YU CHUN KEUNG MEMORIAL COLLEGE NO.2
2003-2004 Form 6 Physics Second Examination
Section A (38)
1.
2.
3.
4.
5.
6.
7.
B
A
B
A
B
D
C
8. A
9. C
10. A
11.
12.
13.
14.
15.
16.
17
D
B
D
A.
D
D
C
18. B
19. B
20. B
21.
22.
23.
24.
25.
26.
27.
A
A
B
A
A
A
C
28. C
29. A
30. C
31.
32.
33.
34.
35.
36.
37.
D
A
C
C
D
D
B
38. A
Section B
1. (a)
(4)
(b) There is intercept at the v2-axis; this implies the spacecraft is moving when
1
= 0 or r   . Thus the spacecraft will escape to infinity.
(1+1)
r
The speed is given by v2 = 4.0  106 or v = 2000 m s1
(1)
(c) The spacecraft is orbiting around the planet in circle. Considering the centripetal
force on the spacecraft:
1
mv 2
r
r2
1
or,
v2 = GM  
r
1
Thus, in the v2-   graph obtain in (i), the slope is equal to GM
r
slope
4  1013
Hence,
M=

 6.0  10 23 kg
11
G
6.67  10
GMm
2.
*
*
*
*
*
*

(0.5+0.5)
The light source used should be strong and properly shielded so that no stray
light falls on the screen.
To obtain sharper fringes, monochromatic light source can be used.
The double-slit used should be very narrow.
The slit separation should be very small (0.5 mm) so that the light from the
two slits can overlap.
The distance between the screen and slits should be at an appreciable distance of
about 1  2 m such that the separation of fringes is observable and the intensity
is not too low.
The light filament or source slit should be parallel to the double slits.
(any four)
3.
(a) Stationary waves are produced when the progressive wave reflected from a
boundary interferes with the wave travelling towards the boundary. The
resulting wave will be such that nodes are formed at both ends of the wire.
(2)
(Draw figures like part of p.229-230, Figures 13.44 and 13.45)
(2)
m
The conditions for stationary waves to occurs are given by L 
2
Where L is the length of wire and m = 1, 2, 3, …
(1)
(b) Appropriate experiment and description
(1+1
4.
(a) consecutive horizontal dark and bright bands are observed. A is dark, B is
bright and C is dark.
(1)
(b) The film is very thin at A and the optical path difference is almost zero. (1)
Since there is a phase change of  occurs at one reflected ray. Thus destructive
interference occurs and a dark band is observed.
(1)
(c) C is the second dark band. So the optical path difference at C must be .
Thus,
(where d is thickness of film,
2n' d  
n’ is the refractive index of film)
(1)

d
 0.227 m
(1)
2n '
(d) The thickness of the film does not increase uniformly down the film. (1)
(e) The order of dark bands and bright bands will be reversed. The first band will
be bright.
(1)
There is no phase change of  due to reflection and maximum occurs when
optical path difference is 0, , 2, …
(1)
(f)
If white light is used. Bright band will be coloured.
(1)
The top of the band will be violet and the bottom part will be red.
(1)
It is because at difference level, the thickness of film is difference, constructive
interference of different colours will occur.
(1)
(p.220)
Section C
1.
a 

D y
D: distance from slits to screen
(a) For Young’s double-slit experiment,
where
a: slit separation;
y: fringe separation; : wavelength
2
From the data for yellow light,
a  5.50  10 7
 
 2  10 3 m
3
D y 0.275  10
(1)
' D 4.0  10 7
For violet light: fringe separation y' 

 0.20 mm (1)
a
2  10 3
" D 6.0  10 7
(b) For red light: fringe separation y" 

 0.30 mm (1)
a
2  10 3
(c) Let x be the distance of m-th order maximum fringe from the central maximum
x
mD
a
x' 
For violet light,
m'  ' D
a
x" 
For red light,
m" " D
a
Purple fringe formed when a red overlaps with a violet fringe,
x '  x"
i.e.
m' '  m" "
or
m"  ' 2


m' " 3
(1)
Thus the distance where purple fringe formed is:
When m'  3 and
m"  2 ,
x' 
m' ' D 3' D

 6  10  4 m
a
a
When m'  6 and
m"  4 ,
x' 
m' ' D 6' D

 12  10  4 m (1)
a
a
(1)
(d)
central
maximum
0.6
1.2
1.8
violet
fringes
red
fringes
(mm)
(3)
purple
fringes
2. (a)
L
(2)
8th
4th
(b)
displacement
(2)
L
t/s
0
1
2
3
3
4
(c)
3. (a)
(b)
speed =
0.6
 2.4 m s 1
0.25
(1)
f 
1 1
  0.5 s
T 2
(0.5)

v 2. 4

 4.8 m
f 0. 5
(0.5)


3
I  I o cos 2 30  0.42 I o
(1+1)
Intensity emerging from first polarizer 
I' 

Io
2
(1)

Io
2
cos 2 30  0.21 I o
2
4. (a)
(1)
(1)
receiver
(2)
water
image of
transmitter
(1)
Wave traveled to receiver via two paths: (1) and (2).
Ray (2) is the reflected ray from the surface of river.
Rays (1) and (2) interfere. The resultant intensity depends on the path
difference between (1) and (2), which is determined by the height of tide of the
river.
(1)
(b) At high tide, path difference is zero; since there is a phase change of  at the
reflected ray, thus destructive interference occurs. Resultant intensity received
is zero.
(1+1)
(c)
25 m
25 m
O
B
5m
A
AB  25 2  5 2  25.295 m
OAB  50.990 m
(1)
Thus, path difference = OAB  OB = 0.99 m = 3.3  <
Maximum occurs at path difference equals to
Thus the path difference will give 3 maxima.
1
,
2
3
,
2
7

2
5
,
2
(1)
7
,....
2
(1)
(c) No, individual amplitude will not be affected
(1)
since their plances of polarization are at right angles, no interference occurs.
(1)
d sin   m
5. (a) (i)
(1)
10 6 sin   1  589  10 9
(ii)
  36.086
9999 or 10000
(1)
(1)
4
(b) For destructive interference to occur,
path difference between 1st and 5001st  5000  12 
path difference between 5000th and 10000th  5000  12 
st
th
Thus, path difference between 1 and 10000 = 10001
 10001 
(c) d sin '  
  1.0001
 10000 
(1)
(or 10000) (1)
(1)
10 6 sin '  1.0001  589  10 9
'  36.090
'  36.090  36.086  0.004
(d) Let  be the new diffraction angle
d sin   m'
(1)
(1)
10 6 sin   1  589.6  10 9
  36.129
    36.129  36.086  0.043
(1)
(1)
The two lines may be resolved.
Because separation between them is 10 tens the angular width.
(1)
(1)
5