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10英文中學會考試題預習專欄
堅道英文書院主編
(廿五)
PHYSI6T (25)
物理科
Diverging lansest
Follow the same procedure ze in the case of the converging lens, except that the ray parallel to the axis, instead of passing through the opposite foosl point after refraction, is deverged so as to appear to come from the near focal point. In this case the Image is always virtual, lies to the left of the lens, reduced in size, and is erecte
nmiest
三期星日二廿月四年〇七九一公年九十五圞民華中
(接第七張第二頁)
1,770英文中學會考試題預習專欄
经前英文書院主編
化
學科
(廿五)
113. The following compound contsiu selalo nydrogen:
which may be displaced by addiam mataly
A.
Aоne.
B. I only
C. S, W, Xi
D. S, T
F. X only
Inomerism is exhibited by the molsoniaz Formuli of the compounds
A
Method
where u in the object are rende
vis the image distance and 1020) OMG,
the image is virtual and is therefo negative.
is the focal length and is 20/2.om
1/u - 2/20 - 1/10
1/u - 1/10 + 1/20
* 3/20
Answer: The object should be placed 65 on
front of the mirror
B. Ray Diagram Method
Problem Proosdures
Always dṛaw a ray diagram, if for no other reason thán to oheak the calculation from the spe formula, both for the location and the nature of the image. Numerical problems involving Lenses are but myplication of the formula.
\3/u + 3/v = 1/t
and real de positive.
virtual is nega
D:
15. The following compounds will react with bromina
in the absence of lighti
A
S.-only
B. S, T
C. S, T, U
D. W only.
E. Tonly
16. The following compounds will react with
phosphorous pentachloride
BW only
C..
Z onl
none.
The following compounds will react with concentrated sulphuric acids ATL W.
E.
T only
To form a virtual image with a concave mirror, the object must lie between the focal point and the mirror.n
For an image to be 20 cm to the right of the mirror, the ray approaching the heat end of the image from the principal focus: must have been reflected through the focus, whereupon it must have been parallel to the axis before reflection.
The ray approaching the head end of the image from the centre of curvature must have "experienced no change in direction.
Consequently, by tracing these two rays back to their origin, the head and of the object is located as in the diagram, at approximately six and two-thirds on in front of the concave mirror. The diagram checks the calculations..
2
Solution
Formula Metkod
\\3/u + 1/v = 1/r
here u is the object distance and le 24 cm.
via the image distance
f is the focal length of the lens,
since the lens has a virtual focus,
I should be negative
+ 1/v = -1/8
-178 -3/24
·--(1/6)
The image is virtual, erest,
19 6 cm to the right of the surface,
arication
image distance
6/24
bject distance
INDHGr! The image 19 as lárme as the
-object, or 0.3 cm.
Aspen Méthod
Proces is the name as in 24-1, Omitted
This problem is an application of the single Burface rafraction, and will be explained in naxtsizBuo.
Lave-of" kƏ FASTRAS
Refraction
Light is a wave motion. If the speed of a wave motion is changed, the wavefront may change direction. It bends when travels obliquely from one medium into
Graphical Solutions
Converging Lonses :
Draw a diagram with object to the ler and on the ixia
of the lane. Specify the focus. From the head and of the object, select the ray parallel to the axis and direct it through the opposite focal point after refraction.
3. Pass a ray from the head end of the object directly through the centre of the lens. ray de undeviated by refraction.
This
The point of intersection of these two rays, OP. their produced, after refraction is a reproduction (image) of the point of their origin (object),
Exercise 25
25-1. A candle flame 3 cm high is placed at a
listance of 3 metres from a wall. How far Iru. the wall must a condave mirror be placed in order that it may form an image of the flame
om high on the walls What must be the radiu of curvature of the mirror?.
one conuave, the other convex, of radius of curvature 20 cm. are placed coaxially 40 cm apart, with their polished surfaces facing each other, and an object is placed mid-way between them. Find the position of the image formed by reflection first at the conver, then at the concave surface.
25-2. Two mirrors, of
25-3. A slab of glass of thickness 5 om ande
refractive index 1.5 is held a few cantina trea in front of a concave mirror of radius of curvature 40 om,, the faces of the slab being perpendiuclar to the principal axis of the mirror, How far from the mirror must a small
object be placed if its reflected image coincides with the object?"
25-4. Find the position of the image formed by
convex lens of 20 cm focal length, when an object is placed (a) 30 on from it, (b) 15. from 11.
Given that the refractive index for air-giana 11,60 and for air-water is 1.33, calculate the critical angle for water-glassy
another medium of different optical density, Kefraction. is the change i direction of travel of light energy when it crosses the boundary between two medium.”
LAN 1. The refracted ray lies in the same plane as
the incident ray and the normal at the point
f incidence.
LAW 2 The ratio of the sine of the angle of incidence
to the sine, of the angle of refraction is a constant for light of a given frequency passing from one given medium to another. 2015.
s also known as Snell's bawi
known.
Keractive Index
Refractive ilder is a term used to describe aptroal density, the ability of a transparent medium to cause refraction by changing the speed of light.
speed of light in medium Kelativa refractive index.
speed of light in medium
sin 01
Bin
Examples or Retraction
(1) Real and apparent depth.
E
**Rays of light incident from a substance' which is optically denser than air will be refracted away from the normal in air. Thus atick appears to be bent and an object appears t be more shallow. than it actually be.
(2) Refraction through priem.
Light is always deviated when it passes through a prism. White light is found to emerge coloured (dispersion)
(3) Object: view through a transparent sheet
The distal surface of a glass sheat appears. to be a little nearer the eye then it really is, and the extent of this shift may be calculated as
real thicknes apparent thioknese
refractive index -
and the object viewed through, the glass appear: to be brought nearer the eves by an amount equal to this shift.
Total Internal Reflection
When light passes from a genser into a rarer. medium, it is evident that the retracted ray is bent away from the normal but the angle of refraction cannot exceed 90 degrees. for at thin angle there
C. W only
D. I only
18, Which of the following compounds are read
oxidized by ohemical reagents s
W only
B. I only
CU, H, X
D. 8, 1, W,
19. Which of the following compounds po
carbon-carbon triple bönd?
S
B.
U
D. V
B. W
Which of the following compound does the analytical data correspond to?...
3.7 g. of the compound yielded 6.6 gá, carbon dioxide and 2.7 ga. of water vapour.
D. X
SOLUTIONS
B. L, N,
11.
C. N only
D. Q only.
12. 13.
#14.ME.
19. B. 5, T
E. O, P,
6. C. P
8.
16
·27, E
A 3
20.
corresponds a certain angle of incidenos. i, given
by Snell's Law
M, ein 1 - Mzein r
where 1 19
Bin 1 = μl g/ M, x sin r
My! My x min 90o
110 30 Mg Mg
-M zl Ma
as the critical angle of incidence
t is found that when the angle of incidence exceeds the critical angle, the result is total reflection bɛok into the initial medium, Actually there is partial reflection corresponding to angles of incidence less than the critical angle, but for all angles greater than 1 the reflection in complete
Thin Lenses in air
The method used in spherical mirrors can be employed in lenses. There are two spherical surfaces, The image formed by the first one becomes the object for the second one. It is more convenient, in the osse of thin lenses having the two surfaces rather close together with the same first and third media to solve the problem algebraically for a sinela formula
+ 1/4 - 1/%
where u is the object distance.
vis the image distance
fis the focal length
solve the problem of each surface separately, Similarly, the formulas for magnification at endi surface, when combined, yield,
total magnification.
tal maga
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