 Fig.1 Ocean waves carry a large amount of energy. Light wavesand sound waves help us to understand most o the properties of waves such as reflection, and refraction. Here we start with a special property of waves known as diffraction.  Fig. 2. Pictures of diffraction.Diffraction
Diffraction is the bending of waves at the edges or gaps. When a wave passes an edge the corners of the wave front ben. If the gap is smaller than the wave length , the bending at the two edges become noticeable and the resulting fronts progress as curves. If the gap is large the bending appears only at the edges and is hardly noticeable. Electromagnetic waves. Fig.4.Light consists of transverse waves . Fluctuating of electric and magnetic fields. Interference. This is a phenomenon that occur when similar waves clash or combine. Interference is the combination of two or more waveforms to form a single wave. Consider a sound wave. Here the air molecules vibrate. If many sounds come, can a single molecule vibrate in many ways ?  Fig.4. Wave 1 and wave 2 combine to form the wave shown by the dots.
 A beautiful design formed due to interference of water waves. Stationary Waves Fig.6. Creating a stationary wave with a rope. n= nodes. an- antinodes. Dennis is waving a rope up and down. But a progressive (travelling) wave does not appear. This kind of a wave is a standing wave or a stationary wave. This wave is due to the interference of two identical waves travelling in opposite directions. The black represents the incident wave while the blue indicates the reflected wave. The points that move up and down are the nodes. Shown as ‘an’. n- shows the nodes. Two point source interference. Dark lines indicate crests while the thin lines show the troughs. They originate at S1 and S2, two point sources. Simple letters a,b, c ,d and e indicate some particles in the medium. When the dark lines meet (crests) particle is in upward motion. Can you find them? High light for answer- c and d. Is there a marked point where thin lines (troughs) meet? Answer:- e The particles c,d and e are in the same state of motion; moving up or down. This is constructive interference. They are on the red lines, which are the antinodal lines. What happens if a crest meets a trough? Are there such particles? Yes ‘a’ nd ‘b’. At these points forces cancel and that is destructive interference. These are on the blue lines and are known as nodal lines. Solving problems. Example 1. There is a float x equidistant from two sources. The point y is 30 m from S1 and 21 m from S2. If the wave length is 6m describe the movement of the floats. Answer. The float X will be on the 0 antinodal line. It will move up and down. Waves to point Y from S1 = 30/ 6 = 5 waves. Waves to point Y from S2 = 21/6 = 3 .5 waves The point is on a nodal line. It will not show movement. Path difference Consider the point ‘a’. Count the number of waves from here to S1 and S2. Let us tabulate the number of wavefronts to the marked letters.
Difference given here in red are in the units of wave length. Problem solving. Example 1. There is a float x equidistant from two sources. The point y is 30 m from S1 and 21 m from S2. If the wave length is 6m describe the movement of the floats. Answer. The float X will be on the 0 antinodal line. It will move up and down. Waves to point Y from S1 = 30/ 6 = 5 waves. Waves to point Y from S2 = 21/6 = 3 .5 waves The point is on a nodal line. It will not show movement. Two source interference in light waves.  Fig.6. Monochromatic light from two slits form a fringe of bright and dark lines. Fig.7. Thomas Young 1773 -1829 (Eng.) did an experiment similar to this, in 1801, and was able to determine the wave length of sunlight. This also provided evidence to contradict Sir Isaac Newtons corpuscular theory of light. Example 2.  Fig.8. Method to get the wavelength. Determine the wavelength using the data given here. S1 to S2 = o.00025 m. …….. D. Sources to screen = 9.78 m. …………...L. Fringe gap 4 bright lines = 0.102m……….x Solution Method 1. Using the blue triangle sinθ = pd/d………………………………...pd=sinθ*d. In the yellow triangle sineθ = x/L As x is very small compared to L , sine θ become same as tanθ. Then tanθ= x/L = 4λ/d 4λ =d (x/L) 4λ =d (x/L) 4λ =0.00025(0.102/ 9.78) λ= 6.52x10-7 Method 2. Young’s equation………………..λ = x * d / (n * L) λ = 0.102 x 0.00025/ 4 x 9.78 ……… = 6.52x10-7m Doppler effect We can easily recognize the difference in pitch due to the changing of distance between a source and the listener. Fig.9. Changing of a siren sound. The formula to be used in calculations. When approaching:-
Example An ambulance travelling at 50ms-1 uses a siren of frequency 450Hz. Find the apparent frequency to a listener when it is approaching. Speed of sound = 330ms-1. Amb. speed = 50ms-1, Freq =450Hz sound 330m/s When Appoaching:- fa= 330 x 450 / 330 - 50 = 148500 /280 = 530.36 Hz.As an ocean wave approaches land, they slow down and the wave length becomes shorter . This propels the crests higher as the energy accumulates. At a particular point they become unstable and rolls off.  Fig.9. Breaking of a sea wave. If you watch the movement of a float due to waves it would be possible to observe that it moves in the direction of the wave with the crest and goes back with the trough. So the water moves in a circle while the wave proceeds. As the wave approaches the shore, vertical diameter decreases producing an elliptical motion . Normally we consider only the vertical movement; going up and down. An ocean wave normally carries the energy received from the wind.  Jenny observes during a second 20 crests passing through rock in the sea.The gap between two crests was 7m. The wave height was 1m. Find the following:- 1.1 The wavelength, 1.2 The frequency. 1.3 Velocity 1.4 Amplitude 1.5 Near a breakwater wall she noted that there was no linear movement of the wall inspite of level rising up and down in some places. What would have caused this condition? Four students gave these answers. a. Incident wave merging with the reflected wave. b. Formation of a standing wave, c. Interference of the incident wave and the reflected wave. d. Complete annihilation of the incident wave and the reflected wave. A. All answers are correct. B. All answers are wrong. C. Only a and d. are corret. D. Only a,b and c are correct. 6x5=30 marks Q.2.0 A light wave of 6.5xx10-7 wavelength passed through two slits forming a bright and dark fringes on a screen 3m away. Find the spacing between fringes. Help if necessary highlight along this line. Use formula λ =d (x/L) 10 marks Q.3.0
The two speakers, Sp1 and Sp2, three meters apart are producing a note, 264 Hz. Mary walking from P to Q observes that the volume is changing. Using the data given, find the following:- Speed of sound as =330 ms-1. Distance between speakers 3m. Frequency of wave 660Hz.
5x4=20marks. Q. 2. A speaker in a merry-go-round emits a signal and an instrument attached records it as of frequency 500 Hz.The linear velocity at the circumference is 20 ms-1. Speed of sound 330ms-1. Find the following:- 1.The frequency of the note Jenny hears when the speaker is at the four ends marked as , W,N,E and S? ( four answers) 2. What would happen to these frequencies if Jenny runs towards the merry-go-round. sx5=25 marks. For Answers click Answerpages | ||||||||||||||||||||||||||||||||
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Fig.3
Fig. 5. Two sources producing waves. 
