This document provides information about waves, including definitions of key wave properties like amplitude, wavelength, frequency, and speed. It describes longitudinal and transverse waves, and how they can be represented graphically. Specific topics covered include progressive waves, representing longitudinal waves, past exam questions, using a cathode ray oscilloscope to analyze waves, the wave equation, intensity of waves, and phase differences between waves.
4. Wave motion
● Wave motion is energy
transferred through moving
oscillations or vibrations .
● These can be seen in
vibrations of ropes,
springs and ripple tanks .
● The oscillations/vibrations
can be perpendicular or
parallel to the direction
of wave travel .
5. ● Waves c a n a l s o be
de mons t r a t e d by r i ppl e
t a nks .
● Ri ppl e t a nks ma y be
us e d t o de mons t r a t e
t he wa ve pr ope r t i e s of
r e f l e c t i on, r e f r a c t i on
a nd diffraction .
6. 1. Displacement, x : is the
distance of a point on
the wave from its
equilibrium position .
2. Amplitude, A : is the
maximum displacement of
a particle in the wave
from its equilibrium
position .
General Wave
Properties 3. Wavelength, λ:
( S2018_V22_Q4)
• di s t a nc e move d by
wa ve f r ont / e ne r gy dur i ng
one c yc l e .
or
• mi ni mum di s t a nc e be t we e n
t wo wa ve f r ont s .
or
• di s t a nc e be t we e n t wo
a dj a c e nt wa ve f r ont s .
7. 3. Period, T: is the time
taken for one complete
oscillation of a point
in a wave.
● uni t : s e c onds , s .
4. Frequency, f : i s t he
numbe r os c i l l a t i ons
pe r uni t t i me .
● uni t : pe r s e c ond, s - 1
or he r t z , Hz .
𝑓𝑓 =
1
𝑇𝑇
Both transverse &
longitudinal waves,
can be represented
by two graphs:
1. The di s pl a c e me nt - di s t a nc e
gr a ph ( t i me c ons t a nt )
2. The di s pl a c e me nt - t i me
gr a ph ( di s t a nc e c ons t a nt )
9. From t he gr a ph, we c a n
de duc e :
i . The a mpl i t ude of t he wa ve :
A = 4. 5 c m
i i . The wa ve l e ngt h:
λ = 4 c m
Worked example
Fr om t he gr a ph, we c a n
de duc e :
i . The a mpl i t ude of t he wa ve :
A = 4. 5 c m
i i . The pe r i od:
T = 0. 4 s
i i i . The f r e que nc y:
𝑓𝑓 =
1
0.4 𝑠𝑠
= 2. 5 Hz
10. Progressive
Waves
A progressive waves
transfers energy from one
place to another as a result
of oscillations/vibrations.
(W2014_V23 & W2009_V23)
Example :
1. The energy from sound
reaches our eardrums to
vibrate .
2. the electromagnetic waves
from the sun carry the
energy to the Earth for
the survival of living
things .
● The r e a r e t wo t ype s of
pr ogr e s s i ve wa ve s ;
Tr a ns ve r s e
wa ve s
Longi t udi na l
wa ve s
pr ogr e s s i ve
wa ve s
11. Transverse
waves
Transverse waves is when
the particles of the medium
vibrate at right angles to
the direction of travel of
the energy (W2014/V23)
• Examples of transverse
waves are :
1. Electromagnetic waves
e. g. radio, visible
light, UV.
2. Vibrations on a guitar
string
• Transverse waves can be
polarized .
transverse wave.mp4
12. Longitudinal
waves
Longitudinal waves is when
the displacement of
particles/vibration/
oscillation is parallel to
the direction of energy
(W2017/V23)
• Examples are sound waves &
ultrasound waves
• Longitudinal waves cannot
be polarized .
• The wavelength λ of a
l ongi t udi na l wa ve i s t he
di s t a nc e be t we e n
s uc c e s s i ve c ompr e s s i ons
a nd r a r e f a c t i ons .
longitudinal wave.mp4
13. Representing waves
• Longitudinal waves
shows repeating
pattern of compressed
and rarefaction .
• This gives rise to high
and low pressure
regions .
• This can be difficult
to draw so often
longitudinal wave is
represented as it if
were a sine wave.
18. Cathode- Ray
Oscilloscope
• A cathode - Ray Oscilloscope is
a laboratory instrument used
to display, measure and
analyse waveforms of
electrical circuits .
• Figure shows a microphone is
connected to the input of the
c. r . o.
• The microphone converts the
sound waves into a varying
voltage that has the same
frequency as the sound waves .
• This voltage is displayed on
the c. r . o. screen .
• An oscilloscope is
basically a voltmeter that
shows you how voltage
varies with time .
• I ns t e a d of ge t t i ng a
di gi t a l r e a dout ( a s on a
mul t i me t e r ) i t gi ve s you a
gr a ph.
19. • It plots a voltage against
time graph on the screen .
• The y- axis is voltage ( s o
you c a n s e e how ma ny vol t s
a r e a c r os s t he c ompone nt ) .
the Y- plate sensitivity
(Y - gain) i n vol t s / di vi s i on.
e . g. V di v⁻¹ or V c m⁻¹
• The x- axis is time ( t he
vol t a ge i s s t e a dy ( D. C. ) or
va r yi ng ( A. C. ) .
the time base settings is i n
t i me / di vi s i on
e . g. ms di v ⁻¹ or s c m⁻¹
• The di s t a nc e be t we e n pe a ks
or t r oughs , L i s me a s ur e d
us i ng t he s c a l e on t he
c . r . o. di s pl a y.
20. Determine f or t he s ound:
i . The t i me pe r i od
T = Lx
= 4. 0 c m x 2. 0 ms c m⁻¹
Ans we r : 8. 0 x 10⁻³ s
i i . The f r e que nc y
Ans we r : 125 Hz
Worked example
The t i me ba s e s e t t i ng f or
t he c . r . o. us e d t o obt a i n
t he t r a c e i n f i gur e be l ow i s
2. 0 ms c m⁻¹ .
21. Worked example
Figure below shows the
trace on an oscilloscope
screen when sound waves are
detected by a microphone .
The time - base is set at 1
ms div ⁻¹ . The y- ga i n i s s e t
t o 20 mV di v⁻¹ .
De t e r mi ne
i . t he f r e que nc y of t he
s ound wa ve s a nd
Ans we r : 250 Hz
i i . t he a mpl i t ude of t he
os c i l l os c ope t r a c e .
Ans we r : 70 x 10⁻³ V
23. The wave
equation
The wave equation links
the speed, frequency and
wavelength of a wave.
We can find the speed
wave, 𝑣𝑣 us i ng:
A wa ve wi l l t r a ve l a
di s t a nc e of one whol e
wa ve l e ngt h, λ i n a t i me
e qua l t o one pe r i od, T.
Howe ve r ,
He nc e
𝑣𝑣 = 𝑓𝑓𝜆𝜆
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑓𝑓 =
1
𝑇𝑇
𝑣𝑣 =
𝜆𝜆
𝑇𝑇
=
1
𝑇𝑇
𝜆𝜆
24. Worked
example
● The wave in the diagram
below has a speed of
340 ms–1.
● What is the wavelength of
the wave?
● Answer : 0. 095 m
Intensity
The intensity of a wave is
the rate of energy
transmitted per unit area
at right angles to the wave
velocity.
𝐼𝐼 =
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
=
𝑃𝑃
𝐴𝐴
25. • Unit for intensity is
Watts per square metre,
Wm⁻²
• The i nt e ns i t y of a wa ve
ge ne r a l l y de c r e a s e s a s i t
t r a ve l s a l ong.
• The r e a r e t wo r e a s ons f or
t hi s :
1. The wa ve ma y ‘ s pr e a d
out ’ ( a s i n t he e xa mpl e
of l i ght s pr e a di ng out
f r om a c a ndl e ) .
2. The wa ve ma y be a bs or be d
or s c a t t e r e d ( a s whe n
l i ght pa s s e s t hr ough t he
Ea r t h’ s a t mos phe r e ) .
26. ● As a wa ve s pr e a ds out ,
i t s a mpl i t ude de c r e a s e s .
● For a wa ve of a mpl i t ude ,
A a nd f r e que nc y f , t he
i nt e ns i t y I i s
pr opor t i ona l t o A² f² .
I ∝ A² → I = kA²
I ∝ f² → I = kf²
Worked
example
● A wa ve f r om a s our c e ha s a n
a mpl i t ude of 5. 0 c m a nd a n
i nt e ns i t y of 400 W
m⁻² .
i . The a mpl i t ude of t he wa ve i s
i nc r e a s e d t o 10. 0 c m.
Ca l c ul a t e t he i nt e ns i t y now.
Ans we r : 1600 W
m⁻²
i i . The i nt e ns i t y of t he wa ve i s
de c r e a s e d t o 100 W
m⁻² .
Ca l c ul a t e t he a mpl i t ude now.
Ans we r : 2. 5 c m
27. ● Wave f r om a poi nt s our c e
s pr e a d out e qua l l y i n a l l
di r e c t i ons .
● Sur f a c e wa ve i s a s s ume d t o
be s phe r i c a l whe r e a r e a i s
e qua l t o 4πr ² .
● The r e f or e t he i nt e ns i t y,
𝐼𝐼 =
𝑃𝑃
4π𝑟𝑟^
Spherical
wave
● I nt e ns i t y of t he wa ve
de c r e a s e s wi t h i nc r e a s i ng
di s t a nc e f r om t he s our c e .
● At 3 t i me s t he di s t a nc e ,
t he s a me s ound e ne r gy wi l l
be s pr e a d ove r 9 t i me s t he
a r e a , a dr op t o 1/ 9 t he
i nt e ns i t y.
30. Phase
● A term used to describe the
relative positions of the
crests or troughs of two
different waves of t he s a me
f r e que nc y i s phase .
● W
he n t he c r e s t s or t r oughs a r e
a l i gne d, t he wa ve s a r e in
phase .
● W
he n t he c r e s t of one wa ve
a l i gns wi t h t he t r ough of
a not he r , t he y a r e
i n antiphase .
31. Phase
differences
● When crests and troughs
are not aligned the waves
are said to have phase
differences .
● The pha s e di f f e r e nc e t e l l s
us how much a point or a
wave is leads or lags
behind another .
● Pha s e di f f e r e nc e i s
me a s ur e d i n fractions of a
wavelength, degrees or
radians .
● The pha s e di f f e r e nc e c a n
be c a l c ul a t e d f r om:
1. t wo di f f e r e nt poi nt s on
t he s a me wa ve or
2. t he s a me poi nt on t wo
di f f e r e nt wa ve s .
32. Two different
points on the same
wave
● The phase difference between
two points :
○ In phase i s 360o or 2π r a di a ns
○ In anti - phase i s 180o or π
r a di a ns
● I n f i gur e s how t wo poi nt s A
a nd B, wi t h a s e pa r a t i on of
one whol e wa ve l e ngt h λ,
vi br a t e i n pha s e wi t h e a c h
ot he r .
● The pha s e di f f e r e nc e be t we e n
t he s e t wo os c i l l a t i ng pa r t i c l e s
a t A a nd B i s 360° . ( You c a n
a l s o s a y i t i s 0° . )
● The s e pa r a t i on be t we e n poi nt s C
a nd D i s qua r t e r of a wa ve l e ngt h
– t he pha s e di f f e r e nc e be t we e n
t he s e t wo poi nt s i s 90° .
33. ● In ge ne r a l , whe n t he
s e pa r a t i on be t we e n t wo
os c i l l a t i ng pa r t i c l e s on a
wa ve i s x, t he n t he pha s e
di f f e r e nc e ϕ be t we e n
t he s e pa r t i c l e s i n de gr e e s
c a n be c a l c ul a t e d us i ng
t he e xpr e s s i on:
ϕ =
𝑥𝑥
𝜆𝜆
𝑥𝑥 360°
35. The same point on
two different
waves.
● The diagram below shows the
red wave leads t he bl ue
wa ve by ¼ λ.
● I n c ont r a s t , t he bl ue wa ve
i s s a i d t o lag be hi nd t he
r e d wa ve by ¼ λ.
● Pha s e di f f e r e nc e ϕ c a n be
me a s ur e d i n:
fractions
degrees
radians
video: path diff 1.mp4
36. Past year
questions
(S2015_V12_Q25)
Past year
questions
(S2006_Q25)
The frequency of a certain wave is
500 Hz and its speed is 340 ms⁻¹ .
W
ha t i s t he pha s e di f f e r e nc e
be t we e n t he mot i ons of t wo poi nt s
on t he wa ve 0. 17 m a pa r t ?
A s ound wa ve ha s a s pe e d of
330 ms ⁻¹ a nd a f r e que nc y of 50 Hz .
W
ha t i s a pos s i bl e di s t a nc e
be t we e n t wo poi nt s on t he wa ve
t ha t ha ve a pha s e di f f e r e nc e of
60° ?
39. Prediction on
particle movement
● To determine the movement of
a particle in transverse
wave, we have to sketch
another wave i n t he wa ve
di r e c t i on.
The di a gr a m s hows a t r a ns ve r s e
wa ve on a s t r i ng wi t h t wo
poi nt s P a nd Q ma r ke d. The wa ve
i s movi ng i n t he di r e c t i on
s hown. W
ha t wi l l ha ppe n ne xt t o
pa r t i c l e P a nd Q?
Example
40. The di a gr a m be l ow s hows a
wa ve on a s t r i ng wi t h
pa r t i c l e s P, Q, R, S a nd T
on t he wa ve . The wa ve i s
movi ng i n t he di r e c t i on a s
s hown.
Worked example
a ) W
ha t c a n you s a y a bout
t he mot i on of t he
pa r t i c l e s the next
moment?
b) W
ha t c a n you s a y a bout
t he mot i on of t he
pa r t i c l e s a t t hi s
instant ?
Fi r s t l y, you ha ve t o know t ha t
t he wa ve a bove i s transverse
wave ( t he direction of the
vibration i s perpendicular t o
t he direction of wave motion ) .
He nc e , pa r t i c l e s wi l l
onl y vi br a t e up a nd down onl y.
41. a) P: down, Q: up, R: up,
S: down, T: down
b) P: at rest , Q: up, R: at
rest , S: down, T: down
● Ta ke not e of P and Q at
crest and trough
respectively .
● He nc e a t t hi s pa r t i c ul a r
i ns t a nt , P i s a t i t s
hi ghe s t poi nt a nd R i s a t
i t s l owe s t poi nt , he nc e
bot h momentary at rest .
Solution
45. Doppler effect
of Sound
Doppler effect is when the
observed frequency is
different to source
frequency when source
moves relative to observer
(S2017_V21_Q5)
● The s i r e n of a n a mbul a nc e
a ppe a r s t o c ha nge i n pi t c h
whe n i t pa s s e s you.
● The pi t c h i s hi ghe r a s t he
ve hi c l e s a ppr oa c he s you
a nd l owe r a s i t r e c e de s .
● Thi s i s a n e xa mpl e of t he
doppler effect .
46. ● Consider a s our c e of s ound
wa ve s wi t h a c ons t a nt
f r e que nc y a nd a mpl i t ude .
● The s ound wa ve s c a n be
r e pr e s e nt e d a s c onc e nt r i c
c i r c l e s whe r e e a c h c i r c l e
r e pr e s e nt s a c r e s t or
t r ough a s t he wa ve f r ont s
r a di a t e a wa y f r om t he
s our c e .
Sour c e s t a t i ona r y
47. The source is
stationary
In figure shows a source of
sound emitting wave with a
constant frequency 𝑓𝑓𝑠𝑠 ,
t oge t he r wi t h t wo obs e r ve r s
A a nd B.
● I f t he s our c e i s
s t a t i ona r y, wa ve s a r r i ve
a t A a nd B a t t he s a me
r a t e .
● So bot h obs e r ve r s he a r
s ounds of t he s a me
f r e que nc y 𝑓𝑓𝑠𝑠 .
● W
e c a n a l s o s a y whe n t he
t r uc k i s s t a t i ona r y, t he
wa ve l e ngt h of t he s ound
i s t he s a me i n f r ont of
a nd be hi nd of t he t r uc k.
A
B
48. The source is
moving
If the source is moving
towards A and away from B
the situation will be
different .
● The di a gr a m s hows t he
wa ve l e ngt h of t he s ound
a r e s qua s he d ( s hor t e ne d)
t oge t he r i n t he
di r e c t i on of A a nd
s pr e a d a pa r t ( l a r ge r ) i n
t he di r e c t i on of B.
● The s ound a ppe a r s a t
a higher f r e que nc y t o
t he obs e r ve r A ( movi ng
t owa r ds t he obs e r ve r A)
● a nd lower f r e que nc y t o
obs e r ve r B ( movi ng a wa y
f r om t he obs e r ve r B)
A
B
Doppler Effect and Its Application.mp4
49. An equation for
observed frequency
● There are two different
speeds involved in this
situation :
1. Speed of the source, 𝑣𝑣𝑠𝑠
2. Spe e d of t he s ound wa ve s
t r a ve l t hr ough t he a i r , 𝑣𝑣
( whi c h i s una f f e c t e d by
t he s pe e d of t he s our c e )
● Spe e d of a wa ve v de pe nds
onl y on t he me di um i t i s
t r a ve l l i ng t hr ough
50. We know t ha t whe n t he
obs e r ve r i s s t a t i ona r y, t he
f r e que nc y r e c e i ve d by t he
obs e r ve r i s t he f r e que nc y
e mi t t e d by t he s our c e .
And t he wa ve l e ngt h obs e r ve d
by t he obs e r ve r i s :
𝑣𝑣 = 𝑓𝑓𝑠𝑠 𝜆𝜆0
𝜆𝜆0 =
𝑣𝑣
𝑓𝑓𝑠𝑠
51. Calculating
doppler effect
When a source of sound waves
moves relative to a
stationary observer, the
observed wavelength 𝜆𝜆𝑜𝑜 i s :
The obs e r ve d f r e que nc y 𝑓𝑓𝑜𝑜 , i s
gi ve n by:
● The wa ve ve l oc i t y f or
s ound wa ve s i s 340 ms - 1.
● The ± de pe nds on whe t he r
t he s our c e i s movi ng
t owa r ds or a wa y f r om t he
obs e r ve r .
I f t he s our c e i s
movi ng towards , t he
de nomi na t or i s 𝑣𝑣 – 𝑣𝑣𝑠𝑠
I f t he s our c e i s
movi ng away, t he
de nomi na t or i s 𝑣𝑣 + 𝑣𝑣𝑠𝑠
𝑓𝑓𝑜𝑜 =
𝑣𝑣
𝜆𝜆𝑜𝑜
= 𝑓𝑓𝑠𝑠
𝑣𝑣
𝑣𝑣 ± 𝑣𝑣𝑠𝑠
𝜆𝜆𝑜𝑜 =
𝑣𝑣 ± 𝑣𝑣𝑠𝑠
𝑓𝑓𝑠𝑠
52. A t r a i n wi t h a whi s t l e
t ha t e mi t s a not e of
f r e que nc y 800 Hz i s
a ppr oa c hi ng a n obs e r ve r
a t a s pe e d of 60 ms ¯¹ .
W
ha t f r e que nc y of not e
wi l l t he obs e r ve r he a r ?
( s pe e d of s ound i n a i r
= 330 ms ¯¹ )
Worked example Solution
He r e t he s our c e i s
a ppr oa c hi ng t he obs e r ve r .
𝑓𝑓𝑜𝑜 = 𝑓𝑓𝑠𝑠
𝑣𝑣
𝑣𝑣 − 𝑣𝑣𝑠𝑠
𝑓𝑓𝑜𝑜 = 800 𝐻𝐻𝐻𝐻
330 𝑚𝑚𝑚𝑚−1
330 𝑚𝑚𝑚𝑚−1 − 60 𝑚𝑚𝑚𝑚−1
𝑓𝑓𝑜𝑜 = 980 𝐻𝐻𝐻𝐻
53. A pl a ne ’ s e ngi ne e mi t s a not e
of c ons t a nt f r e que nc y 120 Hz .
I t i s f l yi ng a wa y f r om a
s t a t i ona r y obs e r ve r a t a s pe e d
of 80 ms ⁻¹.
( s pe e d of s ound i n a i r = 330 ms ¯¹ )
Ca l c ul a t e :
i . The obs e r ve d wa ve l e ngt h of
t he s ound r e c e i ve d by t he
obs e r ve r .
i i . I t s obs e r ve d f r e que nc y.
( s pe e d of s ound i n a i r =
330 ms ⁻¹)
Worked example Worked example
The s ound e mi t t e d f r om t he
s i r e n of a n a mbul a nc e ha s a
f r e que nc y of 1500 Hz . The
s pe e d of s ound i s 340 ms ⁻¹ .
Ca l c ul a t e t he di f f e r e nc e i n
f r e que nc y he a r d by a
s t a t i ona r y obs e r ve r a s t he
a mbul a nc e t r a ve l s t owa r ds
a nd t he n a wa y f r om t he
obs e r ve r a t a s pe e d of
30 ms ⁻¹ .
Ans we r : 3. 4 m
Ans we r : 97 Hz
Ans we r : 270 Hz
57. ● Visible l i ght i s j us t a
s ma l l r e gi on of t he
e l e c t r oma gne t i c s pe c t r um.
● Al l e l e c t r oma gne t i c wa ve s
ha ve t he f ol l owi ng
pr ope r t i e s i n c ommon:
1. The y a r e a l l t r a ns ve r s e
wa ve .
2. The y c a n a l l t r a ve l i n
a vacuum.
3. I n a va c uum, a l l E. M
wa ve s t r a ve l a t c ons t a nt
s pe e d, 3. 0 x10⁸ ms ⁻¹.
Electromagnetic
Spectrum
4. Si nc e t he y a r e t r a ns ve r s e ,
a l l wa ve s i n t hi s s pe c t r um
c a n be r e f l e c t e d,
r e f r a c t e d, di f f r a c t e d,
pol a r i s e d a nd i nt e r f e r e nc e .
58. EM spectrum wavelengths and
frequencies
● The electromagnetic spectrum is arranged in a specific
order based on their wavelengths or frequencies .
59. The speed of
light
● The wavelength λ a nd t he
f r e que nc y f of t he wa ve s
a r e r e l a t e d by t he
e qua t i on:
c = f λ
● Thi s i s t he s a me a s t he
wa ve e qua t i on: t he wa ve
s pe e d v = c.
● The s pe e d of l i ght i n a
va c uum i s c ons t a nt ,
c = 3. 0 x10⁸ ms ⁻¹
● The hi ghe r t he frequency,
t he hi ghe r t he energy of
t he r a di a t i on.
● Howe ve r , whe n l i ght
t r a ve l s f r om a va c uum
i nt o a ma t e r i a l me di um
s uc h a s gl a s s , i t s s pe e d
decreases but i t s
f r e que nc y remains the
same.
● The r e f or e i t s wavelength
must decrease .
60. Radiation Wavelength range/ m Frequency range/ Hz
radio waves >10⁶ to 10⁻¹ < 3 x 10⁹
microwaves 10⁻¹ to 10⁻³ 3 x 10⁹ - 3 x 10¹¹
infrared 10⁻³ to 7 x 10⁻⁷ 3 x 10¹¹ - 4.3 x 10¹⁴
visible
7 x 10⁻⁷ (red) to
4 x 10⁻⁷ (violet)
7.5 x 10¹⁴ - 4.3 x 10¹⁴
ultraviolet 4 x 10⁻⁷ to 10⁻⁸ 7.5 x 10¹⁴ - 3 x 10¹⁶
X-rays 10⁻⁸ to 10⁻³ 3 x 10¹⁶ - 7.5 x 10²⁰
𝛾𝛾-rays 10⁻¹⁰ to 10⁻¹⁶ 3 x 10¹⁸ - 3 x 10²⁴
61. 1. Ca l c ul a t e t he f r e que nc y
i n M
Hz of a r a di o wa ve
of wa ve l e ngt h 250 m.
The s pe e d of a l l
e l e c t r oma gne t i c wa ve s
i s 3. 0 x 10⁸ ms ⁻¹.
Ans we r : 1. 2 M
Hz
2. Ca l c ul a t e t he
wa ve l e ngt h i n nm of a n
X- r a y wa ve of f r e que nc y
2. 0 x 10¹⁸ Hz .
Ans we r : 0. 15 nm
Worked example Past year
questions
(S2017_V11_Q26)
W
hi c h l i s t s hows e l e c t r oma gne t i c
wa ve s i n or de r of i nc r e a s i ng
f r e que nc y?
64. ● Polarisation i s t he
pr oc e s s by whi c h t he
os c i l l a t i ons a r e ma de t o
oc c ur i n one pl a ne onl y.
● I t i s a pr ope r t y t o
di s t i ngui s h be t we e n
t r a ns ve r s e a nd
l ongi t udi na l wa ve .
● A l i ght wa ve t ha t i s
vi br a t i ng i n mor e t ha n one
pl a ne i s r e f e r r e d t o a s
unpol a r i s e d l i ght .
● Li ght e mi t t e d by t he s un,
by a l a mp i n t he c l a s s r oom
or by a c a ndl e f l a me a r e
e xa mpl e s of unpol a r i z e d
l i ght .
Polarisation
65. ● The vi br a t i ons a r e s a i d
t o be pl a ne pol a r i s e d i n
e i t he r ve r t i c a l pl a ne or
hor i z ont a l pl a ne .
● Si nc e l ongi t udi na l wa ve s
vi br a t e a l ong t he i r a xi s
of pr opa ga t i on, i t i s
not pos s i bl e t o pol a r i z e
a l ongi t udi na l wa ve .
● Pol a r i s a t i on i s a c hi e ve
by us i ng ve r t i c a l or
hor i z ont a l s l i t .
66. ● A light wave is an
electromagnetic wave
that travels through the
vacuum of outer space .
● Electromagnetic waves
are transverse waves
consisting of electric
and magnetic fields that
oscillate perpendicular
to the direction of
propagation .
● When we talk about the
direction of polarisation
of light wave, we refer
to the electric field
component .
67. ● Waves c a n be pol a r i s e d
t hr ough a t r a ns pa r e nt
pol yme r ma t e r i a l , s uc h
a s pol a r oi d.
● The pol a r oi d ha s l ong
c ha i ns of mol e c ul e s
a l i gne d i n one
pa r t i c ul a r di r e c t i ons .
Polarization by
use of a
Polaroid Filter
● Any e l e c t r i c f i e l d
vi br a t i ons a l ong
( pa r a l l e l ) t he s e c ha i ns
of mol e c ul e s a r e
a bs or be d.
● M
e a nwhi l e , e l e c t r i c
f i e l d vi br a t i ons a t
r i ght a ngl e s t o t he
c ha i ns of mol e c ul e s a r e
t r a ns mi t t e d.
● The e ne r gy a bs or be d i s
t r a ns f e r r e d t o t he r ma l
e ne r gy i n t he Pol a r oi d.
69. (plane polarised wave)
• Consider a single piece
of Polaroid where the
reference direction is
vertical .
• When a beam of
unpolarized light is
directed at the Polaroid,
a beam of vertically
polarized light rays is
transmitted .
70. ● When a be a m of unpol a r i z e d
l i ght i s di r e c t e d a t t he
pol a r oi d whe r e t he
r e f e r e nc e di r e c t i on i s
ve r t i c a l , a be a m of
ve r t i c a l l y pol a r i s e d l i ght
r a ys i s t r a ns mi t t e d.
● Thi s pl a ne pol a r i s e d
l i ght i s i nc i de nt on a
s e c ond pol a r oi d whos e
t r a ns mi s s i on of a xi s i s
ve r t i c a l .
● The s e c ond pol a r oi d i s
c a l l e d a n a na l ys e r .
● The i nc i de nt l i ght pa s s e s
s t r a i ght t hr ough.
What would happen
when you view
unpolarised light
using two Polaroids?
71. ● Now r ot a t e t he a na l ys e r
t hr ough 90° , s o i t s
t r a ns mi s s i on a xi s i s
hor i z ont a l .
● Thi s t i me , t he a na l ys e r
wi l l a bs or b a l l t he
l i ght .
● The a na l ys e r wi l l a ppe a r
black .
72. ● Turning t he a na l ys e r
t hr ough a f ur t he r 90°
wi l l l e t t he l i ght
t hr ough t he a na l ys e r
a ga i n.
● W
ha t ha ppe ns a t a ngl e s
ot he r t ha n 0° a nd 90° i s
di s c us s e d l a t e r .
What is Polarisation.mp4
73. Application of
polarisation
Since longitudinal waves
vibrate along their axis of
propagation, it is not
possible to polarise a
longitudinal wave.
Polaroid sunglasses
● Li ght r e f l e c t e d f r om t he
s ur f a c e of wa t e r , or gl a s s ,
i s pa r t i a l l y pol a r i s e d i n a
pl a ne pa r a l l e l t o t he
r e f l e c t i ng s ur f a c e .
● Thi s me a ns t ha t a f t e r
r e f l e c t i ng t he s un' s wa ve s
os c i l l a t i on a ngl e s , t he y
a r e no l onge r r a ndom i n a l l
di r e c t i ons but ha ve a
pr e f e r r e d di r e c t i on on
a ve r a ge .
74. ● In t he c a s e of a
hor i z ont a l s ur f a c e —
l i ke
a l a ke or a r oa d—
t he
pr e f e r r e d di r e c t i on i s
hor i z ont a l .
● The pol a r i z e d f i l t e r s on
t he s e l e ns e s
pr e f e r e nt i a l l y bl oc k t he
hor i z ont a l c ompone nt of
l i ght os c i l l a t i on whi l e
t r a ns mi t t i ng t he
ve r t i c a l c ompone nt .
partially polarised.mp4
● The r e s ul t i s a da r ke r
i ma ge but wi t h be t t e r
c ont r a s t .
75. The i nt e ns i t y of t he
t r a ns mi t t e d l i ght de pe nds
on t he a ngl e 𝜃𝜃.
Cons i de r t he i nc i de nt
pl a ne pol a r i s e d l i ght of
a mpl i t ude i s A₀.
The c ompone nt of t he
a mpl i t ude t r a ns mi t t e d
t hr ough t he pol a r oi d a l ong
i t s t r a ns mi s s i on a xi s i s
A₀ c os 𝜃𝜃.
Malus’s law
I nt e ns i t y of l i ght i s
pr opor t i ona l t o t he
a mpl i t ude s qua r e d.
𝐼𝐼 ∝ 𝐴𝐴2
76. So t he i nt e ns i t y of l i ght
t r a ns mi t t e d wi l l be :
𝐼𝐼 = 𝐼𝐼0 𝑐𝑐𝑐𝑐𝑐𝑐2
𝜃𝜃
W
he r e ;
𝐼𝐼 = t r a ns mi t t e d i nt e ns i t y
𝐼𝐼0 = i nc i de nt i nt e ns i t y
𝜃𝜃 = a ngl e be t we e n t he a xe s
of t he pol a r i s e r a nd
a na l ys e r
𝐴𝐴0 sin 𝜃𝜃
𝐴𝐴0 cos 𝜃𝜃
𝐴𝐴0
77. ● The change in intensity against the angle of transmission
a xi s i s a c os i ne s qua r e d gr a ph.
● Va r i a t i on of t r a ns mi t t e d i nt e ns i t y I wi t h a ngl e θ.
● Fr om gr a ph, ma xi mum i nt e ns i t y whe n θ = 0° , 180° a nd s o on,
● a nd z e r o whe n θ = 90° , 270° a nd s o on.
78. The half rule
When unpolarised light
passes through the first
polariser, half the
intensity of the wave is
always lost (
' G
U
) .
Worked example
page 236
● The Pol a r oi d woul d a l l ow onl y pl a ne
pol a r i s e d l i ght t o ge t t hr ough,
wi t h t he e l e c t r i c f i e l d vi br a t i ons
a l ong t he t r a ns mi s s i on a xi s of t he
Pol a r oi d.
● Al l ot he r os c i l l a t i ng e l e c t r i c
f i e l ds f r om t he i nc omi ng
unpol a r i s e d l i ght wi l l be bl oc ke d
by t he l ong c ha i ns of mol e c ul e s of
t he Pol a r oi d.
● Some of t he i nc i de nt l i ght e ne r gy
i s t r a ns f e r r e d t o t he r ma l e ne r gy
wi t hi n t he Pol a r oi d.
1. Expl a i n wha t ha ppe ns t o
unpol a r i s e d l i ght i nc i de nt a t
a Pol a r oi d.
79. 2. Pl a ne pol a r i s e d l i ght of
i nt e ns i t y 12 W
m⁻² i s
i nc i de nt a t a Pol a r oi d.
Ca l c ul a t e t he i nt e ns i t y of
t he t r a ns mi t t e d l i ght whe n
t he a ngl e be t we e n t he
pl a ne of pol a r i s a t i on of
t he i nc i de nt l i ght a nd t he
t r a ns mi s s i on a xi s of t he
Pol a r oi d i s
i . 45°
Ans we r : 6 W
m⁻²
i i . 60°
Ans we r : 3 W
m⁻²
3. Pl a ne pol a r i s e d l i ght i s
i nc i de nt a t a Pol a r oi d.
Ca l c ul a t e t he a ngl e 𝜃𝜃,
whi c h gi ve s t r a ns mi t t e d
l i ght of i nt e ns i t y 30 %
t ha t of t he i nc i de nt
i nt e ns i t y of l i ght .
Ans we r : 57 ⁰
80. Unpolarised l i ght i s
i nc i de nt on a pol a r i s e r .
The l i ght t r a ns mi t t e d by
t he f i r s t pol a r i s e r i s
t he n i nc i de nt on a s e c ond
pol a r i s e r .
The pol a r i s i ng( or
t r a ns mi s s i on) a xi s of t he
s e c ond pol a r i s e r i s 30°
t o t ha t of t he f i r s t .
Worked example The i nt e ns i t y i nc i de nt on
t he f i r s t pol a r i s e r i s I .
W
ha t i s t he i nt e ns i t y
e me r gi ng f r om t he s e c ond
pol a r i s e r ?
81. Step 1
Fr om t he ha l f r ul e , whe n
t he l i ght pa s s e s t hr ough
t he f i r s t pol a r i s e r ha l f
of i t s i nt e ns i t y i s l os t .
Solution
St e p 2
M
a l us ’ s l a w i s us e d t o
f i nd t he i nt e ns i t y of t he
pol a r i s e d l i ght a f t e r t he
s e c ond pol a r i s e r .
St e p 3
Combi ne t he i nt e ns i t y
dr ops .
82. Unpolarized l i ght wi t h a n
i nt e ns i t y of 𝐼𝐼0 = 16
W
m⁻² i s i nc i de nt on a pa i r
of pol a r i s e r s .
The f i r s t pol a r i s e r ha s
i t s t r a ns mi s s i on a xi s
a l i gne d a t 50o f r om t he
ve r t i c a l .
Worked example The s e c ond pol a r i z e r ha s
i t s t r a ns mi s s i on a xi s
a l i gne d a t 20o f r om t he
ve r t i c a l .
Ca l c ul a t e t he i nt e ns i t y
of t he l i ght goi ng
t hr ough t he pa i r of
f i l t e r s .
83. Solution
Step 1
From the half rule, when the
light passes through the first
polariser half of its intensity
is lost .
The light incident on the first
polarizer is unpolarized, so the
angle is irrelevant .
Step 2
Through the second polariser .
Step 3
Combine the intensity drops .
84. Exam
- style
questions
a) State what is meant by
plane polarised light .
[ 1]
● A pl a ne pol a r i s e d wa ve i s
a t r a ns ve r s e wa ve wi t h
os c i l l a t i ons ( of t he
e l e c t r i c f i e l d) i n j us t
one pl a ne .
b) Ve r t i c a l l y pl a ne pol a r i s e d
l i ght i s i nc i de nt on t hr e e
pol a r i s i ng f i l t e r s .
The t r a ns mi s s i on a xi s of t he
f i r s t Pol a r oi d i s ve r t i c a l .
The t r a ns mi s s i on a xi s of t he
s e c ond f i l t e r i s 45° t o t he
ve r t i c a l a nd t he
t r a ns mi s s i on a xi s of t he
l a s t f i l t e r i s hor i z ont a l .
Show t ha t t he i nt e ns i t y of
l i ght e me r gi ng f r om t he
f i na l f i l t e r i s not z e r o.
[ 4]
85. ● The i nt e ns i t y of
t r a ns mi t t e d l i ght f r om
t he f i r s t pol a r i s i ng
f i l t e r = I ₀ ( t he s a me a s
t he i nc i de nt i nt e ns i t y)
[ 1]
● The i nt e ns i t y of l i ght
f r om t he s e c ond f i l t e r
wi l l be :
I = I ₀ c os ² θ = I ₀ c os ² 45°
= 0. 50 I ₀ [ 1]
● The i nt e ns i t y of l i ght
f r om t he l a s t f i l t e r wi l l
be :
I = I ₀ c os ² θ =
[ 0. 50 I ₀] c os ² 45° =
0. 50I ₀ × 0. 50 = 0. 25I ₀ [ 1]
● The f i na l t r a ns mi t t e d
i nt e ns i t y i s not z e r o,
but 25% of t he or i gi na l
i nt e ns i t y.
