave will travel at the speed of light, c = 3 x 10 8 m s -1 Since, the time taken for one complete wavelength (  s = ) to pass a point is the period (T) of the wave. The frequency of the wave is the number of waves past a point in one second, f = 1 / T

23. Thus, v = c,  s =, and  t = T = 1/f Then, becomes, ie, The Wave Equation

24. Example A ray of green light has a wavelength of 540nm in a vacuum. Find its frequency.

25. Example A ray of green light has a wavelength of 540nm in a vacuum. Find its frequency. Example

26. A radio station broadcasts at a frequency of 720kHz. Find its wavelength.

27. Example A radio station broadcasts at a frequency of 720kHz. Find its wavelength.

28. Class Problems Conceptual Questions: 17 Descriptive Questions: 8 Computational Questions: 5-7, 10-11

29. 9.5 THE ELECTROMAGNETIC SPECTRUM

30. The Spectrum of Visible Light The visible spectrum is the part of the electromagnetic spectrum that we can see It is a very small section of the E-M spectrum

31. The Spectrum of Visible Light The approximate wavelengths of different colours is given below The spectrum is a continuum, the colours merge into one another

32. The Electromagnetic Spectrum Like the visible spectrum, the whole E-M spectrum is a continuum; the different regions merge and overlap

33. The Electromagnetic Spectrum Note that the scales of frequency and wavelength are logarithmic, not linear

34. The Electromagnetic Spectrum The frequency and wavelength ranges vary since there is no hard and fast boundaries, but generally accepted ranges are given on p. 17 of the Key Ideas textbook

35. Class Problems Conceptual Questions: 18-19 Descriptive Questions: 10-11 Computational Questions: 13-14, 16-18, 20-22

36. 9.6 APPLICATION: LASER AIRBORNE DEPTH SOUNDER

37. Laser Airborne Depth Sounder (LADS) LADS (Laser Airborne Depth Sounder) is a system that uses an aircraft to fly over a body of water to automatically measure the depth of the water.

38. Laser Airborne Depth Sounder (LADS) The LADS system works by emitting a laser at the water When the laser ‘hits’ the water, it is partially reflected and transmitted, the reflected laser then returns to the plane

39. Laser Airborne Depth Sounder (LADS) The transmitted laser continues through and is reflected of the bottom of the water and back up to the plane Since the speed of light in water is known, the time between the reflected and transmitted ray can be used to determine the depth. Appendix 9 of the Key Ideas textbook gives more detail on the speed of light in media other than a vacuum

40. Laser Airborne Depth Sounder (LADS) The extra distance travelled by the second pulse (transmitted ray) is to the bottom of the water and back, i.e., twice the depth of the water. Therefore the depth of the water is half of the extra distance travelled.

41. Example An aircraft carrying a LADS system is flying horizontally above a lake. The laser pulses are reflected of the surface of the water and of the bottom of the lake return to the aircraft 3.33  s and 3.55  s after being emitted. If the speed of light in fresh water is 2.25 x 10 8 ms -1, calculate the depth of the water where the sounding was taken.

42. Example The extra distance travelled by the second pulse is 49.5m, therefore the depth of the lake is 24.75m deep.

43. Scanning at angles other than 90° If the LADS was only directed straight down, then only the depth directly beneath the plane would be found The plane actually scans at angles to the side up to 14° This width is approximately 240m and is called a swath

44. Scanning at angles other than 90° The LADS can do this because water is very rarely perfectly flat, thus some of the light will be reflected back to the plane The LADS system uses a Nd:YAG laser which produces an infrared beam (1064nm) The energy is emitted in pulses of 5ns, at a rate of 168 pulses per second

45. Scanning at angles other than 90° The green scanning beam is produced by frequency doubling the laser, it has a wavelength of 532nm This wavelength is the least absorbed wavelength by coastal water The scan has a resolution of about 10m, and scans at a rate of 54km 2 per hour to a depth of 50m

46. The Necessity To Use A Powerful Laser Lasers used in the LADS system are 1Megawatt. This is a very powerful laser! Why? Suspended sediment in the water can scatter the laser reducing the amount of energy reaching the bottom. Water tends to absorb the light that passes through it. The seabed also absorbs light. Only a small fraction of the energy actually return to the aircraft as it scans from side to side and the surface of the water and seabed is quite rough. The laser used is very power full and could blind someone below, therefore the laser is spread out over a distance to reduce the risk of injury below.

47. Class Problems Descriptive Questions: 5, 12 Computational Questions: 24, 26-28