Download Section Three: Light and Matter

Section Three: Light and Matter
Topic One: Electromagnetic Waves
electromagnetic waves are transverse waves that have both an electric and a magnetic component. A changing
magnetic field (B) would produce a changing electric field (E) that is at right angles to the magnetic field and that a
changing electric field will produce a changing magnetic field. The net result of the interaction of the changing E and
B fields is the production of an electromagnetic wave that moves away from the wave source at the speed of light.
electromagnetic spectrum includes radio waves, microwaves, infrared radiation, visible light, x-rays and gamma rays.
They differ in frequency f and wavelength λ but all travel at the speed of light.
Electric Fields and Waves
Shake the end of a stick back and forth in still water and you will produce waves on the water surface.
Similarly shake a charged rod back and forward in empty space and you will produce electromagnetic waves in space.
This is because the shaking charge can be considered a varying electric current and a varying magnetic field surrounds
such an electric current. No medium is required.
The vibrating fields move out from the vibrating
source and at any point in space the electric field
is perpendicular to the magnetic field and both
are perpendicular to the direction of the wave.
Light is energy that is emitted by accelerating
charges - often electrons in atoms. This energy
travels in a wave that is partly electric and partly
magnetic. A changing electric field E will produce a changing magnetic field B..
This electromagnetic wave is transverse in nature because both the E and B fields are perpendicular to the direction of
propagation of the wave. These waves are waves of fields not matter and can propagate (move) in empty space.
Accelerating electric charges give rise to electromagnetic waves.
Reception of EM Wave by an Antenna
One kind of antenna consists of one or more conducting rods. The
electric field in the EM wave exerts a force on the electrons in the
conductor causing them to move back and forth at the frequencies
of the wave.
Note: A horizontal antenna radiates (or receives) waves that are
polarised in the horizontal plane.
A vertical antenna radiates (or receives) waves that are
polarised in the vertical plane.
Polarisation
The plane of polarisation of an electromagnetic wave is the plane defined by the direction of travel and the direction
of the oscillating electric field.
Unpolarised light oscillates in many directions
perpendicular to the direction of travel. These
oscillations can be resolved into vertical and horizontal
components. Unpolarised light can be passed through a
polariser to produce plane polarised light in which the
oscillations are confined to a single plane.
A polariser (or polarising filter) can also be used as an analyser to
determine the polarisation of light. If the light passes unaltered,
the polarising filter is aligned with the polarisation plane of the
light. If the light is completely blocked, the polarising filter is at
right angles to the plane of polarisation of the light.
APPLICATION: LADS (Laser Airborne Depth Sounder)
This device is an Australian system which aims to accurately chart water depth in the regions of sea surrounding
Australia – of course it can be used to chart fresh water lakes as well.
As a device to measure depths in water it is not unique. This was done previously using sound waves emitted vertically
down from slow moving ships near the coast.
LADS uses the same physical idea in that it involves the measuring the time difference between emission and reception
of a pulse of waves that have been both reflected from the top water surface and it bottom. If the speed of the wave in
the water is known then Time difference × speed = depth
2
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LADS uses optical rather than sonar methods.
A pulsed beam of laser light is directed down from a plane – from an infra red laser. The pulse width is very short to
improve accuracy. This light reflects from the surface to the plane, which is carrying the source. An electronic device
doubles the frequency of the laser light so that a beam of green visible light is also produced. The infra red light is
almost all reflected from the water surface, the green light penetrates the surface and is reflected from the bottom of the
sea, or lake. Both the infra red and green light travel at the same speed in air.
The plane flies quite low above the water surface and
by measuring the time delay between a pulse
being reflected from the surface and one
(green)
from the bottom the time for double the depth for the green light
to travel at a known speed, in water, can be calculated.
The speeds of green light, in water are about
2.24 × 108 m s-1 – it depends on whether salt water or fresh water is used.
In addition problems arise because of light energy being absorbed
or scattered by particles in the water so that
high-energy lasers are needed.
(infra red)
(green)
water surface
infrared
ocean floor
green
Topic Two: Interference of light
Key Terms
interference patterns
Displacement
are produced when two (or more) coherent sources produce waves of the same frequency
and amplitude which superimpose in the same region of space (ie they overlap).
sources are coherent
when there is a constant phase relationship between the waves emitted by the sources.
crest
is the highest point of that portion of
+ ve
W a v e le n g t h
a transverse wave above the
C r e st
equilibrium position.
trough
is the lowest point of that
portion of a transverse wave
below the equilibrium position.
A m p l it u d e
T ro u g h
destructive interference occurs if the amplitude of the
- ve
resultant of two interfering waves is smaller than the amplitude of either wave.
constructive interference occurs if the amplitude of the resultant of two interfering waves is larger than the
amplitude of either wave.
diffraction
occurs when waves bend and spread out as they pass an obstacle or through a narrow
opening. The amount of diffraction depends on the wavelength of the waves relative to the
size of the obstacle. In the case of a narrow opening, the amount of bending increases as
the size of the opening decreases.
diffraction grating
consists of a large number of closely space parallel slits which diffract light incident on
the grating. The diffracted light will exhibit a pattern of light and dark spaces due to
constructive interference and destructive interference.
polarization
is a property of light that indicates that light is a transverse wave phenomenon. A
transverse wave is a wave in which the oscillations or vibrations of the wave are at right
angles to the direction of motion of the wave. An electromagnetic wave in which the
electric vector is vibrating in only one plane is said to be plane polarized.
Speed, Frequency and Wavelength
The back and forth vibratory motion (often called oscillatory motion) of a swinging pendulum or mass suspended on a
spring produces a sine curve.
This curve depicts a wave:
Wavelength (λ) is the distance from one crest to another crest, more generally the shortest distance between successive
identical parts of the wave that are in phase. λ is measured in metres.
Amplitude (A) is the distance of the maximum displacement point of the wave from the mean position. (metres.)
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Frequency (f) is the number of to-and-fro vibrations in time. f is measured in cycles per second or hertz.
where Period (T) is the time taken to complete one oscillation.
Wavemotion involves a transfer of energy from one point to another, with no net transfer of matter.
frequency =
1
period
Wave Equation
v = fλ
here: v = speed or velocity of the wave (in m s-1)
f = frequency in hertz (Hz)
λ = wavelength in metres (m).
Transverse Waves
A wave where the motion of the oscillations of the wave are at right angles (perpendicular) to the direction of travel.
v
It consists of a series of crests and troughs.
Longitudinal Waves
The medium moves back and forth parallel to the direction in which the wave travels. The wave consists of a series of
compressions and rarefactions.
Note: Transverse waves and longitudinal waves transfer
R = R a r e fa c t io n C = C o m p r e s s i o n
the energy along the medium. The matter does not
transfer but moves to and fro about a fixed position.
C
R
C
R
C
R
C
R
This relationship holds for all kinds of waves whether they
are water waves, sound waves, radio waves, or light waves.
λ
P a r t ic le
The speed of light is a universal constant, c = 3.0 × 108 m s-1 in a vacuum.
V i b r a t io n
Electromagnetic Waves
Light is energy that is emitted by accelerating electric charges – often electrons in atoms. This energy travels in a wave
that is partly electric and partly magnetic. Such a wave is an electromagnetic wave.
Light is a small portion of the broad family of electromagnetic waves that include radio waves, microwaves and x-rays.
The range of the electromagnetic waves is called the electromagnetic spectrum. The range of wavelengths in the visible
spectrum is shown below. (1 nm = 1.0 × 10-9 m)
Wave Fronts
A wave front is a line on a surface on which all
points are in phase at any instant. This means that
when one point is at the crest, so are the others.
Huygens’ Principle
“Every point on a wave front can be considered as a source of tiny wavelets that spread out in the forward direction at
the speed of the wave itself. The new wave front is the envelope (ie. the common tangent to) of all the wavelets.”
Note: This principle is useful in predicting the future position of a wave front when an
earlier position is known.
Suppose that AB is a wave front for waves created from a point source S (ie. the wave is
travelling away from S at a speed v). In the wave front AB which is travelling away from
a source S with a speed v. To find the wave front after a short time t tiny circles, centred
on wavefront AB, radius r = vt represent the Huygens’ (imaginary) wavelets.
The tangent to all these wavelets, the curve CD, is the new position of the wavefront.
Interference
A monochromatic light source emits a wave of a single colour and therefore has single frequency and wavelength in a
vacuum(in air). e.g. Red light has wavelength of about 7.5×10-7m and hence a frequency of about 4×1014Hz.
Two sources are coherent if the waves created by them have a constant phase relationship.
An interference pattern is observed only when the sources are coherent. If two tiny light bulbs are used as sources, an
interference pattern would not be seen. The light emitted by one light globe would have a random phase with respect to
the second light bulb. Two such sources, whose output waves bear no fixed relationship to each other, are called
incoherent sources.
Wave Addition Wave Addition is the superposition of wave forms (interference)
C re st +
C re st
Crest + Trough
+
T ro u gh
+
+
=
C on stru ctive
I n te r fe r e n c e
T ro u g h
+
=
Destructive
Interference
=
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Young’s Double Slit Interference
Convincing evidence for the wave nature of light was provided by Thomas Young who was able to measure the
wavelength of visible light using an interference effect.
If light from a single monochromatic source falls onto a screen containing two closely spaced slits S1 and S2, bright and
dark fringes are seen on the screen behind.
Young realised that the bright fringes of
light resulted from light waves from
both holes (or slits) arriving crest to crest (constructive interference - more light).
Similarly, the dark areas resulted from light waves arriving trough to crest
(destructive interference - no light).
Young’s experiment can done in the laboratory with two closely spaced holes (or
slits) with a sodium vapour lamp, light from which is passed through a single slit
providing the monochromatic coherent source (a laser is even better because no
first single slit is required only a double slit).
The arrangement above is equivalent to two rocks being thrown into a lake, or
when sound from two loudspeakers interferes.
These two slits have circular waves leaving them because of the diffraction effect.
Using the diagrams we can see how an interference
pattern is produced on the screen.
The waves of wavelength λ are shown entering the
slits S1 and S2 which are a distance apart d. The waves
spread out in all directions after passing through the
slits. The diagrams show only three different angles.
(a)
(b)
Diagram (a) shows that at the centre of the screen (θ = 0°) the waves from each slit travel the same distance and are in
phase. ie. a crest of one wave arrives at the same time as the crest of the other wave. Hence the amplitude of both
waves add to form a larger amplitude and this is known as CONSTRUCTIVE INTERFERENCE and there is a bright
spot in the centre of the screen.
Diagram (b) shows that at an angle θ the lower wave travels an extra distance of one whole wavelength, and the waves
are in phase. Note that from the shaded triangle the extra distance equals d sinθ. At the point on the screen constructive
interference occurs again since the paths of the two rays differs by one wavelength (or any whole number of
wavelengths).
Diagram (c) shows that at this angle of θ the lower wave travels an extra distance
equal to one-half wavelength, so that the two waves arrive at the screen fully out of
phase. At this point on the screen since the extra distance travelled is one half
wavelength (or 3/2λ, 5/2λ and so on) the two waves are exactly out of phase when
they reach the screen: the crest of one wave arrives at the same time as the trough of
the other and so produce zero amplitude. This is DESTRUCTIVE INTERFERENCE
and the screen is dark.
The pattern on the screen is a series of bright and dark lines (or fringes) as shown
below.
Young’s Double Slit Interferometer
The coherent sources produce a pattern of uniformly spaced bright (constructive interference) and dark (destructive
interference) fringes.
The point O is a maximum intensity or reinforcement since the wavelets from the sources S1 and S2 meet at O at the
same time. The waves travelling from S1 and S2 to the screen will have a path difference.
If the path difference (the extra distance travelled by one wavelet to reach the same point on the screen) is:
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an odd number of half wavelengths then DESTRUCTIVE INTERFERENCE
an even number of half wavelengths then CONSTRUCTIVE INTERFERENCE.
A constant phase relationship between the sources is a necessary condition for observable interference to occur, they
are COHERENT (implies same λ).
To achieve
SCREEN
maximum contrast the amplitudes must be equal.
Single
Double
Slit
A single source S produces wavelets that meet S1 and S2
at the same time (since SS1=SS2). Therefore the sources
S1 and S2 have a constant same phase relationship (in
fact the same phase) so they are coherent.
From the diagram below
Slit
P
x
S1
S
d
M
θ
θ
S2
L
O
C
S2 C = dsin θ (path difference)
sinθ ≈ tanθ ≈ (since θ small)
x
=
L
dx
PD = S2 C =
L
If consider a position of constructive interference then
dx
= whole number of λ
L
= n λ (where n = 0,1,2,3,4, ...)
PD = S 2 C =
For a particular constructive band (fringe)
The path difference for the next band (fringe) is
dx 1
= nλ
L
dx
S2C = 2 = (n + 1)λ
L
S2C =
Rearranging these two equations so that x is the subject of the formulae gives:
x1 =
Note:
nλL
d
x2 =
(n + 1)λL
d
W = x2 − x1 =
(n + 1)λL − nλL
d
d
so W =
λL
d
W is the fringe separation or band width.
since λ is small we need L to be large
and d small to see W.
W is proportional to λ if L and d are
constant.
WRED > WVIOLET since λRED>λVIOLET.
The diagram of the intensity pattern of interference
fringes produced by the double slit experiment. The arrow marks the central fringe.
APPLICATION: CD (Compact Disc) player
The CD player uses a moulded plastic disc for the reproduction of sound. The CD works digitally using a binary code.
The CD has a spiral groove engraved on its surface (using a pulsed high-powered laser). The spiral goes from the
inside of the disc out. In the grooves there are a succession of pits
ie
The CD can be duplicated using a process where the engraved moulding is coated with a thin layer of either
Aluminium or Silver (both good reflectors of light as well as good conductors of electricity). This metal layer is then
coated with a transparent, acrylic plastic.
When the disc is placed in the CD player and turned on, the narrow beam of light from the low power, solid state (and
hence robust) laser can be alternatively reflected from the upper and lower humps on the disc - the disc is read from
underneath;
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λ/4
2 × λ/4 = λ/2,
thus there is a λ/2 total path difference between light reflected from the two, "depths" on the disc due to destructive
interference when path difference = λ/2
annul
zero path difference
therefore reinforce.
λ/4
The digital signals received are transformed into electrical pulses, which can then be transformed into pressure
variations that we detect as sound. The diagram below shows how the system is set up
Reading a CD.
The fine beam of a laser, focused even more finely with lenses, is directed at the under surface of a rotating compact
disc. The beam is reflected back from the areas between the pits but reflects much less than the pits. The reflected
light is reflected as shown, reflected by a half-reflecting mirror MS. The strong and weak reflections correspond to the
0s and 1s of the binary code representing the audio or video signal.
The advantage of a CD over other methods of sound
reproduction is that nothing mechanical (like a needle with a
record) actually touches the disc and also light from the pits is
such that dust causes no sound distortion.
The reflected laser beam is also used to control the spin rate
of the disc. The rate varies from 200 revs/minute when
near the centre to 500 revs/minute near the circumference.
This ensures a constant linear speed. A standard CD
stores about 74 minutes of sound.
Sample Problems: WAVES: Interference
1.
Consider the arrangement for Young’s double slit experiments using monochromatic light.
(a)
What effect would decreasing the distance between the two slits have on the distance between the centres of the
bright bands?
W=
2.
λL
If d is decreased then W is increased (inverse proportionality with λ and L constant).
d
What is the distance between the slits for a double slit arrangement which, illuminated with yellow light of
wavelength 6.0×10-7m, produces and interference pattern 1.5m away with bright bands 0.30 cm apart?
L = 1.5 m, W = 0.30 cm = 3×10 m, λ = 6.0×10 m
-3
3.
-7
λL
⇒ 3 × 10 −3 =
6 × 10 −7 × 1.5
d
d
6 × 10 −7 × 1.5
d=
= 3 × 10 − 4 m
−3
3 × 10
Monochromatic light passes through two parallel slits 0.03cm apart and then falls on a screen 120cm further
away. The first bright band is 0.16cm from the centre of the interference pattern. Find the wavelength of the
light.
-4
-2
d = 0.030 cm = 3×10 m, L = 1.2 m, W = 0.16×10 m
4.
(a)
W=
W=
λL
⇒ 0.16 × 10 − 2 =
λ × 1 .2
d
3 × 10 - 4
0.16 × 10 − 2 × 3 × 10 − 4
λ=
= 4 × 10 − 7 m
1 .2
Young’s double slit experiment.
In a Young’s double slit experiment using sodium light, nine interference fringes were found to occupy 3.6 mm
when viewed through a micrometer eyepiece with an image plane was 240 mm from the slits. Calculate the
separation of the slits, assuming the wavelength of sodium light to be 6×10-4 mm.
10 bands occupied 3.6mm (9 fringes)
λL
6.7 × 10 −7 × 2.4 × 10 −1
L = 240 mm, d = ?, λ = 6×10-7 m,
W=
⇒ 3.6 × 10 − 4 =
d
d
W = 3.6/10 = 0.36 mm
−4
d = 4 × 10 m
Page 6 of 17
(b)
When the sodium lamp was replaced by a white light source with a green filter, and the eyepiece moved 120
mm further from the slits, five green fringes were found to occupy 2.3 mm in the eyepiece. Find the mean
wavelength of the light passed by the filter.
5 fringes (6W = 23 mm)
λx
λ × 360 × 10 −3
−4
W
=
⇒
×
=
3
.
8
10
so λ = 4.2 × 10 −7 m
L = 240 + 120 = 360 mm, λ = ?,
-4
d
×
4
10
W = 2.3/6 = 3.8×10-4m
5.
What is meant by the term coherent light sources? Why are they necessary for interference effects to be
observed with light and how are they produced?
Coherent sources have a constant phase relationship (same λ) and better resolution if same amplitude (same
intensity). Light is not normally coherent (except laser light).
6.
In a Young’s double-slit interferometer, the separation of the slits is 0.1 mm. What is the band width of the
d = 0.1 × 10 -3 m
interference pattern produced 50 cm from the slits?
(a) with red light of wavelength 700nm
WR =
(b) with blue light of wavelength 450nm
WB =
λL
d
λL
d
L = 0.5m
⇒ 35mm
I
λ R =7 00n m
W R =3.5 cm
λ B =4 50n m
W B =2.2 5cm
⇒ 22.5mm
x
Page 7 of 17
Diffraction Gratings
Transmission Diffraction Gratings
We have discussed diffraction before with regard to water waves as well as light and we now have seen that it refers to
the spreading or bending of light around edges.
A diffraction pattern exists around any sharp object illuminated by a point source. This resembles the interference
fringes of a double slit and is due to the waves being diffracted around the object.
Diffraction Grating
A large number of equally spaced parallel slits is called a diffraction grating, although it may be more appropriate to
use the term interference grating.
The grating can be made by precision machinery by scratching very fine parallel lines on a glass
plate. The untouched spaces between the lines serve as the slits.
Photographic transparencies of the original grating serve as an inexpensive grating.
d Transmission Grating: is a glass slab on which a large number of lines have been scratched so that the
spacings of scratches is very small (eg. 10,000 scratches per cm).
The light is passed through the transparent part thus supplying a many slit source.
Reflection Grating is a mirror surface on which a large number of lines have been scratched.
The light is reflected from the smooth reflecting surface, eg. compact disc.
Both operate similarly except that the reflection grating is preferred as less energy from the incident light is absorbed.
These are also the most expensive. Reflection gratings are also useful as ultra-violet and infrared light are absorbed by
glass.
The analysis of the diffraction grating is much like that of Young’s double-slit experiment.
We assume parallel rays of light are incident on the grating.
Since this light is parallel there is a need for it to be focussed by a convex lense or even the lense in our eye.
We also assume that the slits are narrow enough so that diffraction by each of them spreads light over a very wide
angle on a screen and interference can occur with light from all other slits.
Light rays that pass through each slit without deviation (θ=0°) interfere
constructively to produce a bright line at the centre of the screen.
Constructive interference also occurs at an angle θ such that the rays from adjacent
slits travel an extra distance of Δl = mλ, where m is an integer.
For constructive interference
d sin θ = Δ l = m λ where (m = 0, 1, 2, ...)
These are the principal maxima and this is the criterion to have a maximum brightness.
Note previously in interference:
S2C = path difference = d sin θ where m is the order of the pattern
eg
m
m
=
=
1
2
dsinθ=mλ=λ
dsinθ=mλ=2λ etc.
When white light is passed through the grating, each wavelength
produces constructive interference at a different angle and so a series of spectra are produced.
(a)
a line spectrum with two wavelengths of
light, 400 nm and 700 nm.
(b)
Diffraction produced by white light.
the shorter wavelengths (blue) are diffracted through smaller
angles than the longer wavelengths (red), and the 1st order
spectrum will show less dispersion (spread) than the 2nd order.
the intensity of the 1st order spectrum is greater than the
intensity of the 2nd order because of diffraction causing greater
dispersion.
Page 8 of 17
Sample Problems: WAVES: Diffraction
Light of wavelength 5.0×10-7m passes through two parallel slits and the falls on a screen 4.0m away. The bright
bands of the interference pattern are found to be 2.0cm apart. Find the distance between the slits. The same two
slits are now illuminated by light of a different wavelength and it is found that the fifth minimum for this light
occurs at the same point on the screen as the fourth minimum for the previous light. Find the wavelength of the
second source of light.
1.
W=
2.
λx
d
⇒d=
λx
W
5 12 W2 = 4 12 W1 since W ∝ λ
=
5.0 × 10-7 × 4
= 10 − 4 m
-2
2 × 10
5.5λ 2 = 4.5λ1
λ2 =
4 .5
× 5 × 10 −7 = 4 × 10 −7 metres
5 .5
A diffraction grating has 6×105 lines per metre. Find the angular spread in the second order spectrum between
red light of wavelength 7.0×10-7m and violet light of wavelength 4.5×10-7m.
for red
dsinθ = nλ
for violet dsinθ = nλ
⎛ 2 × 7 × 10 -7 ⎞
⎟ = sin -1 0.838 = 56.9°
so θ = sin ⎜⎜
−6 ⎟
⎝ 1.67 × 10 ⎠
⎛ 2 × 4.5 × 10 -7 ⎞
⎟ = sin -1 0.539 = 32.6°
so θ = sin -1 ⎜⎜
−6 ⎟
⎝ 1.67 × 10 ⎠
-1
Angular separation is 24.3°
3.
A He-Ne gas laser which produces monochromatic light of a known wavelength λ=6.328×10-7m is used to
calibrate a grating in a spectroscope. The first order diffraction line is found at an angle of 20° to the incident
beam.
How many lines per metre are there on the grating?
dsinθ = nλ dsin20° = 6.328 × 10 -7 so d =
4.
The red light from three separate sources is mixed and passed through a diffraction grating of 3×105 lines per
metre. The wavelengths of the three lines are:
Hydrogen 6.56×10-7m
(a)
6.328 × 10 -7
= 1.85 × 10 −6 metres so 5.4 × 10 5 lines/metre
sin20°
Neon 6.50×10-7m
Argon 6.97×10-7m
Calculate the deviations for the first order diffraction lines of each of these sources.
⎛ 6.56 × 10 -7 ⎞
⎟ = sin -1 0.197 = 11.4°
for hydrogen θ = sin -1 ⎜⎜
−6 ⎟
×
3
.
33
10
⎝
⎠
⎛ 6.50 × 10 -7 ⎞
⎟ = sin -1 0.195 = 11.3°
for neon θ = sin -1 ⎜⎜
−6 ⎟
×
3
.
33
10
⎝
⎠
⎛ 6.97 × 10 -7
for argon θ = sin -1 ⎜⎜
−6
⎝ 3.33 × 10
5.
6.
7.
⎞
⎟⎟ = sin -1 0.209 = 12.08°
⎠
Explain why no interference bands are observed when light from two adjacent lamps fall on a screen.
Two sources are not coherent. The photons are produced randomly and there is not a constant phase
relationship.
A beam of light is polarised by passing it through a piece of ‘polaroid’. In what way does the polarised beam
differ from the original non-polarised beam?
The polarised beam has one of its planes of vibrations removed or “filtered” out where as the unpolarised
beam has all planes of vibration present.
Describe how you would use a disc of polarising material to show that direct sunlight is not polarised whereas
diffuse light from the blue sky is partially polarised.
When a polarising filter is held up to the blue sky and rotated there is a decrease in intensity due to the light from
the blue sky being partially polarised.
Page 9 of 17
8. Calculate the first and second order angles for light of wavelength 400 nm and 700 nm if the grating contains
10,000 line/cm.
The grating contain 104lines/cm=106lines/m, which means the separation between slits is d=(1/106)m=1.0×10-6m. In
first order (m=1), the angles are:
m λ
1 × 4.0 × 10 -7
sin θ 400 =
sin θ 700 = 0.700
=
= 0.400
1.0 × 10 - 6 m
d
so θ 400 = 23 . 6 ° and θ 700 = 44 . 0 ° .
In second
order
sin θ
so θ
400
400
m λ
2 × 4.0 × 10 - 7
sin θ 700 f 1
=
= 0.800
1.0 × 10 - 6 m
d
second
order
does not exist for
= 53 . 0 ° but the
=
λ = 700nm
because
sin θ cannot
exceed
1. No
higher
orders
will appear.
9. Spectra Overlap
White light containing wavelengths from 400 nm to 750 nm strikes a grating containing 4000 lines/cm. Show that the
blue at λ=450 nm of the third order spectrum overlaps the red at 700 nm of the second order.
The grating spacing is d = (1/4000) cm = 2.50 ×10-6 m. The blue of the third order occurs at an angle θ given by:
mλ
3 × 4.50 × 10 -7
= 0.540
sin θ =
=
2.50 × 10 - 6 m
d
Red in second order occurs at:
mλ
2 × 7 .00 × 10 -7
sin θ =
=
= 0.560
-6
d
2.50 × 10 m
which is a greater angle; so the second order overlaps into the beginning of the third-order spectrum.
Topic Three: Photons
KEY IDEAS
photoelectric effect
a phenomena which can only be explained if light has particle-like characteristics. The
light particles are called photons.
threshold frequency(f0)
is the minimum frequency at which electrons are ejected from a surface.
work function (W)
is the minimum energy required to break an electron free from the attractive forces
which hold the electron to the surface of a metal.
de Broglie wavelength
is the wavelength associated with a particle of mass m travelling at speed v. The
wavelength is given by λ=h/(mv) where λ is the de Broglie wavelength of the particle.
Photoelectric Effect
UV
A metal has outer electrons that are weakly bonded to the atoms and so are free to move
Radiation
throughout the metal.
These electrons may be removed when they absorb a small amount of energy either:
METAL
when the metal is heated → thermoelectric effect
Photo
UV
Electrons or when the metal is irradiated with suitable electromagnetic radiation →
Radiation
photoelectric effect.
electrons
Experimental Evidence (Observations of the photo-electric effect)
1.
Photo-electrons are released only if the frequency of EM radiation is above a
E
certain minimum frequency. This frequency is called the threshold frequency (ƒ0).
2.
If ƒ>ƒ0 the number of photo electrons released per unit time [photoelectric current] is
proportional to the intensity of the radiation.
3.
Photoelectron emission is instantaneous and is not dependent on the intensity of
radiation.
f
4.
The maximum kinetic energy of the photoelectrons increases linearly with the frequency
of the incident light and is not dependent on intensity.
5.
No electrons are emitted if the frequency is below the threshold frequency no matter velocities
vary
how intense the light.
6.
For frequencies above the threshold frequency electrons are emitted with a range of
velocities up to a maximum.
zinc
K
f
0
f>f0
Page 10 of 17
Wave Theory (Classical Theory)
Maxwell’s theory of electromagnetic radiation postulated that light was a continuous wave of fluctuating E and B
fields. This theory makes predictions about the photoelectric effect.
Electromagnetic (EM) Theory predicts:
1.
The energy absorbed by an electron would be dependent on the intensity of radiation i.e. the maximum energy
Kmax of the emitted electrons would increase as the intensity of light increased. (Observed Kmax ∝ (ƒ-ƒ0))
2.
If low intensity radiation were used there would be a time delay before electrons emitted.
(Observed instantaneous emission)
3.
Electrons will absorb low frequencies more easily than high frequencies.
(Observed high frequency releases more photoelectrons)
4.
More electrons are emitted if we increase the intensity of the light.
(Observed no emission if ƒ>ƒ0 - with no explanation for the threshold frequencyƒ0).
Photon Concept of Light
A model of light where we view light not as a continuous wave but in discrete bundles or packets. A quantum or a
photon is the name given to this discrete packet.
E = hf where h = Planck’s Constant = 6.63 × 10-34 Js
A photon is a discrete bundle of energy with E∝f.
These quanta of light fall on the photo-electric surface and one “free” electron completely absorbs one photon. If given
enough energy this electron will be emitted. The smallest energy to remove an electron is called the work function (W).
W = hf 0
where f0 = threshold frequency
The photoelectrons (photoemission) occurs when a photon (hf) is absorbed by a high energy electron.
If f<f0 then no emission.
K (max) = 21 mv(max) = hf -W
2
= hf − hf 0
(for the fastest moving electrons this is just a statement of energy conservation).
ΔE=hf0
0
0
ΔE=hf
0
W
W
-ve
-ve
W
-ve
The Photoelectric Cell
EM
Radiation
CATHODE
The photoelectric cell has a curved metal cathode (cadmium or many other metals) and
a thin pole anode sealed in an evacuated glass tube.
When EM radiation strikes the cathode electrons are released and may move to the
anode, so allowing an electric current to flow through the photo-cell.
Photo cells are often used as switches where the photo cell is only “on” while light
strikes the cathode, it is “off” at all other times, lift doors, self opening doors.
ANODE
The photocell may also be used to measure the energy of photoelectrons.
In this application the potential difference between the cathode and anode is reversed so
that the photoelectrons must do work (lose energy) in order to move to the anode.
As the reverse potential is increased, fewer electrons reach the anode and so the
photoelectric current decreases (indicating the photo electrons are released with a range
of energy). The smallest reverse potential which stops all photo-electrons from reaching
the anode, (null current I = 0 amps) is called the stopping voltage VS.
It can be seen that Energy of photoelectrons ≤ eVS (ie maximum energy of photoelectrons
eVS = hf − W
Kmax = eVS.
Experiments show that the Kmax is dependent on the frequency of the nature of the metal
surface.
K max
eVs
metal1
metal2 metal3
grad h
f0
f
Page 11 of 17
EXAMPLE : Photon Energy
1. Calculate the energy of a photon of blue light, λ=450nm. Since
(6.63 × 10 )(3.0 × 10 ) = 4.4 × 10
=
−34
8
4.4 × 10 −19
= 2.7 eV
1.6 × 10 −19
E = hf =
hc
Hence
(
100 ) 500 × 10 −9
E Eλ
=
=
= 2.5 × 10 20
n=
−34
8
6.63 × 10
3.0 × 10
hf
hc
−19
J or
f = c λ , we have
λ
4.5 × 10 − 7
Photons from a Light Bulb
2. Estimate how many visible light photons a 100W light bulb emits per second.
Let's assume an average wavelength in the middle of the visible spectrum, λ≈500nm.
The energy emitted in one second (=100 J) is E = n hf where n the number of photons emitted per second and f = c/λ.
(
(
)
)(
)
This is an overestimate since much of the 100J of electric energy input is transformed into heat rather than light. If the
efficiency is between 1% and 10%, then the number of photons emitted is on the order of 1019, still an enormous number.
Photoelectron speed and energy
3. What is the maximum kinetic energy and speed of an electron ejected from a sodium surface whose work function
W0 = 2.28 eV when illuminated by light of wavelength (a) 410 nm; (b) 55 nm.
(a) For λ = 410 nm.
E = hf =
hc
λ
=
(6.63 × 10 )(3.0 × 10 )
−34
8
4.1 × 10 −7
= 4.85 × 10 −19 J or 3.03eV
K (max) = hf -W = 3.03eV − 2.28ev = 0.75eV or1.2 × 10−19 J
K (max) = 21 mv (max) where m = 9.1 × 10 −31 kg
2
Since
(b) For λ = 550 nm.
v=
2 K max
m
= 5.1 × 10 5 ms −1
hf = 3.6 × 10-19 J = 2.25eV
Since this photon energy is less than the work function, no electrons are ejected.
The photoelectric effect, besides playing an important historical role in confirming the photon theory of light, also has
many practical applications.
eg. burglar alarms, automatic doors, smoke detectors, photographic light meters and film sound track.
X-rays
X-rays: incident high energy electrons produce high frequency electromagnetic radiation.
X rays
electrons accelerated
by large p.d.
metal
Filament F
Hood H
Anode A
produces a current to heat the wire and emit electrons.
used to focus electrons
electrons accelerated away from filament with a large acceleration due to high PD 40kV to 400kV
between the Anode and the Cathode.
Target T
tungsten target on which the electrons impinge.
The process of X-ray production is very inefficient, about 99% of the energy being lost as heat.
Page 12 of 17
X-rays and Photons
If the accelerating potential is V the electrons from the Cathode reach the target T with an energy Ve
K = 12 mv = Ve
2
(Work = potential difference × charge). ie.
A few electrons penetrate the atom and interact with the electric field with the electron being slowed down.
ie.
K1 = 12 mv1
2
incident
K 2 = 12 mv2 final
The energy available becomes
Note: energy lost by one electron
2
K = K 1 − K 2 = 12 mv1 − 12 mv 2
2
1
2
mv1 − 12 mv 2 = hf
2
2
2
where hf is the energy of one X-Ray photon.
The process of X-ray production is such that most of the energy of the electrons heats the target and must be removed
from the anode by a cooling liquid usually a silicon oil.
½mv 22
proton
+
electron
path
½mv 12
X ray
photon
The diagram shows a bremsstrahlung photon produced by an electron decelerated by interaction with a target atom.
Because there will be a range of final velocities due to the degree of interaction (slowing down) we have a range of Xray photons produced. This gives us a continuous range of X-ray photons.
However this range has a maximum value which is determined when all of the available electron energy goes into the
photon when the electron has a final velocity v2 of zero.
2
K 2 = 12 mv 2 = 0 so
where λmin = c .
f max
K 1 − K 2 = 12 mv1 − 12 mv 2 = hf
2
2
becomes
K max = 12 mv1 = hf max
2
Considering the Intensity Curves
X-Ray Quality
If the tube voltage is increased we increase the accelerating potential and so
the incident electrons have more energy.
The maximum energy is higher and also the number of photons increases so a
higher intensity of X-rays is produced.
X-Ray photons with a higher frequency and a lower wavelength have a
greater penetrating power. A beam of higher frequency X-ray photons is said
to be “harder” or of higher quality than those where f is less. Lower
frequency X-ray photons are said to be "softer".
X-Ray Quantity
If the filament current is increased then more electrons are released and there
are more electrons available for photon production and so a higher intensity
of X-Ray photons is produced.
Note:
there will be no increase in energy of individual photons so no
increase of fmax.
Page 13 of 17
Problem set example: Photons
1. Calculate the energy of a photon of
(a) red light (f=5×1014Hz),
(b) violet light (f=7×1014Hz).
-34
14
E = hf = 6.63 × 10 -34 × 7 × 1014
E = hf = 6.63 × 10 × 5 × 10
= 4.6 × 10 -19 Joules
= 3.3 × 10 -19 Joules
2. Find the EK of the most energetic electron ejected from a metal by a radiation of frequency 7×1014Hz if the work
function of the metal is 2V.
W = 2V = 2 × 1.6 × 10 -19 = 3.2 × 10 -19 J
E K(max) = E i - W = 6.63 × 10 -34 × 7 × 1014 − 3.2 × 10 −19 = 4.67 × 10 -19 − 3.2 × 10 −19 = 1.44 × 10 -19 joules
3. What is the threshold frequency of a metal that has a work function of 3.3volts?
W 5.28 × 10 -19
=
W = 3.3V = 3.3 × 1.6 × 10 -19 = 5.28 × 10 -19 J W = hf 0 so f 0 =
= 7.97 × 1014 Hz
h 6.63 × 10 -34
4. Will light of frequency 4×1014Hz eject electrons from a metal whose work function is 5V? Explain your answer.
W = 5V = 5 × 1.6 × 10 -19 = 8 × 10 -19 J
E i = hf = 6.63 × 10 -34 × 4 × 1014 = 2.65 × 10 -19 J E i is less than W so no emission
5. Light of a frequency above the threshold frequency falls on a given metal. As the intensity of the light is increased:
(a)
what happens to the number of photoelectrons ejected from the metal?
More electrons are emitted since more photons are falling on the metal.
(b)
what effect does it have on the maximum kinetic energy of the electrons?
None. Only a more copious emission of electrons.
5. The energy needed to release an electron from sodium metal is 2.3eV.
−19
(a) What is the energy in joules? W = 2.3eV = 3.68 × 10 Joules
(b) Calculate the threshold frequency of sodium and hence determine whether or not orange light of wavelength
6.0×10-7m will be able to release electrons from the sodium surface.
W
c 3 × 10 8
14
-7
f0 =
= 5.55 × 10 Hz then λ = 6 × 10 m so f = =
= 5 × 1014 Hz
−7
λ 6 × 10
h
6. The stopping voltage of a photoelectric cell is 1.0volts when light of wavelength 5.5×10-7m falls on the cathode
surface.
VS = 1.0Volts = 1.6 × 10-19 Joules E =
hc
λ
=
6.625 × 10 −34 × 3 × 108
so = 3.61 × 10 −19 Joules
−7
5.5 × 10
(a) What is the maximum kinetic energy of the emitted electrons?
E max = 1.6 × 10 −19 Joules
(b) What is the work function of the cathode? W = 2.01 × 10
−19
(c) What is the threshold frequency of the metal of the cathode?
The work function of a metal is 6×10-19J.
Joules
f0 =
W
= 3.03 × 1014 Hz
h
5. In a photoelectric experiment using a clean sodium surface, the maximum energy of the emitted photons was
measured for a number of different incident frequencies, with the following results.
Frequency (1014Hz)
Energy (eV)
11.8
2.60
10.6
2.11
9.9
1.81
9.1
1.47
8.2
1.10
6.9
0.57
Plot the graph of these results and hence find:
Page 14 of 17
(a)
Planck’s constant and then the threshold frequency of sodium.
P h o t o - e le c t r ic E f f e c t
y =
7 E -3 4 x - 4 E -1 9
5 E -1 9
Energy(max) joules
4 E -1 9
3 E -1 9
2 E -1 9
1 E -1 9
0
-1 E -1 9
0
4 E + 1 4
8 E + 1 4
1 E + 1 5
2 E + 1 5
-2 E -1 9
-3 E -1 9
-4 E -1 9
F r e q u e n c y(h e r tz )
6. A photosensitive surface C is illuminated with monochromatic light of successively different wavelengths. The
diagram below illustrates that potential differences have to be applied between C and A to prevent electrons released by
the light from crossing to the collector A and registering as a current on the ammeter G. The values of potential
difference which just stop electrons reaching A are as follows:
C
Wavelength (10-7m)
3.66
4.05
4.36
4.92
5.46
5.79
A
A
Potential Difference (V)
1.48
1.15
0.93
0.62
0.36
0.24
Plot a suitable graph and hence determine:
(a) the threshold frequency,
(b) the work function of the surface A,
(c) the value this experiment finds for Planck’s constant,
(d) the maximum kinetic energy (in joules) with which electrons are ejected from the surface by the light of wavelength
3.66×10-7m.
Wavelength(metres) Frequency(hertz) Energy(joules) Stopping Voltage (volts)
3.66E-07
8.20E+14
2.368E-19
1.48
+
4.05E 07
7.41E 14
1.84E-19
1.15
+
4.36E 07
6.88E 14
1.488E 19
0.93
4.92E-07
6.10E+14
9.92E-20
0.62
5.46E-07
5.49E+14
5.76E-20
0.36
+
5.79E 07
5.18E 14
3.84E 20
0.24
S to p p in g V o lta g e
y = 7 E -3 4 x - 3 E -1 9
Energymax(joules)
3 E -1 9
2 E -1 9
1 E -1 9
0
-1 E -1 9 0
2E + 14
4E + 14
6E + 14
8E + 14
-2 E -1 9
-3 E -1 9
Fr e q u e n c y( h e r tz )
Page 15 of 17
7. A potential difference of 10000V is applied across an x-ray tube. Calculate the minimum wavelength for the x-rays
produced.
Energy available = Ve = 1.6 × 10 -15 Joules
If all this kinetic energy is converted into a photon what is its wavelength.
E = 1.6 × 10 -15 = hf =
hc
λ
so λ min = 1.24 × 10 −10 metres
8. What potential difference must be applied across an x-ray tube to produce x-rays with a minimum wavelength of 10-11m?
hc 6.6 × 10 −34 × 3 × 10 8
V × 1.6 × 10 -19 =
=
so V = 1.24 × 10 5 volts
−11
10
λ
Electrons are accelerated in an x-ray tube by a potential difference of 50000V.
(a) What energy do they have on reaching the target?
Energy = Ve = 8 ×10 -15 joules
(b) What is the shortest wavelength of the continuous x-ray spectrum produced?
E=
hc
λ
6.6 × 10 −34 × 3 × 10 8
= 2.48 × 10 −11 metres
8 × 10 −15
so λ =
(c) If the electrons are stopped in a distance of 10-5m, calculate their average acceleration while being stopped.
2E K
=
m
Energy = mv ∴ v1 =
1
2
2
(
2 × 8 × 10 -15
= 1.33 × 10 8 ms -1
-31
9.1 × 10
v 2 = 0ms −1 so v 2 - v1 = 2as 0 - 1.33 × 10 8
2
2
)
2
= 2 × a × 10 -5 then a = 8.79 × 10 20 ms − 2
Topic Four: Wave Behaviour of Particles. DeBroglie Matter Waves
When light is being transmitted it exhibits wave-like properties. When light is interacting with matter it has particlelike characteristics.
E = hf
p=
hf h
=
c λ
or
λ=
h
h
=
p mv
Waves associated with moving matter are known as de Broglie waves or matter waves.
h 6.6 × 1034
e g (1) mass = 50 kg, velocity = 10 ms-1 λ = =
= 1.33 × 10 −36 m
p
500
e g (2) m = 9.1×10-31 kg (mass of an electron),
K = 90 eV, v = 5.6×106 ms-1
λ=
h
6.6 × 10 34
=
= 1.3 × 10 −10 m
p 9.1× 10 31 × 5.6 × 10 6
This in is the X-Ray region of the electromagnetic spectrum.
Electron Diffraction
If consider the intensity pattern obtained from wave interference eg. X-Rays diffracted by crystal lattice then electron,
matter waves could have wavelength in the X-Ray ranges and a crystal could be used as a diffraction grating to produce
a diffraction pattern.
Davisson and Germer produced electron diffraction in 1927 and the relationship
d sin θ = nλ
was satisfied using
λ=
h
p
Page 16 of 17
APPLICATION: THE ELECTRON MICROSCOPE.
Sight is one of the most important human skills, which enables us too observe the structure of materials – natural and
artificial because they reflect light.
With the eye alone it is impossible to distinguish between two objects less than about 0.1 mm apart. First magnifying
glasses and then microscopes have been developed to enable us to distinguish between
(ie RESOLVE) objects that are even closer together.
Light has, however, a wave nature and diffracts at sharp edges of objects or small gaps between them. The shortest
visible wavelength (violet) is of wavelength of 400 nm or 4 × 10-7 m and with the best microscopes the best possible
resolving power is only the approximate value of this wavelength.
Shorter wavelength radiations eg X-rays (0.10 nm or less) would help – there are problems, however, because no
possible ordinary lens has been yet devised and we could not see them with our eyes.
De Broglie, postulated, in 1924, that moving electrons had an associated wavelength λ = h / p (p= momentum). This
gave physicists the idea that perhaps electron beams could be focused using either or both electric (E) and magnetic (B)
fields and thus perhaps an electron microscope could be constructed.
The effective wavelength when the electron is accelerated through of potential of 40 kV is:
W = Ve = 4 × 104 × 1.6 ×10-19 = 6.4 × 10-15 J.
Energy = 1/2 m v2 = 6.4 × 10-15 J
v = 1.19 × 108 ms-1
λ = h / p = 6.63 × 10-34 / ( 3.1 × 10-31 × 1.19 × 108) = 6.12 × 10-12 m. with an equivalent increase in resolving power.
The basic structure of the electron microscope is similar to that of a compound light microscope.
The electron microscope can produce images that have a much greater
magnification than a light microscope and also enables a far greater resolution
to be obtained than that possible with a light microscope. This is because the
diffraction pattern for high-energy electrons has a much lower width than that
for any visible light waves.
Hence instead of being restricted to magnifications of less than
2 000, magnifications of many hundreds of thousands are feasible enabling
such things as actual structures of crystal lattices to be observed.
In the Transmission Electron Microscope produced by a two dimensional
image the objective and eyepiece lenses are magnet fields that exert forces on
the electron (motion of charged particles in electric and magnetic fields) to
bring them into effective focus. Carefully designed current-carrying coils of
wire produce these fields.
Problem set example:
1.What is the momentum of a photon of x-radiation of wavelength 10-11m?
h 6.6 × 10 -34
p= =
= 6.6 ×10 - 23 secN
−11
λ
10
2. Calculate the momentum and energy of a photon of yellow light of wavelength 6.00×10-7m?
p=
h
λ
=
6.6 × 10 -34
= 1.1 × 10 - 27 secN
−7
6.00 × 10
E=
hc
λ
=
6.6 × 10 -34 × 3 × 10 8
= 3.3 × 10 -19 J
−7
6.00 × 10
λ=
3. Calculate the deBroglie wavelength of: (a) a 1000kg car moving at 10ms-1.
(b) an electron accelerated through a potential difference of 100V.
E = 100 × 1.6 × 10 −19 = 1.6 × 10 −17 J v =
2E K
=
m
h 6.6 × 10 -34
=
= 6.6 × 10 -38 metres
p 1000 × 10
2 × 1.6 × 10 −17
h
6.6 × 10 -34
= 5.9 × 10 6 ms −1 so λ = =
= 1.23 × 10 -10 metres
−31
−31
6
p 9.1 × 10 × 5.9 × 10
9.1 × 10
4. At what speed does an electron have a deBroglie wavelength of 2.0×10-8m?
λ=
h
6.6 × 10-34
h
h
so v =
= 3.6 × 104 ms-1
=
=
−31
−8
mλ 9.1 × 10 × 2.0 × 10
p mv
5. An electron has an energy of 60000eV. Find:
(a) its velocity
2E K
2 × 9.6 × 10 -15
=
= 1.45 × 10 8 ms -1
−31
m
9.1 × 10
p = mv = 9.1×10-31 ×1.45 ×108 ms-1 = 1.32 ×10−22 secN
E = 60000eV = 9.6 × 10 -15 J v =
(b) the electron’s momentum
(c) its wavelength.
h
6.6 × 10 -34
λ= =
= 4.99 × 10 −12 metres
− 22
p
1.32 × 10
Page 17 of 17