to each other (see work on vectors)..hence a receiving aerial will
pick up some of the signal.
2.2.8 The PRINCIPLE OF SUPERPOSITION
When two or more waves meet two things are observed to happen.
- They will pass through each other and be unaffected see fig 11.3 Hutchings. (this is just as well practically)
-
At the point(s) where they meet the RESULTANT amplitude will be the VECTOR SUM of the individual waves...
eg..consider the two waves below meet and superpose.
Resultant Wave..
2.2.9 Uses of Superposition.
(1) Double Slit Interference.
(Young's Double Slit Expt).
- Consider two sources of waves placed close to each other.
- Each wave spreads out in all directions and crosses the other giving varying degrees of interference.
-
At particular points a crest of 1 wave will cross the crest of the other and constructive interference will occur.
-
At particular points a crest of 1 wave and the trough of will meet producing destructive interference
- If a line is taken through the pattern areas of maximum and minimum interference will be seen.
If the two sources were Sound from two loudspeakers then a person travelling along this line should hear alternating loud and quiet sounds. If the two sources were Light bulbs then one would expect to see Light and Dark areas. This does not occur in practice due to the uneven nature of each wave from the two sources (lack of COHERENCE) and the fact that for light the sources have to be close together.
2.2.10 Conditions for Observing Interference with Light.
-
The Light must be COHERENT..
This may be achieved by using a laser, followed by the double slit to produce two sources.
OR By using a single slit followed by the double slit. Produces fringes much closer together a travelling microscope is usually needed to view the fringes.
-
Each slit width must be much smaller than the separation of the slits
2.2.11 Equation for Locating Light Maxima link with Path Difference
Consider two coherent sources A and B. a screen is placed distance D away. The distance between the two sources is d. The wave has a wavelength λ.
The difference in the distances between each source and a point on the screen is called the path difference
Path Difference = BP - AP
Position of 1st Maxima.
The first maxima occurs at a distance d from the centre at a position P say.
The light at P must be the result of a peak from A and a peak from B.
This first occurs when the distance from A is ONE WAVELENGTH greater than the distance of P from B.
ie The Path Difference = 1 Wavelength.
Position of 2nd Maxima.
Path Difference = 2 Wavelengths. and so on...
Position of nth Maxima
Path Difference = n Wavelengths
2.2.12 Positioning of Minima.
In this case the path difference is 1/2 wavelength or 1.5 wavelength etc.
2.2.11 Young’s Double Slit Equation:
- This equation is used to calculate the spacing between maxima.
fringe spacing = Dλ
d
Examples:
(1) A red light of wavelength 600nm is incident on a double slit of slit separation 2mm. Calculate the spacing between the fringes on a screen placed 2m away
(a) Describe what would happen to the fringe spacing if the slit spacing was reduced
(b) Describe what would happen if yellow light were used instead.
(2) Estimate the speed of sound produced by the signal generator Young’s double source experiment in the lab.
2.2.12 Standing or Stationnary Waves.
Explains why
- musical insturments give particular sounds.
- long organ pipes produce low sounds and small organ pipes produce high notes.
- how you get different notes out of a guitar string
- aerials have particular lengths
- people like singing in the bath !!
So what is a stationnary wave ?
STATIONARY or STANDING WAVES result from the SUPERPOSITION or INTERFERENCE of 2 wavetrains which have the same amplitude, frequency and wavelength but travelling in opposite directions with the same speed. This occurs at particular frequencies only.
What does this mean ?
- Two waves of same speed and frequency travel towards each other
-
At particular frequencies the wave appears to have stood still ie a stationnnary wave has been produced In this condition we say that accoustic RESONANCE has occured.
Important: Unlike Progressive waves there is no net transfer of momentum or energy by a stationary wave.
2.2.13 Standing wave problems.
Students should be able to recall and draw the following diagrams then work out the relationship between the wavelengths and the length of the pipe/string from basic principles.
(a) Standing Waves on a String. (Transverse) Stringed instrument problems.
Key Ideas:
-
There must be a NODE (zero displacement) at each end of the string (fixed end and vibrating end). If the end of fixed it cannot move so there cannot be any displacement.
- Always draw a diagram showing the situation.
Diagrams:
Fundamental or first harmonic.
Note that this is the largest wave or part of a wave that can be fitted between the fixed ends in this case 1/2 a wave can be fitted in
If L is the length of the string and the length doesn’t change that much then we can say that
L = λ / 2
2nd Harmonic
In this case we look for the next largest wave or part of a wave that we can fit between the fixed end.
L = λ
3rd Harmonic
L = 3λ
2
Example
The speed that a wave travels in a guitar string is m/s. If the string is 70cm long what is the frequency of the lowest note that will be produced
Solution:
(1) using the wave equation v = f λ
f = v / λ
problem we know v but not λ answer draw a diagram !!!!
(2) lowest frequency means largest wavelength ie 1st Harmonic
we know that L = λ
2
Hence λ = 2L
λ = 2 x 0.7
= 1.40m
(3) Using f = 500 the velocity depends on the
λ tension in the string amongst
other things !
f = 500
2 x 0.7
f = 357 Hz
Hence lowest frequency heard is 360 Hz.
(b) Standing Waves in a Pipe. Longitudinal
These are a little more difficult to visualise. Consider a longitudinal wave with compressions (particles compressed together, area of high pressure) and rarefractions (particles pulled apart, area of low pressure). These may be pictured in the following way
Uses wind instruments, singing in the bathroom etc...
Longtudinal Wave
Diagrammatic.
Representation.
Key Ideas
-
At an open end there must be an ANTINODE.
-
At a closed end there must be a NODE.
- Always draw a diagram with the wave superimposed in the tube.
Diagrams: Pipe Open at Both Ends.
Fundamental Frequency. ______________________________
or 1st Harmonic
______________________________
2nd Harmonic.. ______________________________
______________________________
3rd Harmonic. ______________________________
______________________________
Pipe Closed at One End
Diagrams:
Fundamental Frequency. ______________________________
______________________________
2nd Harmonic.. ______________________________
______________________________
3rd Harmonic.. ______________________________
______________________________
Problems with pipes are then solved in the same way as the waves on string problems by finding the wavelength from the dimensions of the pipe the using v = f λ to find the frequency
-
When a standing wave in a pipe occurs there is definite increase in the loudness of the sound the pipe is said to RESONATE.
-
NOTE: Strictly speaking there are small "end corrections" that need to be made to the lengths of the pipes to determine the resonant frequency..this is beyond the scope of this course.
Questions. Hutchings Q13.2 (p217) , W&H Qs 41.44 (hint work out λ first then find L)
41.46, 14.18 (hint draw a good diagram!)
This brief set of notes introduces students to some of the basic concepts of geometric optics ie where light can be draw as straight lines as opposed to being drawn as waves.
2.2.20 Reflection
Plane mirror (ie flat reflecting surface)
normal, line at 90°
to the surface
angle of incidence (i) angle of reflection (r)
Incident ray reflected ray
Law of reflection angle of incidence = angle of reflection
2.2.21 Refraction
When a ray of light passes from one medium to another it bends (refracts).
e.g. Light going from a less In this case the ray bends
dense to a more dense towards the normal.
medium e.g. air to glass
e.g. Light going from a more In this case the ray bends
dense to a less dense away from the normal.
medium e.g. glass to air
2.2.22 Why does the ray of light bend ?
As the ray of light moves from a dense to a less dense medium it slows down. This change in speed causes it to change direction.
Refractive Index.
The ratio of the speed of light in a vacuum (c) to the speed of light in the material (v) is called the ABSOLUTE REFRACTIVE INDEX of the material (n)
n = c / v
Snells law,
n1 sin θ1 = n2sinθ2
where
n1 is the absolute refractive index of material 1 and (here absolute means
n2 is the absolute refractive index of material 2. compared to a vacuum)
2.2.23 Refractive Index (μ)
May be more usefully called the RELATIVE REFRACTIVE INDEX. It is the ratio of one refractive index to the other.
μ = n1 = (c1/ v1) / (c / v2) = c x v2
n2 v1 c
μ = n1 = v2
n2 v1
2.2.24 The Critical Angle.
Consider the diagram shown below with a ray of light passing from a dense material to a less dense material. As you can see the light is refracted for small angles of incidence. At a particular angle of incidence, known as the critical angle, the angle of refraction is 90°, for angles of incidence greater than the critical angle all the light is relflected this is known as total internal relfection. Unlike reflection in mirrors (where there is an intensity loss of about 4%) there is no loss in intensity in total internal reflection.
Calculating the critical angle.
Consider a ray of light moving from a more dense material of absolute refractive index(n1 ) to a less dense material of absolute refractive index (n2)
n1 sin θ1 = n2sinθ2
n1 = sinθ2
n2 sin θ1
μ = sinθ2
sin θ1
but θ2 = 90 ° hence sinθ2 = 1
μ = 1
sin θc
μ = 1
sin θc
2.2.25 Application of the critical angle in optical fibres.
An optical fibre consists of a graded (graded means that the density is not uniform) central core of glass which is surrounded by an outer cladding of material with a lower refractive index of refraction. As you can see from the diagram below most of the light is totally internally reflected down the fibre.
Advantages of using optical fibres
- The raw material is very cheap (glass is made from sand !)
- There is virtually no interference for communications sent this way.
- The signal gets sent at the speed of light reducing signal delay.
- Its vitually impossibly to “tap” into an optical transmission
- Approx 20 000 more messages can be sent through 1 fibre compared to a similar copper cable which can only take one message at any time.
- The signal is sent digitally down the cable reducing signal distortion.
Questions.
(1) Calculate the angle of refraction for a ray of light moving from water to glass if the angle of incidence in the water is 30°. n for water is 1.33, n for air is 1.0003.
(2) Calculate the angle of incidence if the angle of refraction of a ray of light moving from glass to air is 40°. Take n for glass to be 1.58.
(3) Calculate the critcial angle of from glass (n=1.58) to the following materials
(i) air n = 1.0003 and
(ii) turpentine n =- 1.47