Transcript Electromagnetic Waves

Electromagnetic Waves
G3 Two Source Interference of
Waves
G4 The Diffraction Grating
G5 X-Ray Diffraction
Recap: Superposition (interference)
Whenever two waves of the same type meet at the
same point, the total amplitude (displacement) at
that point equals the sum of the amplitudes
(displacements) of the individual waves.
Superposition links
- PheT Sound (see jar file)
- Superposition of two pulses
- With editable wave equations
- Creating a standing wave
Two Source Superposition Experiments
Demo: Superposition of Sound Waves
Signal
generator
(3kHz)
CRO
loudspeakers
microphone
Path Difference
Q
2nd subsidiary
maximum
P
1st subsidiary
maximum
O
Central
maximum
P’
1st subsidiary
maximum
Q’
2nd subsidiary
maximum
S1
S2
At O : Zero path
difference
At P and P’ path
difference = 1λ
At Q and Q’ Path
difference = 2λ
So in the above example…
S2Q – S1Q = 2λ
For constructive interference at any point,
wavefronts must be ‘in phase’ and their path
difference must be a whole number of wavelengths:
path difference = nλ
For destructive interference at any point, wavefronts
are ‘π out of phase’ and their path difference is
given by:
path difference = (n + ½) λ
Note: The two wave sources must be of similar
amplitude for the interference pattern to be
observed. If one of the sources is of much greater
amplitude, the pattern will not be clearly visible.
Young’s Double Slit Experiment
Setup:
Interference
pattern
Lamp
Coloured
filter
Double slit
Screen
Single slit
Coherent light
D
P
λ
S1
T
d
M
θ
θ
O
S2
y
P’
D = distance from slit to screen (few m)
d = slit separation (0.5mm)
y = distance from central maximum to first subsidiary maximum
At point P’, the path difference between waves from S1
and S2 is one whole wavelength. Thus they arrive in
phase, constructively interfere and create a ‘bright
fringe’.
S1T = λ
Distance D is large and the wavelength of the light is
small so θ must also be very small. We can also
approximate that angle S1TS2 is a right angle.
sinθ = λ
d
tanθ = y
D
Using the small angle approximation sinθ = tanθ = θ

λ = y
d
D
or...
y = Dλ
d
Effect of increasing the total number of slits
(See diagram.)
Increasing the total number of slits...
- Increases sharpness of fringes
- Gives rise to other faint fringes between the
bright fringes
- Increases the intensity of the central maximum
The Diffraction Grating
The diffraction grating is (in theory) a piece of opaque
material with many parallel, equidistant and closely
spaced transparent slits that transmit light. In practice
lines are ruled onto glass with diamond leaving
transparent glass in between.
Experiment:
Observe a white light
source through both coarse
(100 lines/mm) and fine
(500 lines/mm) gratings.
Repeat placing red or green
filters in front of the grating.
Observations for white light:
- The diffraction spectra of white light has a central
white band (called the zero order image).
- On either side are bright bands of colour. The first
red band is the ‘first order spectrum’ for that
wavelength. Further out is another identical red
band - the ‘second order spectrum’ etc.
- Bands for the visible spectrum are seen, with violet
being nearest the centre and red furthest.
- A finer grating forms less orders, further apart than
on a coarse grating.
Consider a grating with coherent, monochromatic light
of wavelength λ incident upon it:
A
θ
d
N
θ
B
Monochromatic
light
C
Light
diffracted at θ
to the normal
Section of
grating
If light from any pair of slits reaching a point in the
distance (e.g. focused by the lens in your eye)
causes maximum constructive interference, we
know that the path difference must be a whole
number of wavelengths...
AN = nλ
Hence...
d sinθ = nλ
n = 0, 1, 2, 3 etc
Thus light from all slits constructively interferes with
that from every other at values of θ determined by
the equation, producing bright regions. This gives
rise to the different order spectra for any particular
wavelength of light.
Q1.
a. Determine the values of θ for the first and
second order spectra of light of wavelength
600nm incident upon a fine grating of 600
lines/mm.
b. Which is brightest and why?
c. Explain why a third order spectrum is not
possible.
Q2.
Explain (using the formula), why a finer grating
produces less orders of spectra than a coarse
grating.
n=2
n=1
n=0
Monochromatic
light
n=1
Section of
grating
n=2
Experiment
Use a diffraction grating to determine the
wavelength of laser light.
X-Ray Diffraction
Incident
wavefronts
Diffracted
wavefronts
Plane of atoms
θ
θ
A
d
C
B
If the diffracted waves are in phase, ABC = nλ
i.e.
2d sinθ = nλ
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