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Transcript
C.K.Cheung
Wave
- a form of energy transfer
2 natures of wave:
a/
b/
mechanical wave: require a material medium for propagation.
electromagnetic wave: can travel through vacuum.
2 types of oscillations of wave:
a/
longitudinal wave
Direction of energy
travel
b/
transverse wave
Direction of energy
travel
1
C.K.Cheung
Wavefront:
A surface ( reducing to a line in 2-dim. ) over which the oscillation has the same
phase ( not in phase ) at all points.
Ray
A ray is the path taken by the wave energy.
Note:
A ray is  to a wavefront.
2
C.K.Cheung
Wave velocity
In 1 second:
f=0
f = 1 Hz
f = 2 Hz
.
.
.
.
f = f Hz

f
v=f
waves
( holds for all wave motion )
3
C.K.Cheung
Traveling Wave
Left hand end of string
made to move with
S.H.M. of amplitude = a,
frequency = f
y
x=x
X
X
Very long string
under tension
>0
=0
Let the transverse displacement be represented by y(x, t) :
 y(0, t) = a sin t
( S.H.M.)
At some location x, ( x >0), the displacement y(x, t):
1/
2/
varies with time t in the same way
shows a phase lag 

x=0

x= x
t
t= 0
 y (x, t) = a sin (t -  )
If the wave propagates with constant speed, v :
  x
=kx
k = constant called wave number.
 y( x, t ) = a sin (  t – k x )
(LR)
4
C.K.Cheung
Similarly, for wave traveling from R  L :
The displacement at x would show a phase lead of  :
x=x


x= 0
t
t= 0
 y(x, t) = a sin ( t +  )
 y(x, t) = a sin ( t + kx) ( R  L )
Expression for k:
x=
x=0
 k x = ,
if x = 
2
k=

Hence,

 = 2
y (x, t) = a sin (
2
t
T

2

x)
5
C.K.Cheung
Stationary Wave
1/ waveform does not seem to be traveling along any direction.
2/ formed by 2 identical waves traveling in opposite direction.
A = anti-node
N
A
N = node
A
Consider:
Y1
Y2
 y1 = a sin (t – k x) (L  R)
y2 = a sin ( t +k x ) (RL)
 y = y1+y2 = a sin(t-kx) + sin(t + kx)
y = 2a cos kx sin t
f(x)
f(t)
f(x) = amplitude function
f(t) = oscillating function
6
C.K.Cheung
Graphical Approach to Standing Wave:
Y2
Y1
t=0
t=
T
4
t=
T
2
t=
3T
4
http://www.kettering.edu/~drussell/demos.html
http://library.thinkquest.org/19537/Physics4.html?tqskip1=1&tqtime=0127
7
C.K.Cheung
Hence:
A/m
t=0
X /m
t =T/2
Note:
1/ Oscillation amplitude varies with location. (c.f. traveling wave).
2/ Within same loop, all particles oscillate in phase, but anti-phase with those
in adjacent loops
3/ Distance between nodes or antinodes =

2
4/ A standing wave shows no net energy transfer in space ( but in time, it
shows an interchange of K.E.  P.E.)
5/ Standing wave can also be produced with E.M. waves and with sound waves.
6/ If amplitudes unequal:
8
C.K.Cheung
Coherent Sources:

P
 = constant in time
Conditions for observable interference pattern:
1/ Coherent sources
2/ Comparable amplitudes
Note: In practice, coherent sources are derived from wavefronts of a single
source.
3/ Same polarization (same plane of vibration)
9
C.K.Cheung
Young’s double slits experiment
P
(single frequency)

Monochromatic
light source
x
A
d
B
C
D >> d
Path difference = BC = d sin 
 D >> d   small  sin  ~ tan  =
BC = d (
x
D
x
)
D
For constructive interference:
 BC = m
x
d(
)=m
D
x=
mD
d
m = 0, 1, 2, 3…..
m= 0  central max.
m = 1 ,  1st max.
m = 2,  2nd max
For destructive interference:
1
 BC = (m + )
2
x
1
d(
) = (m + ) 
D
2
m = 0, 1, 2, 3….
1
( m  )D
2
x=
d
10
C.K.Cheung
Hence, on the screen:
P

x
A
d
B
C
D >> d
m = -1
m=0
m = +1
m=0
intensity
x = fringe width
x = fringe width
11
C.K.Cheung
Let x = separation between bright ( or dark ) fringes
(m)D
 x =
d
If m = 1  separation between neighborhood fringes called fringe width x .
Note:
1/ ( x ) for red light > ( x ) for blue light
2/ If d (slit separation ) decreases slightly  ( x ) increases.
3/ If opening of each slit is greatly widened  large number of narrow slits,
each pair of slits producing its own band pattern  patterns overlap with
each other  uniform illumination
4/ If whole apparatus is immersed in a medium of R.I. = n,

  decreases to
,
n
 ( x ) decreases.
(x )(d)
5/  =
, a simple method to measure .
D
6/ 2 point sources ( replace 2 narrow slits ) would give the same fringe system
but slits ( i.e. a line of point sources ) give brighter fringes by reinforcing
the pattern.
7/ The interference is incomplete as m is large because for all fringes except
the central bright one, the amplitude of the 2 wave-trains are not exactly
equal.
8/ When using white light (non -coherent ) source:
m = -1
m=0
m’ =0
m’ = -1
First
max
m = +1
m’ = 0
Central
max
m’ = +1
First
max
12
C.K.Cheung
Each color produces its own set of fringes  spectrum (~ 5 spectra on either side ). Only the
central one is white, because its position is zero path difference for all colors.
9/
The amount of fringes depends on the amount of diffraction occurring at the slits and
this in turn depends on their width. The narrower the slits, the greater will be the number of
fringes due to increased diffraction, but fainter will be the fringes, since less light gets
through.
Interference
pattern
10/ If source is displaced as:
New central max
http://vsg.quasihome.com/interfer.htm
13
C.K.Cheung
Reflection and phase change
1/
Crest
Phase change =  radian
trough
2/
 - changed
3/
No phase change
Hence, when a transverse wave (except water wave) is reflected at a denser

medium, there is a phase change of  radians or .
2
( c.f. longitudinal wave)
14
C.K.Cheung
Optical path
L
v
R.I. = n
nL
C
Time for the light ray to travel through the block = t =
t=(
L
L

C
v
( )
n
( n = (
C
))
v
nL
)
C
= time for light to travel through vacuum of distance (n L)
Definition:
Optical path (O.P.) in a medium = n L
( =  ni Li for several media )
i
15
C.K.Cheung
e.g.
A thin glass film ( R.I. =1.58) is used to cover one slit of a double – slit
arrangement. The central point on the screen is occupied by what used to be
the seventh bright fringe. If  = 550 nm, what is the thickness, t, of the glass
film?
t
Ans: 6.6x10 – 6 m
16
C.K.Cheung
Solution:
 path difference = m
(path difference ) =  (m)
 (nt – t) = 7 
(7)(5.5 x10 7 )
t=
 6.6 x10 6 m
(1.58  1)
Question: How will the fringe pattern shift?
The pattern will shift downwards.
17
C.K.Cheung
‘Blooming’ of lens
In lens system ( e.g. telescope, microscope etc.) reflected light produces:
1/ A background of unfocused light  reduction in clarity of final image
2/ A reduction in intensity of image,  less light transmitted through the
lens.
Improvement
air
MgF2 R.I.=n’
t
Glass ( n > n’ )
Both rays undergo  change of phase
 for destructive interference:

 2(n’ t) =
( for minimum t & assume normal incidence )
2
 t=

4n '
18
C.K.Cheung
Usually, take  for yellow light.
Lens reflected
 lens appears
(white – yellow) light
purple
( red + blue )
‘Bloomed’ lenses produced a marked improvement in the clarity of the final
image in optical instruments.
19
C.K.Cheung
e.g.
85’MC
18.
The surface of a material of refractive index 1.8 is coated with a thin film of
liquid of refractive index 1.5 and thickness 200 nm. White light falls
normally on the thin film. Which of the following light wavelengths (in air)
is not reflected from the thin film?
A.
B.
C.
D.
E.
400
450
600
750
800
nm
nm
nm
nm
nm
A
20
C.K.Cheung
Thin Film Interference
N
D
i
M
i
r
O
C
r
t
thin transparent
film of R.I. = n
E
B
r
X
O.P.D. between ON and OBCM:
O.P.D. = n ( OB+BC) – OD
OD
(
)
sin i
OD
n=
 OC 
OE
sin r
OE
(
)
OC
 n(OE) = OD
 O.P.D.= n(OB+BC) – n(OE)
=n(EB+BC) = n(EB+BX) = n (EX) = n (2t cos r )
21
C.K.Cheung
Hence:
O.P.D.=2ntcos r
For bright fringes:
 O.P.D. + phase change = m
 2nt cos r -
m = 0,1,2,…………

= m
2
 2nt cos r = ( m +
1
)
2
For dark fringes:
 2ntcos r +

1
= (m+ )
2
2
 2ntcos r = m
Note:
For thick films, several values of  satisfy the equation 2nt cos r = ( m +
1
)  for a given value
2
of r ( or i ), the film appears to be generally illuminated.
22
C.K.Cheung
Colour in thin films.
non-coherent extended
source (sky, cloud etc.)
1/
The eye will see the same color of a particular  such that 2nt cos r = ( m +
2/
Different colors are focused to different positions on the retina of the eye.
 a spectrum is seen.
1
) .
2
23
C.K.Cheung
82’IIB
6. A circular wire ring is dipped into soap solution and held vertically. When viewed in a dark
room with monochromatic light of wavelength 6.5  10-7 m reflected normally from the film,
a series of interference fringes are seen. The pattern of fringes at a particular instant is
shown in Figure 5.
(The refractive index of soap solution = 1.33)
dark area
wire ring
held
vertically
dark bands
A
Figure 5
(a) What colour are the fringes?
Orange/red
(b) Why is there a dark area at the top of the ring?
Thickness of film is not uniform. At the top, thickness 

, constructive interference
4n
does not take place.
(c) What is the thickness of the film at point A?
Put

1
 ( m  )
2
2
m = 2  t = 4.89x10-7 m
2 nt +
m = 1, 2,………….
(d) As time goes on and if the film drains downwards and does not break, the fringe pattern
changes from that shown in Figure 5. Describe the changes you would expect to see
(explanations are not required).
(9 marks)
1.
2.
3.
dark area at top increases.
the number of fringes increases
fringes are more closely spaced towards the bottom.
24
C.K.Cheung
Revision on Interference
95’ MC
12.
direction of propagation of wave
A
B
The figure shows a sound wave travelling to the right in air. Air particles A and B are at the
centre of a compression and a rarefaction respectively. Which of the following gives
correctly the directions of motion of A and B at the moment shown?
A.
B.
C.
D.
E.
Particle A
Particle B
to the right
to the right
to the right
at rest
to the left
to the left
at rest
to the right
to the right
to the right
15. A coating material of refractive index 1.25 is used for the blooming of a lens having a larger
refractive index. For normal incidence, if green light is to be transmitted in its greatest
amount through the lens, which of the following thicknesses of the coating would do?
(Given : wavelength of green light in air is 550 nm)
(1) 137.5 nm
(2) 220 nm
(3) 330 nm
A.
B.
C.
D.
E.
(1) only
(3) only
(1) and (2) only
(2) and (3) only
(1), (2) and (3)
25
C.K.Cheung
94’ MC
20.
vibrator
stand
In the above experimental set-up, different stationary wave patterns are produced on an
elastic string by adjusting the frequency f of the vibrator. Which of the following statements
is/are correct?
(1) When f increases, the number of antinodes increases.
(2) When f increases, the speed of the waves on the string increases.
(3) The waves produced in air by the string has the same speed as the waves on the string.
A.
B.
C.
D.
E.
(1) only
(3) only
(1) and (2) only
(2) and (3) only
(1), (2) and (3)
93’ MC
27.
soft
sound
soft
sound
S
L
M
A loudspeaker L produces sound waves with frequency 1 000 Hz. The sound waves are reflected from a wall
S. When a microphone M is moved between L and S, the loudness of the sound detected varies. (Speed of
sound in air = 340 m/s)
Which of the following statements is/are true?
(1) The variation in the loudness of the sound is due to diffraction.
(2) The separation between consecutive positions of soft sound is 0.34 m.
(3) Increasing the sound frequency will make the positions of soft sound closer.
A.
B.
C.
D.
E.
(1) only
(3) only
(1) and (2) only
(2) and (3) only
(1), (2) and (3)
26
C.K.Cheung
32. Which of the following devices is/are used to stored energy?
(1)
(2)
(3)
an inductor
a capacitor
a photocell
A.
B.
C.
D.
E.
(1) only
(3) only
(1) and (2) only
(2) and (3) only
(1), (2) and (3)
27
C.K.Cheung
93’ IIB
8. In Figure 8.1, a layer of material of refractive index 1.40 and thickness 3.0  10-5 m is coated
uniformly on a glass block of refractive index 1.50. It is illuminated by light of wavelength
4.20  10-7 m from an extended source.
Coating,
refractive index = 1.40
Glass, refractive index = 1.50
Figure 8.1
(a) What is the colour of the light?
(1 mark)
Violet / blue
(b) Calculate the wavelength of the light in the film.
’ =
(1 mark)
4.2 x10 7

=
 3x10 -7 m
n
1.4
(c) For normal incidence,
(i)
find the path difference between the lights reflected from the coating surface and
from the coating-glass boundary;
(2 marks)
Optical path difference = 2(1.4)(3x10-5) = 8.4x10-5 m
(ii)
explain what will happen when the two reflected waves recombine.
(2 marks)
5
8.4 x10
Since,
 200
4.2 x10 7
 constructive interference takes place.
28
C.K.Cheung
(d) Figure 8.2 shows part of the interference pattern observed directly from above. Explain
why such a pattern is observed.
(2 marks)
dark ring
diagram NOT to scale
Figure 8.2
Central bright spot is due to constructive interference.
The loci of equal path difference are circular when viewed from above.
Different circular rings are due to slight variation in (optical) path difference that depends
on the angle of viewing.
(e) Sketch the interference pattern if the coating is not on a glass surface but on the top of a
transparent block of refractive index 1.38.
(1 mark)
(f) Explain briefly the difference between the patterns in (d) and (e).
In (d), the phase change due to reflection is
(2 marks)

2
In (e), no phase change due to reflection.
 dark rings in (d) become bright rings in (e) and vice versa.
29
C.K.Cheung
A
Aiirr W
Weeddggee
Traveling microscope
Sodium
lamp
T
θ
t
S1
S2
L
For constructive interference

2t = m
2
1
)
 2t = ( m +
2
1
)
m = 0,1,2,……..
 2S1θ= ( m +
2
For destructive interference

1
 2S1θ+
=(m+
)
2
2
 2S1θ= m 
If the microscope is moved to (m+k) th dark fringe
 2S2θ= (m + k ) 
Hence: 2(S2-S1)θ= k
 θ=
k
2( S 2  S1 )
 T (thickness/diameter)can be found = Lθ
Used in testing any uneven parts of the surface of a glass plate
30
C.K.Cheung
90’ IIB
9.
t
wire
80 mm
Figure 9.1
To measure the diameter of a metal wire, the wire is placed between two flat parallel-sided
glass plates as shown in Figure 9.1, forming a wedge-shaped air film of length 80 mm. The
plates are illuminated normally from above by sodium light of wavelength  and interference
fringes are observed from above.
(a) Draw a labelled diagram of a suitable experimental arrangement which includes the
apparatus of Figure 9.1 to measure the separation between adjacent fringes. (3 marks)
(b) Briefly explain the formation of the fringes and hence write down in terms of  and t, the
condition for a bright fringe to be produced at a particular point where the separation of
the plates is t (see Figure 9.1)
(3 marks)
(c) In this experiment, the wavelength of the light is 600 nm, and the fringes are found to
have a separation of 0.16 mm. Calculate the diameter of the wire.
(4 marks)
(d) If the space in the wedge is filled with water of refractive index 1.33, what will happen to
the fringe pattern observed? Explain your answer briefly.
(2 marks)
31
C.K.Cheung
Newton’s Rings
Traveling microscope
Sodium
lamp
t
Interference between light rays reflected from the air-film between the convex
lens of long focal length and the flat glass plate
For destructive interference:
 2t +

1
=(m+
)
2
2
 2t = m
Loci of equal thickness in air-film satisfying destructive interference are circular
 dark rings can be seen.
32
C.K.Cheung
Note:
1. As a quality check in grinding a lens surface
2. The transmitted pattern is the compliment of the reflected one
3. A localized fringe pattern. (pattern confined /restricted in a particular surface, as in the case
of air-wedge)
4. If the lens is slowly raised up, the fringes will shift inwards
5. The contact point is always dark
6. If the air medium is replaced by water, fringes will shift inwards
33
C.K.Cheung
Optical Instrument
h

D
h’
h
Angular magnification =
’
 ''

'
h'
= D
h
D
=
h'
h
Note:
1. Far point =  for normal eye
2. D = least distance of distinct vision ~ 25 cm
34
C.K.Cheung
Compound Microscope
1. For normal adjustment, final image is formed at D.
2. Final image is virtual & inverted
D
Eyepiece
Objective
Fe
Fo
h’
Object
’
h’’
Final Image
h

D
 ' ' h" / D h" h" h'

 x  me xmo
Magnifying Power = m =
=
h/ D
h h' h

'
35
C.K.Cheung
Eye-ring
Eyepiece
Objective
Object
’
Eye-ring
1.
2.
Eye-ring: the best position of the eye where it will receive as much light as possible from
the object.
Position of eye-ring: image position of objective lens formed by eyepiece.
36
C.K.Cheung
Telescope:
1/ Astronomical telescope
For normal adjustment, the final image is formed at  . The final image is virtual and
inverted.
Length of telescope = fo+ fe


fo
fe

h
’
Construction
line
(may not // to refracted
rays)
m=
h / fe
f
 ''
=
 o

h / fo
fe
37
C.K.Cheung
2/ Galilean telescope
1. For normal adjustment, the final image is at  . The final image is erect (an advantage)
2. The Galilean telescope is shorter than astronomical telescope
3. The eye-ring is situated between the 2 lenses, the eye should be as close as possible to the
eyepiece, and the angle of view is rather limited when compared to astronomical telescope.
( disadvantage )
4. Length of telescope = fo- fe
fo
fe
Construction
line
’

h
h / fe
f
 ''
m=
=
 o
h / fo
fe

'
38
C.K.Cheung
Optical spectrometer
Slit
telescope
collimator
Cross-wire
When in use, the telescope must be adjusted so that parallel light rays entering
the objective are brought to a focus at the cross-wires.
m=2
m=1
m=0
m=1
Diffraction grating
m=2
Steps to be followed before an experiment:
1. View a distant object through the telescope and adjust the image formed on
the cross-wires can be seen clearly through the eyepiece.
2. Align the telescope with the collimator. Look at the illuminated slit through
the telescope. Adjust the distance of the slit from the collimator lens until
the edges of the slit are sharp.
3. Level the turntable by adjusting the three screws underneath.
39
C.K.Cheung
94’IIB
2. (a)
eyepiece
objective lens
object
ho
P
Q
Fo
hi
Fo
FE
FE
final
image
Figure 2.1
A student uses two converging lenses to set up a compound microscope in normal
adjustment. Figure 2.1 shows two light rays, P and Q, form the top of an object falling
on the objective lens of the microscope. The foci of the objective lens are denoted by
FO and the foci of the eyepiece are denoted by FE.
(i)
On Figure 2.1, complete the ray paths for P and Q as they pass through the
microscope, showing how the final image is formed.
(2 marks)
eyepiece
objective lens
object
ho
hi
Fo
P
Q
Fo
FE

FE
final
image
Figure 2.1
40
C.K.Cheung
(ii)
Indicate on Figure 2.1 the visual angle  subtended by the final image at the eye of
an observer using the microscope.
(1 mark)
Refer to diagram above
(iii) Distinguish between linear magnification and angular magnification.
(2 marks)
length of image
Linear Magnification =
length of object
Angular Magnification =
visual angle of image
visual angle of object at D
(iv) Find the angular magnification of the microscope in terms of the height of the
object, ho, and the height of the final image, hi. Show your working. (Take the
least distance of distinct vision to be D)
(2 marks)
Without microscope,  =
With microscope,  =
h1
D
Angular magnification =
h0
D
h

= 1

h0
(b) Figure 2.2 shows four light rays from an object passing through a microscope in normal
adjustment. R and S come from the top of the object, T and U come from the bottom.
R and T pass through the top of the objective lens, S and U pass through the bottom.
R
object
T
S
U
objective
lens
eyepiece
X
Figure 2.2
(i)
On Figure 2.2, X is the best position for the eye to view the image. With reference
to the ray diagram, briefly explain the advantage(s) of choosing X as the viewing
position.
(3 marks)
 All the light from the objective lens ( or the object ) would pass through X.
41
C.K.Cheung


(ii)
The image is then brightest
The field of view is greatest.
Why should the diameter of the beam
2 mm?
at
X be no wider than about
(1 mark)
Matching the width of the beam to the diameter of the pupil.
42
C.K.Cheung
Revision on wave
81’
10.
15 cm
object
A converging lens of focal length 15 cm is used as a magnifying glass with the final image at
infinity. If the least distance of distinct vision is 25 cm, the angular magnification achieved
is
A.
B.
C.
D.
E.
0
1
5/3
15
infinite
M=
h / 15
 5/3  C
h / 25
82’
15. A near-sighted person’s greatest distance of distinct vision is 0.9 m. His sight is improved
by wearing spectacles which increase his greatest distance of distinct vision to 18 m. What
is the magnitude of the focal length of the spectacle lenses?
A.
0.86 m.
B.
0.90 m.
F>v
C.
0.95 m.
C
D.
17.1 m.
F
E.
18.9 m.
84’
15. Newton’s rings produced when a biconvex lens rests on a plane glass plate are observed
using a travelling microscope. If the biconvex lens is very slowly moved vertically upwards
from the lower glass plate, which of the following would be observed?
A.
B.
C.
D.
E.
The central spot remains dark all the time.
The rings disappear immediately.
The rings move towards the centre.
The rings move out from the centre.
The rings are no longer concentric.
C
43
C.K.Cheung
16. Young’s slits are used to produce interference fringes with light of wavelength 600 nm. A
thin sheet of mica of refractive index 1.6 is placed in front of one of the slits and the centre of
the fringe-system is displaced through 8 fringe widths. The thickness of the mica is
A. 120 nm.
B.3 000 nm.
C.4 000 nm.
D.7 700 nm.
E.8 000 nm.
t ( n – 1) = 8 
E
85’
16. Two signal generators are connected to display the formation of beats on a C.R.O. screen.
setting of the two signal generators, the following pattern is observed on the C.R.O.:
For one particular
2.5 cm
If the C.R.O. time base is set at 0.2 ms/cm, the beat frequency is
A. 100 Hz.
B. 200 Hz.
C. 400 Hz.
D. 1 000 Hz.
E. 2 000 Hz.
17.
normal
air
water
ultra-sound
beam
A beam of ultrasound is being emitted from a submarine under water towards the water surface. Which of the
following statements is true?
A. The refracted beam leaving the surface will bend away from the normal.
B. The refracted beam will bend towards the normal.
C. The refracted beam will travel in the same direction as the incident beam.
B
D. Total internal reflection will occur.
E. The refracted beam will travel along the water surface.
18. The surface of a material of refractive index 1.8 is coated with a thin film of liquid of refractive index 1.5 and
thickness 200 nm. White light falls normally on the thin film. Which of the following light wavelengths (in
air) is not reflected from the thin film?
A.
400 nm
B.
450 nm
C.
600 nm
D.
750 nm
A
E.
800 nm
44
C.K.Cheung
19. The image of a distant star produced by an astronomical telescope is a diffraction pattern. If the effective
diameter of the objective lens is reduced by one-half by covering its outer parts with a stop, the area of the
central maximum of the diffraction pattern is
A.
decreased by a factor of 4.
B.
halved.
C.
unchanged.
D.
doubled.
E.
increased by a factor of 4.
20. When parallel light is incident at the Brewster angle in air on the surface of a glass block,
A.
the light is totally reflected.
B.
the reflected light is partially polarised.
C.
the transmitted light is unpolarised.
D. the reflected and refracted wavefronts are at right angles to each other.
E.
the angle of incidence is equal to the angle of refraction.
87’
24.
A
M
S
O
B
A plane mirror M is illuminated by monochromatic light from a slit S. The virtual image of S by reflection and
S itself act as 2 coherent sources and the interference pattern is observed on the screen AOB at a distance from
the mirror. Which of the following statements about the interference pattern on the screen is/are correct?
(1) No interference pattern can be seen in the region OB on the screen.  x
(2) As the mirror M moves downward, the separation of the fringes decreases.  x as d decreases
(3) As the mirror M moves horizontally away from the screen, the separation
of the fringes increases.
decreases
correct, as d decreases
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
(2) and (3) only
D.
(1) only
E
E.
(3) only
88’
26.
displacement
0
a
2a
3a
x
The figure above shows the variation of the displacement of air molecules along the x-axis in a standing sound
wave at a particular time. At what positions will the pressure remain constant with respect to time?
A.
x = 0 and x = 2a only
B.
x = a and x = 3a only
C.
x = 0 and x = a only
B
D.
x = 2a and x = 3a only
E.
x = 0, x = a, x = 2a and x = 3a
45
C.K.Cheung
89’
21. The coated lens of a camera appears purple in daylight. According to the manufacturers, the coating has a
refractive index of 1.25 and the material of the lens has a refractive index of 1.50. If the wavelength of yellow
light in air is 520 nm, the approximate thickness of the coating is
A.
104 nm.
B.
130 nm.
A
C.
180 nm.
D.
210 nm.
E.
260 nm.
22. Light of wavelength  is incident normally on a diffraction grating with p lines per millimetre. the
second-order diffraction maximum is at an angle  from the central position. For a second grating with 3 p
lines per millimetre illuminated normally by light of wavelength 5 /4, the angle between the first-order
diffraction maximum and the central position is . Which of the following relations is correct ?
A.
sin  = (5 sin )/12
B.
sin  = sin (5 /12)
D
C.
sin  = sin (15 /4)
D.
sin  = (15 sin )/8
E.
sin  = sin (15 /8)
23. If the threshold of hearing is 10-12 W/m², a sound level of one microwatt per m² is above threshold by
A.
6 dB.
B.
11 dB.
C.
60 dB.
D.
110 dB.
E.
120 dB.
24. A stationary radar source emits waves of frequency f, and wavelength  which are reflected from an object
moving towards the source at a speed u. The reflected waves reaching a receiver standing near to the radar
source will have an apparent wavelength of
A.
 - 2u/f.
B.
 - u/f.
C.
.
D.
 + u/f.
E.
 + 2u/f.
90’
16.
V
P
shallow
deep
IV
III
II
I
Q
The figure shows wave crests moving in the direction of the arrow towards the interface PQ between a shallow
region and a deep region as shown in the figure above. Which of the lines shown may represent one of the
wave crests in the deep region?
A.
I
B.
II
C.
III
D.
IV
E.
V
46
C.K.Cheung
17.
A disc with a spot on it rotates at a constant speed in a darkened room. A students shines a stroboscopic lamp
on it. When the flashing rate is 8 flashes per second, the disc appears stationary with two spots on it as shown.
When the flashing rate is reduced to 2 flashes per second, the disc would appear to be stationary with
A.
no spot.
B.
one spot.
C.
two spots.
D.
four spots.
E.
eight spots.
18. Which of the following represent the approximate noise levels
(i) in a quiet school library?
(ii) near the airport when an aircraft is taking off and flying overhead?
(i)
(ii)
A.
30 dB
60 dB
B.
60 dB
90 dB
C. 30 dB
90 dB
D.
90 dB
60 dB
E.
60 dB
30 dB
19. The intensity of a sound wave is proportional to the square of the amplitude of the wave. If two waves of the
same frequency are superimposed in phase, the total intensity is proportional to
A. the mean value of the intensities of the two waves.
B. the square of the sum of the two amplitudes.
C. the square of the mean value of the two amplitudes.
D. the square of the difference of the two amplitudes.
E. the sum of the intensities of the two waves.
20. In a simple astronomical telescope, under normal adjustment, which of the following statements is/are correct?
(1) The first image is formed at the focal plane of the objective.
(2) The first image is real and inverted.
(3) The focal length of the objective is longer than that of the eyepiece.
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
(2) and (3) only
D.
(1) only
E.
(3) only
21.
30
o
A boat travels in shallow water, in which waves of all wavelengths travel at a speed of 4.0 m/s.
speed of the boat if the bow wave generated by the boat has an apex angle of 30º?
A.
2.0 m/s
B.
2.3 m/s
C.
4.0 m/s
D.
6.9 m/s
E.
What is the
8.0 m/s
22.
47
C.K.Cheung
The figure shows two pulses travelling to the right along a rope. The right-hand end of the rope is fixed to a
wall. The following figures represent predicted positions of the pulses at later times (where appropriate, the
arrows indicate the direction of a pulse). Which of the following could NOT arise from the initial given
condition?
(1)
(2)
(3)
A.
B.
C.
D.
(1), (2) and (3)
(1) and (2) only
(2) and (3) only
(1) only
E.
(3) only
23.
A
L1
G
L2
B
2 loudspeakers L1 and L2 are connected to a signal generator G. A microphone is moved along the line AB and
the variation in intensity is noted. Which of the following statements concerning the above arrangement is/are
correct?
(1) If the separation of the 2 loudspeakers is less than the wavelength of the sound emitted, no alternation of
maxima and minima can be detected along AB.
(2) If the frequency of the sound waves emitted is increased, the separation between adjacent maxima along AB
will be increased.
(3) If the 2 loudspeakers are vibrating in antiphase, no alternation of maxima and minima will be detected
along AB.
48
C.K.Cheung
A.
B.
C.
(1), (2) and (3)
(1) and (2) only
(2) and (3) only
D.
(1) only
E.
(3) only
24. Sound waves of frequency f are emitted by a source S. When S is moved with speed u (relative to the ground)
towards a stationary observer O, a rise in pitch of f is detected. Which of the following statements is/are
correct?
(1) The speed of sound waves relative to the observer is unaffected by the motion of S.
(2) If both S and O move in the same direction with speed u, no rise in pitch will be detected.
(3) If S is at rest with O moving towards it at speed u, the rise in pitch will also be f.
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
D.
E.
(2) and (3) only
(1) only
(3) only
25. When parallel light is incident at the Brewster angle in air on the surface of a glass block, which of the
following statements is/are correct?
(1)
The refracted light is plane-polarised.
(2)
The reflected light is plane-polarised.
(3) The reflected ray and the refracted ray are at right angles to each other.
A.
B.
(1), (2) and (3)
(1) and (2) only
C.
(2) and (3) only
D.
E.
(1) only
(3) only
26. White light diffracted by a single slit falls on a white screen.
Which of the following statements is/are correct?
(1) The centre of the diffraction pattern is white.
(2) The first minimum is closer to the centre for red light than for blue light.
(3) The central band width is increased as the slit width is increased.
A.
B.
C.
(1), (2) and (3)
(1) and (2) only
(2) and (3) only
D.
(1) only
E.
(3) only
49