1. 4.3 IB Wave Characteristics

2. Wave fronts and Rays Reflection Diffraction Refraction
Name the wave behaviors shown.
Reflection
Diffraction
Refraction
Look at the wave fronts.
Which are similar? Which are different? Explain.

3. Energy of Wave Determines
Loudness - Sound
Brightness - Light
Energy received by observer depends on distance to source.
It’s radiated out in all directions.

4. definition of intensity I
Amplitude and intensity
Intensity is the rate energy is being transmitted per unit area and is measured in (W m-2).
I = power / area
definition of intensity I
1: A 200. watt speaker projects sound in a spherical wave. Find the intensity of the sound at a distance of 1.0 m and 2.0 m from the speaker.
Whatever power is in the wave front is spread out over a larger area as it expands.
The area of a sphere of radius x is A = 4x2.
For x = 1 m: I = P / (4x2) = 200 / (41.02) = 16 W m-2.
For x = 2 m: I = P / (4x2) = 200 / (42.02) = 4.0 W m-2.
Doubling your distance reduces the intensity by 75%!

5. Amplitude and intensity
Intensity is the rate energy is being transmitted per unit area and is measured in (W m-2).
A = 4x2 for a spherical wave, we can rewrite our intensity formula.
Total energy ET of a particle in SHM was ET = (1/2)kxo2, where xo was the amplitude A of the oscillation.
Since P = ET / time, clearly P  ET so that P  A2.
But I = power / area so :
I = power / area
definition of intensity I
I = power / 4x2
Intensity I vs. distance x
I  x -2
I  A 2
Intensity I vs. amplitude A

6. Solving problems involving amplitude and intensity
2: At a distance of 18.5 m from a sound source the intensity is 2.0010 -1 W m-2.
Find its intensity at a distance of 26.5 m.
SOLUTION: We can just use I  x -2 and dispense with finding the actual power as an intermediate step.
Then I1  x1-2 and I2  x2-2 so that
I2 / I1 = x2-2 / x1-2
= x12 / x22
= (x1 / x2)2.
Thus
I2 = I1 (x1 / x2)2
= 2.0010 -1 (18.5 / 26.5)2 = W m -2.

7. Solving problems involving amplitude and intensity
2: At a distance of 18.5 m from a sound source the intensity is 2.0010 -1 W m-2.
(b) Compare the amplitudes of the sound at 18.5 m and 26.5 m.
SOLUTION: We can use I  A 2 and I1 and I2.
Then I1  A12 and I2  A22 so that
I1 / I2 = A12/ A22.
Then
A1 / A2 = I1 / I2
= 10 −1 /
= 1.43.
A1 is 1.43 times A2.

8. Superposition
3: Two waves P and Q reach the same point at the same time, as in the graph.
The amplitude of the resulting wave is:
0.0 mm. B. 1.0 mm. C. 1.4 mm. D. 2.0 mm.

9. 4.3 – Superposition
Fourier series are examples of the superposition principle. any waveform can be represented by summing up sine waves! y 1 4 2 -
y =  yn
n=1 5
y5 = - sin 5t 1 5 T 2T t
y3 = - sin 3t 1 3
y4 = - sin 4t 1 4
y2 = - sin 2t 1 2
y1 = - sin t 1 1

10. Polarization
transverse waves - oscillations are
perpendicular to the propagation of the
traveling wave.
longitudinal waves oscillate parallel to motion.
Transverse waves can have infinitely many modes of oscillation, each perpendicular to the propagation,
longitudinal can only have a single mode.
Because of these allowed modes, the phenomenon of polarization only applies to transverse waves.
We consider EM waves.

11. Light is a transverse wave with two types of oscillation.
Polarization
Light is a transverse wave with two types of oscillation.
EM radiation consists
of perpendicular oscillating electric
and magnetic fields.
© 2006 By Timothy K. Lund
Focusing on electric field.

12. An oscillating electric charge produces EM waves.
Polarization
An oscillating electric charge produces EM waves.
 In light sources like the sun, a glowing gas, or an incandescent filament, the charges can oscillate in any direction, so produce random and continuous orientations of the electric field. c

13. Polarization
Random orientations of electric fields in a light source make up unpolarized light.
A simplified sketch shows only he fields at a single point:
Polaroid, a manmade film can take unpolarized light, and absorb all rays of light whose electric fields are not oriented in a certain line.
The E-fields of the light that passes through are all oriented in a single direction and are polarized.
Unpolarized light
Unpolarized light (simplified view)
Polaroid filter

14. If un-polarized light is passed through the Polaroid film it will absorb all the rays not oriented with the film.
Polaroid film
unpolarized light
linearly-polarized light
POLARIZER
The object used to polarize unpolarized light is called a polarizer.

15. Polarization
Polaroid is made of very long, parallel molecules. A simple model is a slotted disk. Only electric fields oriented in the proper manner are allow to pass. The rest are absorbed.
One would think that the electric field would be allowed to pass through the “molecular chain” only if parallel to them, but only perpendicular E-fields can pass through.
The light that makes it through Polaroid molecules looks like this:

16. Polarization
This is un-polarized light passing through a linear polarizer. (The light is traveling from right to left.)
Linear-polarized is also called plane-polarized.

17. The 1st filter is the polarizer, the 2nd is the analyzer.

18. Two Polaroid filters are placed in a beam of un-polarized light.
What happens to the intensity of the light when the second filter is rotated through 90º?
The first filter plane-polarizes the incoming light.
The second filter, originally oriented to allow passage, is now oriented to completely block the polarized beam.
Polarization

19. Solving problems involving Malus’s law
Analyzer and Polarizer at angles
Polarizing filters are sketched with lines showing the orientation of the E-field allowed through (unlike molecular chains):
 is the angle between the orientations of polarizer P and analyzer A.
Only a component of the E-field can pass.

20. Solving problems involving Malus’s law
The intensity of a wave is proportional to the square of its amplitude.
The intensity of the light that comes out of the analyzer is:
Malus’s law
I = I0 cos2 
I0 is the original intensity of the light and  is the angle of the analyzer.
polarized light E (from polarizer)
direction of allowed E-field (by analyzer) 
E cos  E 
angle through which analyzer has been turned
E-field allowed to pass

21. Solving problems involving Malus’s law
Ex 1: The directions of two sheets of Polaroid are initially parallel.
Calculate the angle through which one sheet needs to be turned in order to reduce the amplitude of the observed E-field to half its original value.
(b) Calculate the effect this rotation has on the intensity.
(c) Calculate the rotation angle needed to halve the intensity from its original value.
Solve cos  = 1 / 2   = 60º.
(b) I = I0 cos2  = I0(1/2)2 = I0 / 4.
(c) Let Io = 1.
I0 / 2 = I0 cos2   cos2  = 1 / 2. Thus
cos  = (1 / 2)1/2   = 45º.

22. Solving problems involving Malus’s law
Use I = I0 cos2 . Then
I = I0 cos2 60º
I = 0.25I0

23. Solving problems involving Malus’s law
I0 cos2 0º = I0
I0 cos2 60º = 0.25I0
I0 cos2 90º = 0
I0 cos2 120º = 0.25I0
I0 cos2 180º = I0

24. Polaroid is not the only way to polarize light
Polaroid is not the only way to polarize light. For example, if light reflects off of the surface of a liquid or passes through a liquid or other solids it can become partially or fully polarized.
Polaroid sunglasses block much of the horizontal light.

25. Polarization by reflection – Brewster’s law.
It turns out that if the refracted and reflected rays make a 90º angle then the reflected ray will be totally linearly polarized.
The particular angle of incidence at which this total polarization occurs is called Brewster’s angle.
Brewster’s law
If refl + refr = 90º then the reflected ray will be completely plane-polarized.
( Or inc + refr = 90º )
PRACTICE: Use the simplified method to draw an unpolarized incident ray becoming completely polarized by reflection.
inc
refl
refr

26. Liquid Crystals can change polarization direction as a function of applied voltage.

27. 3 – D glasses and 3 – D movies.
4:21

28. Hwk Formative assessment 4.3