Upload
duane-marshall
View
222
Download
0
Embed Size (px)
Citation preview
Higher Physics – Unit 3
3.1 - Waves
a
a
λ
λ
crest
trough
Wave Theory
All waves transmit energy.
The energy of a wave depends on its amplitude.
a = amplitude
λ = wavelength
Frequency of a wave: the number of waves per second.
Period of a wave: time taken for one complete wave to pass a point/be transmitted.
T1
f period
(s)
frequency
(Hz)
frequency of wave = frequency of source
Wave speed: is the distance travelled in one second.
Wavelength: is the minimum distance in which a wave repeats.
λ fv
frequency
(Hz)
speed
(ms-1)
wavelength
(m)
In Phase
Two waves are in phase when the same part of both waves arrive at the same point in space (peak/trough) at the same time.
Out of Phase
Two waves are exactly out of phase when opposite parts of both waves arrive at the same point in space (peak and trough) at the same time.
Coherence
Two waves are coherent if they are:
• the same frequency ( same wavelength)
• the same speed
• in phase.
Wave Behaviour
All waves display characteristic behaviour.
They all show:
• reflection
• refraction
• diffraction
• interference.
Reflection
The law of reflection states:
angle of incidence = angle of reflection
mirror
normal
i r
incident ray
reflected ray
Refraction
Waves change direction when passing from one medium to another.
This is because waves move at a different velocity in each media.
ir
ir
Diffraction
Waves bend round corners.
long wavelength
short wavelength
Interference
Interference occurs when two or more waves are superimposed.
The total effect is the sum of the waves.
+ =
constructive interference (in phase)
+ =
destructive interference (exactly out of phase)
Interference is the test for a wave.
In order to prove that any form of energy travels as a wave, an interference pattern must be shown.
Interference Patterns
To show that a form of energy travels as a wave, an interference pattern must be shown.
Thomas Young did this in 1801 for light, proving that light is a wave motion.
s1
s2
two troughs meet
crest lines from s1
crest lines from s2
two crests meet
crest and trough meetTroughs are not drawn, but
they would be between two crests:
constructive interference
constructive interference
destructive interference
+ =
+ =
+ =
d
d
d
d
c
c
c
c
c
All the points of constructive interference join up to form a series of lines.
Similarly, all the points of destructive interference join up to form a series of lines.
The two sources of wave must have the same frequency and be in phase.
Young’s Slits Experiments
laser
central maxima
1st order maxima
2nd order maxima
1st order maxima
2nd order maxima
3rd order maxima
3rd order maximadouble slit
(splits light source into two)
The fringes are more widely spread when the slits are closer together.
maxima
minima
Bright spots are called maxima (waves must be arriving in phase).
Dark regions are called minima (waves arriving exactly out of phase).Path Difference for Maxima
FringePath
Difference
central maxima 0
1st maxima λ
2nd maxima 2λ
3rd maxima 3λ
nth maxima nλ
λ ndifference path
number fringe n where
s1
s2
Waves from s1 and s2 are travelling the same distance, so there is no path difference for the central maxima.
s1
s2
At the first order maxima, waves must arrive in phase.
The path difference must be one complete wavelength.
At the second order maxima, waves must arrive in phase.
The path difference must be two complete wavelengths.
s1
s2
path difference
path difference
Path Difference for Minima
FringePath
Difference
1st minima ½ λ
2nd minima 1½ λ
3rd minima 2½ λ
nth minima (n - ½) λ number fringe n where
λ 21
-ndifference path
At the first order minima, waves must arrive exactly out of phase.
The path difference must be ½ a wavelength.
At the second order minima, waves must arrive exactly out of phase.
The path difference must be 1½ wavelengths.
s1
s2
path difference
s1
s2
path difference
Example 1
Find the wavelength of the signals used in the following experiment, where x represents the 3rd minima.
51 cm
61 cm
x
λ 21
ndifference path
Minima
λ 21
35161
λ 21
210
cm 4λ
Example 2
Find the wavelength of microwaves used in the experiment shown.
75 cm
84 cm
maxmaxmaxmax
Maxima
λ n difference path
λ 3 75-84
n = 3 (since central maxima is n = 0)
λ 3 9
cm 3λ
Diffraction Grating
A diffraction grating is a slide with multiple slits equally spaced. (typically 400-800 per mm).
θ
θ
path difference
s1
s2
s3
d
d
to maxima λ n difference path
ddifference path
θ sin
dλ n
θ sin
sinθ dλ n
For Maxima
Example 1
Calculate the wavelength of light passed through a diffraction grating, when the first order maxima is produced at an angle of 21°.
The grating has 600 lines per mm.
?λ
600101
d3
m101.67 6
1n
21θ θ sin dλ n
21 sin 101.67λ1 -6
21 sin 101.67λ -6
m 105.97λ 7
nm597 λ
Example 2
Light of wavelength 700 nm is shone through a diffraction grating that has 1200 slits per millimetre.
Calculate the angle between the central and first order maxima.
θ sin dλ n nm 700λ
m 10 700 -9
1200101
d3
m108.33 7
1n
?θ
θ sin 108.3310700 1 7-9
7
-9
108.3310700
θ sin
0.84θ sin
0.84 sinθ -1
57.1θ
Measuring Wavelength
Aim
To calculate the wavelength of laser light.
Diagram
laser
screen
diffraction grating
θ
Method
• Measure distance from laser to screen.
• Measure distance between central and first order maxima.
• Using Pythagoras, calculate angle at which constructive interference occurs
• Slit separation is calculated by:
• For first order maxima, n = 1.
slits of no.slide of width
d
Results
θ
1m
0.05m
adjopp
θ tan
10.05
θ tan
0.05tanθ 1
2.8θ
?λ
1n
2.8θ
dθ sin dλ n
Effect of Colour on Gratings
Monochromatic light beams consist of one single frequency of light.
Here are some wavelengths of monochromatic light:
Colour Wavelength (nm)
red 700
green 550
blue 500
violet 400
need to know these values but
** NOT GIVEN IN EXAM **
TASK
Calculate the angle between the central and first order maxima for the four colours of light passing through a diffraction grating with 700 lines per mm.
Red
θ sin dλ n
θ sin 101.4310700 1 6-9
6
-9
101.4310700
θ sin
0.49θ sin
0.49 sinθ -129.3θ
Green
θ sin dλ n
θ sin 101.4310550 1 6-9
6
-9
101.4310550
θ sin
0.38θ sin
0.38 sinθ -122.6θ
Blue
θ sin dλ n
θ sin 101.4310500 1 6-9
6
-9
101.4310500
θ sin
0.35θ sin
0.35 sinθ -120.5θ
Violet
θ sin dλ n
θ sin 101.4310400 1 6-9
6
-9
101.4310400
θ sin
0.28θ sin
0.28 sinθ -116.2θ
Conclusion
The longer the wavelength the the maxima separation.
The shorter the wavelength the the maxima separation.
greater
smaller
White Light Spectra
Diffraction Grating
When a white light source is used to produce an interference pattern with a diffraction grating, the following is seen:
central maxima (white)
1st order maxima (visible spectrum)
1st order maxima (visible spectrum)white light
source
diffraction grating
Dispersion is caused by interference.
All other maxima are seen in the form of the visible spectrum.
Red light makes the greatest angle with the original ray of white light because it has the longest wavelength.
The longer the wavelength the greater the maxima separation.
More than one spectra is produced.
Prism
When light is dispersed using a prism, the following is seen:
prismwhite light
source
Dispersion is caused by refraction.
Violet light makes the greatest angle with the original ray of white light because it has the highest frequency.
The higher the frequency, the greater refraction that takes place.
Only one spectrum is produced.