11.2 DOPPLER EFFECT
The Doppler effect is the change in frequency of a wave for an observer moving relative to the source of the wave.
Stationary source
These columns show speed of sound,
frequency and wavelength
as measured by the observer.
Stationary observer
𝑣 = 𝑓𝜆
Moving source
Stationary observer
s
𝑣=𝑓↑𝜆↓
(The diagram clearly shows a shorter wavelength.)
Stationary source
Moving observer
𝑣+𝑢 =𝑓 ↑𝜆
u
(The diagram clearly shows the wavelength does NOT change.)
So the observer measures a different wavelength when the source is moving, but when the observer is moving
there is NO CHANGE IN OBSERVED WAVELENGTH.
This is why the Doppler effect is defined in terms of change in observed frequency, and not wavelength!
The Doppler effect equations for sound.
Moving source:
𝒗 )
𝒗±𝒖
𝒇′ = 𝒇 (
Moving observer:
𝒗±𝒖)
𝒗
𝒇′ = 𝒇 (
Solve problems on the Doppler effect for sound. (Problems will not include situations where both source and detector are moving.)
Example: VELOCITY of BLOOD. Ultrasound can be used to
detect vascular problems. Note that the blood cells acts
as a moving source, so there are two Doppler effects to consider.
The beam should be as parallel to the blood flow as possible.
The Doppler effect for electromagnetic waves …. use the approximation
Students should appreciate that the approximation may be used only when v << c.
∆𝒇 =
𝒗
𝒄
𝒇
Example: RADAR. Microwaves can be used to measure the velocity of moving objects. Note that the moving object acts
as a moving source, so there are two Doppler effects to consider.
11.3
(Single slit) DIFFRACTION
tan 𝜽 =
L
b
𝜃
=
𝝀
𝒃
but for small angles
𝝀
𝟐
x
𝝀/𝟐
⁄𝒃/𝟐
𝜃 ≈ 𝑠𝑖𝑛𝜃 ≈ 𝑡𝑎𝑛𝜃
𝒃
𝟐𝜃
𝜽=
So
𝝀
𝒃
incident
white
light
𝝀
cQ1.
The central maximum formed by diffraction through a single slit is
A. the same width as the other maxima
C. twice the width of the other maxima
aQ2.
dQ3.
B. the same intensity as the other maxima
D. twice the intensity of the other maxima
A beam of coherent light is incident on a single slit of width b.
After passing through the slit, the light is incident
on a screen at a distance D from the slit. What changes will
increase the width of the central maximum?
D
increase
increase
decrease
decrease
10
Microwaves of frequency 10 Hz will be noticeably diffracted when they
pass through an aperture whose diameter has an order of magnitude of
A.
100 m

B.
10 m
C.
1m
D.
aQ4. The path difference at the first minimum of a single slit
experiment is equal to
A. c/(2f)
aQ5.
b
decrease
increase
decrease
increase
A.
B.
C.
D.
1 cm
B. c/f
C. f/(2c)
D.
f/c
You can increase the fringe separation pattern formed by single slit diffraction by …?
A.
C.
decreasing the frequency of the incident light.
increase the width of the slit.
B.
D.
decreasing the distance between the slit and the screen.
increase the intensity of the incident light on the slit.
dQ6. A single slit diffraction pattern is cast onto a screen. Which one of the following could be
a list of distances that maxima occur from the first minima ?
A. 1 cm, 2cm, 3cm
11.4
B. 1.5 cm, 2cm, 3.5cm
RESOLUTION
𝒔𝒊𝒏𝜽 =
𝒏𝝀
𝒃
C. 0.5 cm, 1cm, 2cm
D. 1 cm, 3cm, 5cm
When light passes through a single
but for a circular
slit use 𝜽 =
aperture use
State the Rayleigh
𝝀
𝒃
𝜽 = 1.22
𝝀
𝒃
criterion for images of two sources to be just resolved.
𝜽 = 1.22
Two sources are just resolved if the
𝝀
𝒃
To increase resolution
You decrease wavelength !!
central maximum of the diffraction pattern of one source falls on the
first minimum of the other source.
Diffraction patterns are well resolved, just resolved or not resolved.
Be careful !! For single slit and circular aperture
but
𝜃
for two sources of light
𝜃
The significance of resolution:
CDs and DVDs store digital information as “bumps” and “pits” etched into a plastic
surface. The data is read by reflecting a laser beam off the surface. The wavelength
of laser light used to read the data and the size of the aperture of the lens used to
receive the laser light places a limit on how close together the bumps and pits can
be placed, that is, places a limit on the resolution of the data.
Electron Microscopes can resolve objects beyond the limits imposed by the wavelength of visible light, by using the wave
properties of electrons. The de Broglie wavelength of an electron is much smaller than the wavelength of a photon of visible
light, so a microscope using an electron beam can resolve objects that are much smaller than those of a light microscope.
Radio Telescopes. Distant objects, such as quasars, emit radio waves with
a wavelength much longer than visible light, which limits the ability of a radio
telescope to resolve radio sources. To get around this limitation, astronomers
use a number of radio telescopes separated by a large distance, called a VLA
(Very Large Array).
Ex 1.
The Galaxy Cygnus A is a strong emitter of radio waves of wavelength 0.15 m. [ θ ≈ 𝑠𝑖𝑛 θ ≈ 𝑡𝑎𝑛 θ ]
The Galaxy is estimated to be 5.0 x 1024 m from Earth.
A radio telescope shows that the radio emission is from two sources separated by a distance of 3.0 x 10 21 m.
Show the diameter of the dish required to just resolve the sources is 0.3 km, too much for one dish. (θ = 6.0 x 10-4 )
2 x 10-4 rad
Ex 2.
What is the resolving power of your eye? (Diameter of pupil approx = 3 mm)
Ex 3.
The brightest star in the winter sky in the Northern Hemisphere is Sirius. In reality, Sirius is a system of two stars
that orbit each other. The Hubble Space Telescope (diameter 2.4 m) is pointed at the Sirius system, which is
7.98 x 1016 m from Earth. (Assume the average light from the stars has a wavelength of 500 nm.)
a) Show the minimum separation needed between the stars in order for the telescope to just resolve them is 2.2 x 1010 m.
b) What is the resolving power of the telescope?
aQ7. If two point sources of light are well resolved their images
A. may not overlap
C. show no evidence of diffraction
bQ8.
Two sources of light are just resolved if the first minimum of one coincides with the other’s …
A. second maximum
aQ9.
B. completely overlap
D. have sharp edges with no blurring
B. central maximum
C. first minimum
D. second minimum
Two galaxies with an angular separation at the observer of 5.0 x 10 −4 radians are observed
with a radio telescope. Both galaxies emit radio waves of wavelength 2.5 x 10−2 m.
The images of the galaxies are just resolved by the telescope. The diameter of the circular
collecting dish of the telescope is
A. 61 m.
B. 50 m.
C. 30 m.
D. 25 m.
dQ10. The resolution of two point sources of light will increase if you decrease their
A. separation
B. intensity
C. wavelength
D. frequency
bQ11. The images of two sources are just resolved. According to the Raleigh criterion …
A.
B.
C.
D.
11.5
the central maximum of the diffraction pattern of one source must coincide
with the central maximum of the diffraction pattern of the other source.
the first minimum of the diffraction pattern of one source must coincide with
the central maximum of the diffraction pattern of the other source.
light from the sources must pass through a circular aperture.
light from the sources must be coherent.
Polarisation
Describe what is meant by polarized light.
Light in which the electric field vectors oscillate in one plane only,
perpendicular to the propagation of the ray.
Describe polarization by reflection.
"Polarizing sunglasses use a polarizing material such as polaroid film to
block horizontally-polarized light, preferentially blocking reflections from
horizontal surfaces. The effect is strongest with smooth surfaces such as
water, but reflections from road and the ground are also reduced."
State and apply Brewster’s law.
In 1812, Sir David Brewster found experimentally that the reflected ray is 100%
polarized when the angle between the reflected ray and the refracted ray is 90°.
Brewster’s angle tan θB =
n2
𝑛1
Exercise: Show that the Brewster angle for light incident on water is 530.
Explain the terms polarizer and analyser.
A polarizer can be used to determine if light is polarized.
It is then called an analyser.
When unpolarised light passes through a polariser its intensity is reduced by 50%.
Calculate the intensity of a transmitted beam of polarized light using Malus’ law.
I = Iocos2θ
Describe what is meant by an optically
•
active substance.
Some substances can rotate the plane of polarized light. We say they are optically active
Sugar solution is optically active. The amount of rotation of the plane
of polarization depends on the concentration of the solution.
Describe the use of polarization in the determination
of the concentration of certain solutions.
Some materials become optically active under stress.
A complicated pattern can be seen when a piece of plastic under stress is placed
between two polarisers at right angles to each other.
Where are the regions of greatest stress?
Outline qualitatively the action of liquid-crystal displays (LCDs).
The use of LCD screens in a wide variety of different
applications/devices can be mentioned.
dQ1. Plane polarized light passing through a polarizer will have its intensity reduced by
A.
100 %
B.
between 0 % and 50 % C.
dQ2. Polarization is only a property of
50 %
D.
A. longitudinal waves
C. reflected waves
between 0 % and 100 %
B.
D.
refracted waves
transverse waves
bQ3. Which of the following best represents the electric field vectors of a plane polarised wave?
A.
B.
C.
D.
bQ4. Vertically polarized light has its intensity halved when it passes through a polarizer.
The angle between the light and the polarizer must be
A.
30 0
B.
45 0
C.
60 0
D.
90 0
cQ5. Unpolarised light passing through a polarizer has its intensity reduced by


A.

C.

0%
50 %
B.
D.
25 %
100 %
bQ6. Consider two polarizing sheets with planes of polarization that are initially parallel. The incoming
light on sheet 1 is unpolarised, (intensity I). The second sheet is rotated until 𝐼0 equals zero.
The rate of change of I is greatest when the
angle of rotation equals
I
𝐼0
A.
C.
10°
80°
B.
D.
45°
90°
cQ7. Unpolarised light has incident angle i and refracted angle r.
The reflected ray will be completely polarized when then
the angle between the incident ray and the refracted ray is
180 0 - i + r
90 0 + 2i
A.
C.
B. 180 0 - r
D. 180 0 - i – r
cQ8. When the angle of incidence is equal to the Brewster angle, which of the following correctly


describes the reflected and refracted ray?


Reflected ray
Refracted ray
A
Completely plane polarised
Completely plane polarised
B
Partially plane polarised
Completely plane polarised
C
Completely plane polarised
Partially plane polarised
D
Partially plane polarised
Partially plane polarised
bQ9. The Brewster angle for light traveling from
water ( n = 1.33 ) to glass ( n = 1.50 ) is
A.
bQ10. A beam of unpolarised light is incident
on a diamond.
The refractive index of the diamond is n. 
At the angle  p , the beam reflected from the
diamond is plane polarized normal to the page.
The angle  p is
A.

C.

tan( n)
1
tan 1
n




B.
tan 1 n
D.
tan
1
n
29 0
B.

48 0

C.
59 0
D.

68 0