1976-02-26 — Page 19

有體育教

真三第張五第日七廿月正年辰丙磨复 WAH KIU YAT PO

報旧僑華

四期星

日六廿月二年六七九一公年五十六國民蕲中

十二屆校際舞蹈節獲獎者圖片特輯

節

NEW SEE ME 5 DE MESA SEJ

甲級獎 嘉諾撒聖心書院上午校「長紅

乙級獎,被公會兆强小學「玄武湖之春

·NEK RKER (EL)

-中级建造上水石湖公立事校「羽果實」

NIE KH6_5_3

NEK HETKLEJ

中級獎:慕光小學(上午)「翡爵一

乙級 謝秀上午校「滋」

中级共凿人事校「各排」

乙級錄,聲公會仁立小學,

乙級女 芬騰湖州公學下午校; 仙境 J

1976中學會考試題預習專欄

物理科 (十六) 何辉明

PHYSICS.

Solution for laat week

(16)

1a) A virtual image is one formed by a

diverging bean and cannot be obtained on a screen.

(1) A plane mirror produces a virtual erect image of same size as the ob- ject.

(ii) A convex mirror produces a virtual image which is always erect and sma- 1ler than the object. (iii) A concave mirror produces a real

inverted image when the object is further than its focal length f. Between f and 21, the image larger than the object; at 2f, it is same. size: further than 21, the image is smaller. And nearer than f, the image is magnified and erect.. b)(i) Since the image formed is 2 real

image, the lens put at O is a con- verging lens for only converging

form real images. lens can (ii)

Jens

36 cm

screen

B

Assume that the distance between 0 and B is \X cm.

distance of the object (candle) from the lens u=(36-x)cm. distance of the image from the lens v=x on.

Focal length of the lens f-8cm.

From Real-is-positive' convention

1

1

+

1

叫

↑

4

* 8

x =

24 or 12-

In order to have a magnified real

the dis- image formed on the screen, tance between 0 and B should be 24cm (iii) Magnification produced by the

lens

m =

11 =

24

-

36 - 24 (iv) From the result obtained in (ii), the distance between the lens and the screen should be *2cm.

(v) The magnification aroduced

m

= {

12

-

36 12

**

1.

ร

(vi) The image formed is 12cm. from the lens, real, and diminished by ? times.

2. Let the object distance be u

From 'Real-is-positive' convention

T

multifying

ery term by v (Since m

From New Cartesian convention

object on left

น

multifying every tern by

(Since m

12

=

(Noted, the following process is sui

able for both Convention)

From the equation as 蛋

When blotting a graph

straight line of slope. ined.

m against

will be obta-

Using the data as the question before,

a graph is shown as below.

magnification

From the graph. Slope

Hence, slope =

品

slope

f+0.2m-

image distance

Where the positive sign confirms that

the lens is converging.

Question for next week

1a) There is a statement

"Paraffin

oil has a greater refractive index than water".

What information does the above st- atement give with regard to

(i) the relative velocities of light

in paraffin oil and in water. (ii) the math of a ray of light when

passing from water into a layer of paraffin oil which floating on top of it?

b) The bottom of a swimming pool is co-.

vered with a piece of glass of 4.5cm thick. Just below the glass is the bottom made of red and green bricks. The refractive index of water is

3 that of glass is. The depth of

water in the pool is m

and

(i) Calculate the refractive index of

glass relative to water.

(11) What is the apparent position of the bricks measured from the upper surface of the glass when it is vi- ewed in water.

(ii) Determine the apparent depth of

water when observation is above water.

(iv) Would the green and the red bricks

appear at the same level?

Pa) Explain why a 45° glass prism of re-

fractive index 1.5 can be used as a total reflecting prism.

30

b) A ray of light is incident normally

on AB and emerges after refraction at the second side AC. If angle A is

Calculate the angle

·30°

of deviation of the ray by the prism (if μb=1,5).

3a) Explain the difference between

transverse and a longitudinal wave, Give An ex^mple of each.

b) The following figures refer to wave- lengths of radiations of five kinds:-

3.Pm, 0.001cm, 2x 10-5cm,

༥་

3x10 cm.

6x10 cm and

These radiations are: x-rays, infra- red, yellow light, radio-wave and ultra-violet. Allocate the wavelength figures in the following table:-

Kinds of radiation

radio wave

infra-red

ultraviolet

yellow light

X-ray

wavelength.

c). If light velocity is 3x10^m/s find

the frequency of x-ray from the infor- mation of bart (b).

1976中學會考試題預習專欄

數學科 (十五) 葉柏芳

TRADITIONAL MATHEMATICS

Section

9.(i) The nth term Tn of the series

1+3+5+7+-••• is 1+2(n-^)

1.e. Tn 70-1

... Sn = 풀(1+1) = 풀(1+ᄅᄆ에서)

the sum to n terms of the ser- ies is a perfect square.

(ii) Since n'

= 1+3+5•••••+(2n-1) sub. n for 25 and 45, we have 252 = 1+3+5+••••+(2x25-1)

152

and 15

= 1+3+5+•••• +49

= 1+3+5+••••+(?x15-1)

=1+3+5+----+29

252-15231+*3+••••+47+49,

which is the sum of certain con- secutive odd numbers.

Given: AB+CD – BC + DAC

To prove: the inscribed circles of

AABD and ABCD touch BD at the same point. Proof: let the inscribed circle of

AABD touch AB, BD, DA at X,Y,Z respectively, and let the inscr- ibed circle of ABCD touch BD at P...

Since AX and A2 are tangents to circle XYZ.

.*. AX AZ (tangent property) Similarly, BX = BY and DY DZ

AB+BD-DA = AX+BX+BY+DY-DZ-AZ

AB+BD-DA 2BY

Similarly, for the inscribed circle of ABCD BC+BD-CD=2BP

.. 2(BY-BP) (AB+CD)+(BC+DA)." But AB+CDBC+DA

... BY-BP =

te, BY=BP

0

(given)

.e. the circles touch BD at the

same point.

Given: AB, AD are tangents to cire

BDE.

BC BD 40

To find (1) BAD

(ii) the length of AB (iii) the area of shaded portion Solution: (1) BD. DC (given)

ABCD is equilateral A. 1.8. /BCD = T

=

C

/ABC ADC = (tangent

property) /BAD = (27 - — 7-2x = x)°

=

(ii) ABAC

ADAC (SAS)

=

ZACB LACD

*

(corr. (s)

C

AB = PC tan = 10 tan 7

10

=

方

(iii) Area of shaded portion

= sector ABD - ABD

༣

ཏུ

x (

x

※ ᄑ - 글 (19)2 x sin(279)

10

x (+

100

9

1003

12

100(47 - 3,3)

36

12. Let M's speed be x mph.

..'N's speed 4x aph. In 4 hours' time M is (20-4x). miles from O and Nis (64-16x) miles from 0.

At this moment MN-10 miles

•*. (20-4x)2+(64-16x)2=102 1.e. 68x2552x + 1099 = O 1.8. (2x-7)(34x-157) = 0

157 1.e. X =

34 314

or

and 4x = 14 or

314

Since 4 *

17

~

17.

7345>

>64, and N would have passed the junction already,

x = 127 is not acceptable.

34

.. M's speed = 34mph and N's speed

# 14 mph.

time for M to pass the junction =

40.

Chr., and time for N to pass the

=

4 32 花

12hr.

junction = =

.". N will first pass the Junction, and

at that moment N will be (20 –

22

x

i... 4 miles away from the junction.