CITY
真四弟張四第日三十月四年子壬屣窥
WAH KIU YAT PO
郭日僑罩
英中會考物理科答案(三)
日期星 日四月六年二七九一股公年一十六國民蠻中 VULE
堅道英文書院·
*.. 2nd image (i.e, the image shown in
the figure)
英中會考數學科答案(三)
OF 236" 191
4--00
堅道苙文書院、
AN 106 300 DA MU NA MOND
Favaios (33)
1
1.
WAVE MOTION, LIGHT AND SOUND
.f
1.
Paper Threa
15
+
-
Mathematics
(Lesson (32)
Suggested solution to mathematics A,
1)(a) tun-
and 189" L-XL 100`
lies in the 4th quadrant.
(6)
SECTION C
4. (a)(1) explain briefly, with the help
of a ray diagram, how you would letermine the focal length of
a converging lens using a plane mirror and one pin.
(ii) A converging lens of focal
length: 10 cm is used to produce a virtual image three times the sise of an object. Name the - sign convention you use, state the correct corresponding len formula, then find the distance of the object from the lens. (b) arrange the following radiation
in the order of increasing wave¬" lenght 1-
A radio wave, an X-ray, yellow light, an infra-red ray and an ultra-violet ray.
Which of the above kinds of elec romagnetic rays could be detected
(i) using your eye,
ii
using a fluorescent aorsan, iii) using a thermopile and a
galvanometer,
(v) through a cardboard screen? (c) Plane waves, produced in a ripple
tank, appeared to be stationary when viewed at a`lowest frequency of 3 flashes of light per second, and had the dimensions indicated' by the diagram below A
Since the linear magnification is 3
V 3u
1
1
+
Jaing
-
-
V
પ
nere
u
บ
(New Ceteeian Convention}
object distance
image distance
f - focal length
_1 1
-Ju
11-
-u
2
f
1
10
20 3
6323
0.
(b) In the order of increasing wave
Length :-
X-ray, ultra-violet ray, yellow light, infra-red, radio wave
5
sin+ è cog/
7 COB-3 sin
5(-2)+0(2)
10).
-4 +3.0
4.22.
·
(-)
-0.4
33
(b) OAB досы
Perimeter of quadrilateral OABC
* 2x + 2AB where r ft. is the
radius of the carcie.
Perimeter
of sector OAC
-2r + are ac
C
(2 + 2AB) 12r+ are Ac) - 1C
2AB
aro AC - 10
σε αυτούς - 1 x 60 x I
180
V
2x cot60"
+ X
10
Sarea
Area of A OF. OR X
·
1. 04 × 03 sìr
Similarly area of ▲ OPQ
× OF × 03 x sin (180 – £
J
1 x CP x 02 sinh
x03 I OK I*** -
Area of A OSK =
Area of 4 CPS =
OP x 08'x win G
* area of the quadrilateral·
-
sine Lon * ^2 + OP_*°02 + C8 x ©r
+ OP x 00]
* sine fon(or + OP} +
OS (OR + OP? |
sin jo? x PROS I PR
-
10
L
ein PR x (02 +_os)
M
sixx PRI Q5 ·í
Area of the given quadrilateral.
in 57°44' xx OS
2 x 14cm x 10cm sir37 44%
- $3.19 sr..
(3) BC cat60TM
50 x cotbú 28.87/
Calculate
-(1). the wavelength of the plane wave, (ii) the velocity of the plane wave.
ab
act
ge!
converging
lena
plane mirror
The focal lens of a converging leng can be determined using the above arrar ment. ▲ pin sewing as the object in placed on one side of the lens and the plane airror on the other, facing the 100s.
When the object is situated at
the foous, rays amitted from any on pont on the object become parallel after passing through the lens. They are ref- lected by the plane mirror, and return to the lens as parallel rays. These will
In
be converged to form a real image at a point in a plane passing through the focue, nomal to the principle axis. particular, raya from a point on tho principle axis will be reflected along its own path (as shown). Hence an' image will be formed at the same position as the object. The adjustment can be done by the method of no parallez. When the object and image are ooincident, the focal length of the lene's equal to the distance between the pin and the lens.
The same can be realized using the lee formula.
let image (without plane mirror).
น
■ -f
1
-
+
(New Catesian Conven➡
11
tion).
V
+
I
V - Cic.
lat image is at infinity
This image is reflected back by the
plane mirror
(1) Yellow light.
(ii) X-ray, (ultra-violet).
(111) infra-red
(iv) yellow light, ultra-violet, X-ray,
* (v) X-ray, radio-wave.
Author's note: (V) In this part of th question the word 'detect' in debat- able. The question askes which of the electromagnetic rays can be dete- ated through a carboard screen but nothing is said about the sensitivity of the detecting instrument, nor the thickness of the cardboard screen. Further, it is not specified whether the detection process is direct or indirect,
Undoubtly X-ray can penetrate through a piece of cardboard of reasonable thickness. Radio wave can also penet- rate through cardboard as it is gener- ally known that radio receives can perform normal functions even in an enclosed cardboard box.
Infra-red can be detected indire- otly by its heating effect on the card board provided that the piece of card- board is not too thick. Light can also be detected if ita intensity is high enough and the cardboard soreep is thin enough, Similarly ultra
violet rays.
In general, electromagnetio waves can penetrate through a card board screen with some loas being introduced. That light and othera cannot pass through indicate that they cannot be fully explained by the
There electromagnetic wave theory.
is the well known corpuscular theory which explains many of the other phen-
onema
of light.
The present standpoint of physic- Late, in the face of apparantly cont- radictory experiments, is to accept the fact that light appears to be dualistic in nature. The phenomena of light propagation may best be explained by the electrmagnetic wave theory, while the interaction of light with matter, in the processes of emission and absorption, is a corpuscular phenomenon.
Applying the corpuscular theory, .ight is taken as stream of photous or energy packets and these cannot pas. through cardboard screen. Similarly infra-red and ultra-violet. X-ray can pass through because its energy is higher. Radio wave is a true electr. magnetic wave and can therefore pasa through cardboard as Maxwell's equations
is valid in all media.
(2 cot60" - ) - 15 r(2 x 1
)=.10
3
10
(23-1)
3
30
- T
30.
33.15
0.322
The radius of the circle is 33.158.
(2)(a) 8.ain25-19 cos2 - 6 sin Scoe Z Dividing the equation throughout by
(assuming cost 0)
2
8 sin2.
• sin cos E
20
GUA2 -
C 50 cot30°
- 86.61
D.
8 tan2
- 9.-é tan
8 tan- & tan-9.0. (4 tan l+3)(2 tan ( -3)
-
- 3
..tant-
Cr tan- -
4
From tank
From tan b
1438*.
T
323°81
5 - 56°191
BD2. 302 + DC2 - 2BC x DC Co#60°
28.872 86.612
2 x 287 x 86.61 x Coa60
- 233.4 + 7502 2500
- 9835.4
*BD:76.40
* The distance between the ends of the
shadows of the flag-pole at these two times is 76.40 ft.
(4) Given: ABCD is a quadrilateral in
which AB CD. AC and BD intersect at X. 〔未完。轉第四張第三頁)
At
present position of 1 must be taken up by a crest, either. 2, 3, 4, or so on. If the frequency of sampling is rapid, the crest will be those near to 1, the Eighest frequency of sampling, the present position of 1 must be taken up by 2. (The lowest frequency' has no significance).
If the highest frequency of sampling or viewing is ở per second-'the beriod
of the Wave is
sec,
,)*Authore note: The word 'lowest' in
this question should be, 'Highest. instead,
12
(1) From the figure. X-
-300
4
---
A wave length
the frequency of observation
4
1 2 cm.
AA
(ii) To find the frequency, consider
*
1
direction
of propagTM ation of
wave
When the wave appeare to be etation- gry, at the next viewing instant. the
14
1
B Hz
T
velocity
8 x 3
24 cm/sec.
Modificetion of the solution Question 2.
2.(a)(ii) a small bob is required to
simulate a point partiole
J
▲ spherical one is preferred 40 it is symmetrical about all axis passing through the centre and therefore air resistance effecte is same on all sides. (vi) The no. of ovcillations to be
timed is arbitary, depanding - on the degres of accuracy.
hey greater the number timed the higher the accuracy. To avoid greater errors the total length of time measured should be large compared with the least count of the stop.watch? and the reaction time of the – experiment.
For this particular one, 10 to 20 oscillations will be appropiate.
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