LIGHT - yuhuaphysics.com Fizik Tg4 B5 2015(FSY4p) n.pdf · • Melukis gambar rajah sinar untuk...

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MODUL • Fizik TINGKATAN 4

•Menghuraikanciri-ciri imejyangterhasilmelaluipantulancahayaDescribe the characteristics of the image formed by reflection of light

•Menyatakanhukum-hukumpantulanState the laws of reflection of light

•Melukisgambarrajahsinaruntukmenunjukkankedudukandanciri-ciri imejyangterbentukolehDraw ray diagrams to show the position and characteristics of the image formed by a

(i) cerminsatahplane mirror

(ii) cermincembungconvex mirror

(iii) cermincekungconcave mirror

•MenghuraikanaplikasipantulancahayaDescribe applications of reflection of light

•MenyelesaikanmasalahyangmelibatkanpantulancahayaSolve problems involving reflection of light

•MembinaalatberdasarkanaplikasipantulancahayaConstruct a device based on the application of reflection of light

•MenerangkanpembiasancahayaExplain refraction of light

•Mendefinisi indekspembiasansebagaiDefine refractive index as

n=sin isinr

•MenentukanindekspembiasansebuahblokkacaataublokperspekDetermine the refractive index of a glass or perspex block

•Menyatakanindekspembiasan,State the refractive index,

n=kelajuancahayadalamvakumkelajuancahayadalammedium /

speed of light in a vacuumspeed of light in a medium

•MenghuraikanfenomenayangdisebabkanpembiasanDescribe phenomena due to refraction

•MenyelesaikanmasalahyangmelibatkanpembiasancahayaSolve problems involving the refraction of light

CAHAYALIGHT5

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5.1 MEMAHAMI PANTULAN CAHAYA UNDERSTANDING REFLECTION OF LIGHT

5.2 MEMAHAMI PEMBIASAN CAHAYA UNDERSTANDING REFRACTION OF LIGHT

Fizik Tg4 B5 2015(FSY4p).indd 203 10/20/15 2:17 PM

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© Nilam Publication Sdn. Bhd. 204


•MenerangkanpantulandalampenuhcahayaExplain total internal reflection of light

•Mendefinisisudutgenting,cDefine critical angle, c

•Menghubungkaitsudutgentingdenganindekspembiasan,n = 1sinc

Relate the critical angle to the refractive index, n = 1

sin c

•MenghuraikanfenomenasemulajadiberkaitanpantulandalampenuhDescribe natural phenomenon involving total internal reflection

•MenghuraikanaplikasipantulandalampenuhDescribe applications of total internal reflection

•MenyelesaikanmasalahberkaitanpantulandalampenuhSolve problems involving total internal reflection

•MenerangkantitikfokusdanpanjangfokusExplain focal point and focal length

•MenentukantitikfokusdanpanjangfokusbagikantacembungDetermine the focal point and focal length of a convex lens

•MenentukantitikfokusdanpanjangfokuskantacekungDetermine the focal point and focal length of a concave lens

•Melukisrajahsinaruntukmenunjukkankedudukandanciri-ciri imejyangterhasilolehDraw ray diagrams to show the positions and characteristics of the images formed by a

(i) kantacembung convex lens (ii) kantacekung concave lens

•Definisipembesaranm= vu

Define magnification as m = vu

•Menghubungkaitpanjangfokus,f,denganjarakobjek,u,danjarakimej,v, iaitu, 1f= 1

u+ 1

vRelate focal length, f, to the object distance, u, and image distance, v, i.e. 1

f = 1

u + 1

v•Menghuraikankegunaankantadalamalatoptikdenganmenggunakanrajahsinar

Describe with the aid of ray diagrams, the use of lenses in optical devices

•Membinaalat-alatoptikyangmenggunakankantaConstruct optical devices that use lenses

•MenyelesaikanmasalahberkaitankantaSolve problems involving lenses

5.3 MEMAHAMI PANTULAN DALAM PENUH CAHAYA UNDERSTANDING TOTAL INTERNAL REFLECTION LIGHT

5.4 MEMAHAMI KANTA UNDERSTANDING LENSES


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Hukum pantulan:The laws of reflection: 1 Sinar tuju, sinar pantulan dan garis normal berada pada satah yang sama.

The incident ray, the reflected ray and the normal to the surface all lie in the same plane. 2 Sudut tuju, i, sama dengan sudut pantulan, r.

The angle of incidence, i, is equal to the angle of reflection, r.

Sudut tuju = sudut pantulanAngle of incidence = angle of reflection

Cara melukis/Drawing method1. Kedudukan imej

ditentukan.The position of its image is located.

2. Jarak imej sama dengan jarak objek, u = v.The image distance is equal to the object distance, u = v.

3. Saiz imej sama dengan saiz objek.The image size is same as the object size.

4. Garis yang menyambung objek dan imej adalah bersudut tegak dengan cermin.The line joining the object and the image is perpendicular to the mirror.

5. Sinar pantulan dipanjangkan di belakang cermin sehingga bertemu di I (imej).The reflected rays are extrapolated to the back of the mirror until they meet at the image, I.

6. Sinar di belakang cermin sebenarnya tidak wujud. Sinar ini diwakili oleh garis putus-putus.The rays behind the mirror do not exist. They are represented by dotted lines.

7. Garis berterusan dari cermin ke mata menunjukkan sinar pantulan.The continuous lines from the mirror to the eye indicate the reflected rays.

• Sudut tuju, i, ialah sudut antara normal dan sinar tuju

The angle of incidence, i, is the angle between the normal

and the incident ray

• Sudut pantulan, r ialah sudut antara normal dan

sinar pantulan

The angle of reflection, r, is the angle between the normal

and the reflected ray

• Garis normal ialah garis yang berserenjang dengan permukaan cermin

The normal is the line that is perpendicular to the mirror

Cermin satahPlane mirror

Ciri-ciri imej cermin satahCharacteristics of the image in a plane mirror

Saiz/Size : Saiz sama/Same size

Keadaan/Condition : Tegak/Upright

Gambaran/Appearance : Maya, songsang sisi/Virtual, laterally inverted

Kedudukan/Location : Jarak objek (u) = Jarak imej (v)

Object distance (u) = Image distance (v)

Rajah sinar: imej yang terhasil oleh cermin satah Ray diagram: The image formed by a plane mirror

NormalNormal

Sinar tujuIncident ray

Sinar pantulanRe�ected ray

i

r

uO

I

v Imej/ImageObjek/Object

i

r

MataEye

u = jarak objeku = object distance

v = jarak imejv = image distance

MEMAHAMI PANTULAN CAHAYAUNDERSTANDING REFLECTION OF LIGHT5.1


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Aktiviti/ActivityGambar rajah menunjukkan seorang budak perempuan melihat cermin satah. Tuliskan imej yang dilihat oleh budak itu.The diagram shows a girl looking at a plane mirror. Write the image seen by the girl.

Istilah pada cerminCommon terminology of mirror

Pusat lengkungan, CCentre of curvature, C

Pusat sfera di mana cermin melengkung terbentuk.The geometric centre of a sphere of which the concave or convex mirror is a part.

Kutub cermin, PPole of mirror, P

Titik tengah pada cermin bulat.The centre point on the spherical mirror.

Paksi utamaPrincipal axis

Garis lurus yang melalui pusat lengkungan bersudut tegak (90°) dengan satah cermin.Straight line which passes through the optical centre at right angles (90°) to the plane of the mirror.

Titik fokus, FFocal point, F

Titik pada paksi utama yang mana sinar tuju yang selari dengan paksi utama:Is the point on the principal axis to which incident rays of light travelling parallel to the principal axis:a. mencapah pada titik

itu selepas dipantulkan keluar daripada cermin cembung.

diverge after reflected outwards on a convex mirror.

b. menumpu pada titik itu selepas dipantulkan ke dalam cermin cekung.

converge to after reflected inwards to a concave mirror.

Jarak objek, uObject distance, u

Jarak antara objek dengan titik P.Distance between object and point P.

Jarak imej, vImage distance, v

Jarak antara imej dengan titik P.Distance between image and point P.

Cermin cekungConcave mirror

R : Jejari lengkung / Radius of curvature

F : Titik fokus / Focal point

f : Panjang fokus = ½ R / Focal length = ½ R

Cermin cekung akan memantulkan cahaya

yang selari ke titik F

A concave mirror will reflect the parallel

light to point F


C F

R

P


f

CF

P

Pusat lengkungan, C

Centre of curvature, C

Titik focus, F

Focal point, F

W, V, U


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Nota penting/Important notes:

Fahamilah konsep ini.Understand this concept.

ObjekObject


C

X

A

Y

T

Bagi cermin cekung, sebarang sinar cahaya yang melintasi pusat lengkungan, C, akan terpantul kembali di sepanjang sinar tuju, kenapa?For a concave mirror, any ray that passes through the centre of curvature, C, will be reflected along its original path (the incident ray). Why is this so?

• PadaX, TXY ialah tangen kepada bulatan.

At X, TXY is the tangent to the circle.

• ∠CXT = 90° (i)

• ∴ Dari (i) di atas, sinar tuju, ACX, akan menghasilkan sinar terpantul XCA.

∴ From (i) above, the incident ray, ACX, will produce the reflected ray XCA.


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Cermin cekungConcave mirror Sinar yang selari dengan paksi utama menumpu pada titik yang dipanggil titik fokus (F) pada paksi utama. Ray parallel to the principal axis converges to a point, called the focal point, on the principal axis.PF = panjang fokus, f focal length, f = jarak antara kutub cermin

(P) ke titik fokus (F) distance between pole of

mirror (P) and focal point (F)

Contoh, f = +10 cmExample, f = +10 cm

Peraturan/Rule 1:Cahaya selari dengan paksi utama akan terpantul ke F

Light parallel to principal axis will reflect to F

Sinar/Ray 1

C FObjekObject

P


Peraturan/Rule 2:

Cahaya yang melintasi pusat lengkungan akan terpantul kembali sepanjang cahaya tuju (garis lurus)

Light passing through the centre of curvature will reflect back along the incident light (straight line)

Sinar/Ray 2

Sinar/Ray 3

ObjekObject

C F P


Peraturan/Rule 3:

Cahaya yang menuju ke P akan dipantulkan dan

mematuhi hukum pantulan

Light passing through to P will reflect and obey the laws of reflection

ObjekObject


C F P

Sinar/Ray 4

Peraturan/Rule 4:Cahaya yang melintasi F akan dipantul. Sinar cahaya yang terpantul

adalah selari dengan paksi utamaLight passing through F will be reflected. The reflected ray will be parallel to the principal axis

ObjekObject

C F P


Peraturan rajah sinar bagi cermin cekung Rules of ray diagram for concave mirror


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C

O

IF P


C F

O

P

I


(ii) Jarak objek: Antara C dan F (f < u < 2f) Object distance: Between C and F (f < u < 2f)

(iii) Jarak objek: Antara F dan P (u < f) Object distance: Between F and P (u < f)

Ciri-ciri imej Characteristics of image

Songsang/Inverted

Nyata/Real

Diperbesarkan/Magnified


Tegak/Upright

Maya/Virtual


Kesimpulan:Conclusion:

Ciri-ciri imej cermin cekung bergantung pada jarak objek (u)

The characteristics of the image in a concave mirror depends on the object distance (u)

ObjekObject

ImejImage

C F P


(i) Jarak objek: Pada C (u = 2f) Object distance: At C (u = 2f)


Songsang/Inverted

Nyata/Real

Sama saiz dengan objek

Same size as the object

Cermin cekung boleh digunakan:Concave mirrors can be used: 1. Doktor gigi untuk

memeriksa gigi pesakit By dentists to examine

the teeth of a patient.

2. Cermin mencukur As shaving mirrors

3. Cermin solek Make-up mirrors

Ciri-ciri imej pada cermin cekungCharacteristics of image for concave mirror


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Cermin cembungConvex mirror

CFP


• Hanyamenghasilkanimejmaya.

Only produce a virtual image.• Menghasilkanmedan

penglihatan yang lebih luas. Provides a wide field of view.• Imejyangtegak. Image always upright.• Saizimejakanmengecil. Size of image will be reduced.

Sinar yang selari dengan paksi utama seolah-olah terpesong dari satu titik di belakang cermin. Rays parallel to the principal axis appear to diverge from a point behind the mirror.

Titik ini yang terletak pada paksi utama dipanggil titik fokus.This point which lies on the principal axis is called the focal point.

PF = panjang fokus, fPF = focal length, f = jarak antara titik

fokus dan kutub cermin (P)

= Distance between focal point and the pole of the mirror (P)

Panjang fokus, f Focal length, f= FP

= 12

× jejari lengkungan radius of curvature

= 12

CP

Bagi sinar cahaya yang selari dengan paksi utama,

cermin cembung akan memantulkan cahaya dari

F selepas pantulan

For light rays parallel to the principal axis, a convex

mirror will diverge light from F upon reflection.

Peraturan/Rule 1:Sinar cahaya selari dengan paksi utama akan memantul dari F

Light rays parallel to the principal axis will reflect from F

Peraturan/Rule 2:

Sinar cahaya yang melintasi pusat lengkungan (C) akan dipantulkan semula sepanjang sinar cahaya tuju (garis lurus). Anda harus dapat menerangkan fenomena ini sekarang.

Light rays passing through the centre of curvature, C, is reflected back along the incident light rays (straight line). Now you should be able to explain why.

Sinar/Ray 1

Sinar/Ray 3 Sinar/Ray 4

Sinar/Ray 2

Peraturan/Rule 3:Sinar cahaya yang mengena P akan terpantul dan mematuhi hukum pantulan

Light going through to P will reflect and obey

the laws of reflection

Peraturan/Rule 4:Sinar cahaya yang melintasi F yang terpantul adalah selari dengan paksi utamaLight rays passing through to F which reflect are parallel to the principal axis

CFPObjekObject


ObjekObject

CFP


ObjekObject

CFP


ObjekObject

CFP


Peraturan untuk rajah sinar bagi cermin cembung Rules for ray diagrams for convex mirror


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MataEye



Medan penglihatan sempitNarrow f ield of view Medan penglihatan luas

Wide f ield of view

Mata/Eye

Ciri-ciri imej cermin cembung adalah 'DUV'Characteristics of the image for convex mirrors is 'DUV'

D = Mengecil/Diminished

U = Tegak/Upright

V = Maya/Virtual

Apabila objek mendekati cermin, saiz imej akan membesarWhen the object is nearer to the mirror, the size of image increases

Sebuah lampu kecil diletakkan pada titik tumpuan cermin untuk menghasilkan sinar yang selariA small lamp is placed atthe focal point of the mirror to produce parallel rays

CFF PObjekObject

ImejImage



Mengecil/Diminished

Tegak/Upright

Maya/Virtual


Mengecil/Diminished

Tegak/Upright

Maya/Virtual

CFF PObjekObject

ImejImage


1 Cermin cembung: Medan penglihatan luas berbanding cermin satahConvex mirror: Has a wider field of view than plane mirror

U > f

U < f

2 Cermin parabola/Parabolic mirror

Boleh digunakan sebagai/Can be used as 1 Cermin memandu/Driving mirrors 2 Cermin sekuriti kedai/Shop security mirrors 3 Cermin selekoh/Blind corner mirrors

Boleh digunakan dalam lampu suluh dan lampu kereta sebagai pemantul

Can be used in torches and car headlamps as reflectors

F

Cermin parabolaParabolic mirror

Mentol pada titik fokusBulb at focal point

Sinaran selari/Parallel raysKelebihan/Advantages:

Boleh mengekalkan keamatan seragam untuk

jarak yang lebih besar

Can maintain a uniform intensity for a greater

distance

Rajah sinar untuk menghuraikan imej bagi cermin cembung Ray diagrams to describe the image by a convex mirror

Aplikasi pantulan cahaya Applications of reflection of light


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Fakta Periskop:Facts of a periscope:1. Dua cermin mestilah

selari dan menghadap satu sama lain

The two mirrors must be parallel and face each other

2. Kedua-dua cermin diletakkan pada sudut 45° dengan sinar cahaya

Both mirrors are set at an angle of 45° to the path of the light rays

Pembinaan alat yang berdasarkan aplikasi pantulan cahaya The construction of devices which are based on the applications of reflection of light

Periskop/Periscope

ObjekObject

CerminMirror

CerminMirror

MataEye


Tegak, sama saiz dan maya

Upright, same size and virtual

Pembiasan ialah/Refraction is:Pembengkokan sinar cahaya di sempadan apabila ia bergerak dari satu medium

ke medium lain yang mempunyai ketumpatan yang berbeza./the bending of a light

ray at the boundary as it travels from one medium to another medium of different densities.

Hukum pembiasan:Laws of refraction:(a) sinar tuju, sinar biasan

dan normal berada dalam satah yang sama. the incident ray, the

refracted ray and the normal all lie in the same plane.

(b) nisbah sin i kepada sin r adalah suatu pemalar,

the ratio of sin i to sin r is a constant,sin isin r

= pemalar/constant

Pembiasan cahaya Refraction of light

Tiada pembiasan No refraction

Sudut tujuAngle of incidence= 0°

• Sinarmerambatdariudarakedalamkaca.

The ray travels from air into glass.• Sinarmerambatdarimedium

kurang tumpat ke medium

lebih tumpat .

The ray travels from a less dense

medium to a denser medium.

• Suduttuju,i, lebih besar daripada sudut pantulan, r.

The angle of incidence, i, is greater than the angle of refraction, r.

• Dalamkaca,sinarterbiasmendekati

normal kerana halaju berkurang . In glass, the ray is bent towards the

normal because the velocity decreases .

• Sinar merambat dari kaca ke dalamudara.

The ray travels from glass into air.• Sinarmerambatdarimediumyang

lebih tumpat ke medium

kurang tumpat .

The ray travels from a denser medium

to a less dense medium.

• Sudutbiasan,r, lebih besar daripada sudut tuju, i.

The angle of refraction, r, is greater than angle of incidence, i.

• Sinarterbias menjauhi dari normal kerana halaju bertambah .

The ray is bent away from the

normal because its velocity increases .

UdaraAir

UdaraAir

iUdara/Air

KacaGlass

r

NormalNormal

NormalNormal

i

r

KacaGlass

KacaGlass

UdaraAir

UdaraAir

iUdara/Air

KacaGlass

r

NormalNormal

NormalNormal

i

r

KacaGlass

KacaGlass

UdaraAir

UdaraAir

iUdara/Air

KacaGlass

r

NormalNormal

NormalNormal

i

r

KacaGlass

KacaGlass

MEMAHAMI PEMBIASAN CAHAYAUNDERSTANDING REFRACTION OF LIGHT5.2


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r

i

iNormalNormal

VakumVacuum

KacaGlass

r

• Sinarcahayamerambatdarimedium kurang tumpat ke

medium lebih tumpat;

A light ray travels from less dense medium to denser medium;

• Terbias mendekati normal/Refracts towards the normal;

• Suduttuju,i lebih besar daripada sudut pembiasan, r

Angle of incidence, i is bigger than angle of refraction, r

• Sinarcahayamerambatdarimedium lebih tumpat ke medium kurang tumpat;

A light ray travels from denser medium to less dense medium;

• Terbias menjauhi normal/Refracts away from normal;

• Suduttuju,i lebih kecil daripada sudut pembiasan, r;

Angle of incidence, i is smaller than angle of refraction, r;

Indeks pembiasan dan halaju cahaya Refractive index and speed of light

MengikutHukumSnell,indekspembiasan,n = sin i (vakum atau udara)

sin r (medium)

According to Snell's Law, refractive index, n = sin i (vacuum or air)

sin r (medium)

BahanSubstance

Indek pembiasanRefractive index

n

Pepejal/SolidIntanDiamondKacaGlassAisIce

2.42

1.46

1.31

Cecair/LiquidAirWaterAlkoholAlcoholBenzenaBenzene

1.33

1.36

1.50

Gas/GasUdaraAirKarbon dioksidaCarbon dioxide

1.000293

1.00045

1 Tentukan indeks pembiasan kaca.Determine the refractive index of the glass.

Penyelesaian/Solution:

n = sin 60°sin 35°

= 1.51

2 Halaju cahaya melintasi medium ialah 1.6 × 108 m s-1. Kirakan indeks pembiasan medium tersebut. The speed of light passing through a medium is 1.6 × 108 m s-1. Calculate the refractive index of the medium.

Penyelesaian/Solution:Halajudalamvakum/Speed in vacuum, c = 3.0 × 108 m s-1

Halajudalammedium/Speed in medium, v = 1.6 × 108 m s-1

n = Halajucahayadiudara,c

Halajucahayadalammedium,v = Speed of light in air, c

Speed of light in medium, v

= 3.0 × 108 m s-1

1.6 × 108 m s-1 = 1.88

35°

60°

KacaGlass

UdaraAir

Latihan/Exercises


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3 Berapakah halaju cahaya dalam medium dengan indeks pembiasan 2.2?What is the speed of light in a medium with a refractive index of 2.2?

Penyelesaian/Solution:n = 2.2,c=Halajudalamvakum/Speed in vacuum, c = 3.0 × 108 m s-1

n = Halajucahayadiudara,c

Halajucahayadalammedium,v = Speed of light in air, c

Speed of light in medium, v = 3.0 × 108 m s-1

v = 2.2

v = 3.0 × 108 m s-1

2.2 = 1.36 × 108 m s-1

Objek kelihatan bengkok dalam airObjects appear to bend in water

Dalam ketara, d, ialah jarak imej dari permukaan air.The apparent depth, d, is the distance of the image from the surface of the water.

Dalam nyata, D, ialah jarak objek dari permukaan air.Real depth, D, is the distance of the object from the surface of the water.

Fenomena yang melibatkan pembiasan cahaya Phenomena due to refraction of light

1 Sinar cahaya dari objek dibiaskan di sempadan air-udara dan membengkok menjauhi normal.

The light ray from the object is refracted at the water-air boundary and bends away from the normal.

2 Apabila sinar cahaya masuk ke dalam mata, ia seolah-olah datang dari imej, I, yang lebih tinggi daripada kedudukan objek.

When the light ray reaches the eye, it appears to come from image, I, which is higher up.

3 Oleh itu objek seolah-olah berada di kawasan yang lebih cetek .

The object therefore appears to be shallower .

UdaraAir

NormalNormal

NormalNormal

AirWater

Imej, IImage, I

ObjekObject

Dalam ketara, dApparent depth, d

Dalam nyata, DReal depth, D


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n = Dd

,

di mana n ialah indeks biasan mediumwhere n is the refractive index of the medium

Dalam nyata dan dalam ketaraReal depth and apparent depth

1 Sinar cahaya yang merambat dari ikan terbias menjauhi normal apabila ia merambat dari air

(medium lebih tumpat) ke udara (medium kurang tumpat).

Rays of light coming from the fish are bent away from the normal as they passing through from water

( denser medium) to air ( less dense medium).

2 Apabila sinar cahaya sampai ke mata, ia seolah-olah datang dari ikan maya yang berada di atas ikan sebenar.

When the light rays reach the eye, they appear to come from a virtual fish which is above the real fish.

3 Dalam ketara ialah jarak dari permukaan air ke imej, I .

The apparent depth is the distance from the surface of the water to the image, I .

4 Dalam sebenar ialah jarak dari permukaan air ke objek, O .

The real depth is the distance from the surface of the water to the object, O .

5 Hubunganindeksbiasan,n, dalam nyata dan dalam ketara ialah:The relationship between of refractive index, n, to real depth and apparent depth is:

n = Dalam nyataDalam ketara

n = Real depth

Apparent depth

6 Dari rajah di atas, ikan nampaknya muncul di bahagian bawah kolam pada kedudukan 50 cm dari permukaan air. Berapakah kedalaman nyata ikan itu?From the diagram above, the fish at the bottom of a pond appears to be 50 cm from the water surface. What is the real depth of the fish?

(PembiasanIndeksair/Refractive index of water = 1.33)

Penyelesaian/Solution:

n = Dalam nyataDalam ketara n =

Real depthApparent depth

1.33 = Dalam nyata

0.5 m 1.33 =

Real depth

0.5 m

Dalam nyata = 1.33 × 0.5 m Real depth = 1.33 × 0.5 m = 0.67 m = 0.67 m

O

UdaraAir

NormalNormal

AirWaterDalam ketara, d

Apparentdepth, d

Dalam nyata, DReal depth, D

I


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7 Dalam rajah di sebelah, lukiskan rajah sinar dari titik Z ke mata untuk menunjukkan objek kelihatan lebih tinggi.In the diagram on the right, draw a ray diagram from point Z to the eye to show how the object appears higher up.

Pantulan dalam penuh cahaya ialahTotal internal reflection of light is

Fenomena yang berlaku hanya apabila cahaya merambat dari medium yang lebih tumpat ke medium yang

kurang tumpat dan sudut tuju melebihi sudut genting.

A phenomenon which occurs only when light travels from a denser medium to a less dense medium and the angle of incidence is

greater than the critical angle.

NormalNormal

ImejImageObjekObject Z

1 Apabila cahaya merambat dari kaca (medium yang lebih tumpat ) ke udara

(medium yang kurang tumpat ), ia akan terbias menjauhi normal.

When light travels from glass ( denser medium) to air

( less dense medium), it will bend away from the normal.

2 Sudut biasan, r, adalah lebih besar daripada sudut tuju, i.

The angle of refraction, r, is larger than the angle of incidence, i.

3 Apabila sudut tuju, i, meningkat, sudut biasan, r, meningkat sehingga r = 90°.

When the angle of incidence, i, increases, the angle of refraction, r, increases until r = 90°.

4 Apabila sudut biasan, r, menjadi 90°, sinar yang dibiaskan merambat di sepanjang sempadan di antara kaca dan udara. Pada peringkat ini, sudut tuju, i, di dalam

medium yang lebih tumpat dipanggil sudut genting, c .When the angle of refraction, r, is 90°, the refracted ray moves along the boundary between glass and air. At this stage, the angle of incidence, i, in the denser medium is

called the critical angle, c .

KacaGlass

UdaraAir

i

r

i

r

KacaGlass

UdaraAir

i = cKacaGlass

UdaraAir

MEMAHAMI PANTULAN DALAM PENUH CAHAYAUNDERSTANDING TOTAL INTERNAL REFLECTION OF LIGHT5.3


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Kesimpulan Conclusion

❶ Cahaya yang merambat dari medium lebih tumpat ke medium

kurang tumpat akan dibiaskan menjauhi normal.Light rays which travel from a denser medium to a less dense medium

are refracted away from the normal.

❷ Apabila sudut pembiasan sama dengan 90°, sudut tuju disebut

sudut genting .When the angle of refraction is equal to 90°, the angle of incidence is

called the critical angle .

❸ Apabila sudut tuju lebih besar daripada sudut genting,

pantulan dalam penuh berlaku.

When the angle of incidence is more than the critical angle, total internal reflection occurs.

Sudut genting, c, ialah sudut tuju di dalam medium yang lebih tumpat apabila sudut biasan, r, dalam

medium yang kurang tumpat ialah 90°

The critical angle, c, is the angle of incidence in the denser medium when the angle of refraction, r, in the less

dense medium is 90°

5 Apabila sudut tuju, i, adalah lebih besar daripada sudut genting, c, cahaya tidak

lagi dibiaskan tetapi semuanya dipantulkan ke dalam semula dan ini

mematuhi hukum pantulan .

When the angle of incidence, i, is larger than the critical angle, c, the light is no longer

refracted but is reflected internally and obeys the laws of reflection .

6 Fenomena ini dinamakan pantulan dalam penuh .

This phenomenon is called total internal reflection .

7 Dua syarat untuk pantulan dalam penuh cahaya berlaku ialah:The two conditions for total internal reflection to occur are:

(a) Cahaya merambat dari medium yang lebih tumpat ke medium kurang tumpat .

The light travels from a denser medium to a less dense medium.

(b) Sudut tuju , i, adalah lebih besar daripada sudut genting, c .

The angle of incidence , i, is greater than the critical angle, c .

i

KacaGlass

UdaraAir

KacaGlass

Pantulan dalam penuhTotal internal re�ection

UdaraAir

NormalNormal

r1

r2

i1i2


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Hubungan antara sudut genting, c, dengan indeks pembiasan, n The relationship between critical angle, c, and refractive index, n

UdaraAir

cKacaGlass

1 Indekspembiasankacaadalah:The refractive index of glass is given as:

n = sin i (dalam udara/in air)

sin r (dalam kaca/in glass)

n = sin 90°sin c

n = 1

sin c

Kira sudut genting kaca yang mempunyai indeks pembiasan, n = 1.5. Calculate the critical angle for glass with refractive index, n = 1.5.

Penyelesaian/Solution

sin 90°sin c

= n

sin c = 1n

= 1

1.5

= 0.6667 c = 41.8°

Rajah menunjukkan satu sinar cahaya, P, ditujukan ke dalam blok kaca. Sudut genting kaca ialah 42°. Ke arah manakah cahaya bergerak dari titik Q? The diagram shows a light ray, P, directed into the glass block. The critical angle of the glass is 42°. In which direction does the light move from point Q?

Jawapan/Answer: CPenerangan: Sudut tuju = sudut genting = 42°, maka cahaya merambat di sepanjang sempadan kaca-udaraExplanation: Incident angle = critical angle = 42°, thus light travels along to the boundary of glass-air

42˚

D

P

Q

Blok kacaGlass block

A BC

Contoh/Examples

Latihan/Exercises


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Logamaya Mirages

Fenomena semula jadi berkaitan pantulan dalam penuh Natural phenomena involving total internal reflection

Udara sejuk(lebih tumpat)Cool air (denser) Udara panas

(kurang tumpat)Warm air (less dense)

Permukaan panasHot surface

1 Di padang pasir, tanah (pasir) adalah sangat panas pada waktu siang, jadi lapisan udara di permukaan

pasir itu dipanaskan dan menjadi lebih ringan serta kurang tumpat .

In the desert, the land (sand) is very hot during day time, so the layer of air in contact with it gets

heated up and becomes lighter and less dense .

2 Hasilnya,lapisanudaradibahagianatasitukurangpanasdan lebih tumpat daripada lapisan di bawahnya.As a result, the successive upper layers of air are not as hot and are denser than those below it.

3 Sinar cahaya dari objek yang jauh, dibiaskan daripada medium lebih tumpat ke medium kurang tumpat.

A ray of light from a distance object, gets refracted from a denser medium to a less dense medium.

4 Cahaya yang dibiaskan membengkok menjauhi normal.

The refracted rays bend away from the normal.

5 Dekat permukaan tanah, pantulan dalam penuh berlaku.

At the surface of the ground, total internal reflection occurs.

6 Logamaya disebabkan oleh pantulan dalam penuh cahaya di lapisan udara yang berketumpatan

berbeza .

The mirage is caused by the total internal reflection of light at layers of air of different densities .

Aplikasi pantulan dalam penuhApplications of total interal Reflection

Periskop/Periscope

45°

45°

Ciri-ciri imej: saiz sama, tegak

Characteristics of the image: same size, upright

Kegunaan periskop:Uses of the periscope:

Untuk melihat objek di belakang halangan. Dalam kapal selam, periskop

digunakan untuk memerhatikan kapal-kapal di permukaan laut.

To view objects behind obstacles. In submarines, periscopes are used to observe ships

on the surface of the sea.


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Kelebihan menggunakan prisma dan bukan cermin dalam periskop

The advantages of using prism rather than mirror in periscope

• Imejyangdihasilkanolehpantulandalampenuhlebihterangdaripadaimejyangdihasilkanolehcermin.

Images produced by total internal reflection are brighter than those produced by mirrors.

• Prismabolehmemantulkan100%cahayapadapermukaannyasebagaipantulandalampenuh.

Prism can reflect 100% of the light at its surface by total internal reflection.

• Cerminsatahmenghasilkanimejberganda(cahayadipantulkandaripadapermukaanbelakangdanhadapancerminakanmenyebabkan imej kelihatan kabur).

A plane mirror produces multiple image ( as light is reflected from the back and front surfaces of the mirror a number of times which blurs the image seen).

• Lapisanbersalutperakbolehmengelupas,olehitumengurangkankeamatanpantulancahaya.

The silver coating in the mirror can also flake, therefore reducing the reflected light intensity.

• Prismakacatahankaratdaripadaairgaram.

Glass prism is very resistant to corrosion from salt water.

Gentian optikOptical fibres

Kelebihan menggunakan kabel gentian optik dan bukan kabel tembaga

The advantages of using fibre-optic cable rather than copper cable

• Lebihnipis(sepertirambutmanusia)danlebihringan.

Much thinner (like human hair) and lighter.

• Bolehmembawalebihbanyakmaklumat(isyarat)padasatumasa.

Can carry much more information (signal) at one time.

• Merekamenghantarisyaratdengankehilanganyangsangatsedikitpadajarakyangbesar.

They transmit signal with very little loss over great distances.

• Tiadagangguandarimedanelektromagnetdalamsambungan'lebihjelas'.

No interference from electromagnet field result in ‘clearer’ connections.

• Tiadarintanganelektrik.

No electrical resistance.

• Tiadabahayakejutanelektrikjikakabelterputus.

No hazard of electrocution if cable breaks.

1 Suatu gentian optik terdiri daripada teras dalam yang mempunyai indeks biasan kaca yang tinggi dan dikelilingi

oleh lapisan luar yang mempunyai indeks biasan yang lebih rendah.

An optical fibre consists of inner core of high refractive

index glass and surrounded by the outer cladding of lower refractive index.

2 Apabila sinar cahaya merambat ke dalam teras dalam pada satu hujungnya, ia akan merambat secara zig zag di

sepanjang gentian dalam satu siri pantulan dalam penuh

.

When a light ray moves into the inner core from one end, it will propagate along the fibre in a zigzag path undergoing a

series of total internal reflections .

Lapisan luarOuter cladding

Teras dalamInner core

NormalNormal

Sinar cahayaLight ray

Perambatan cahaya dalam gentian optikPropagation of light in an optical �bre

Perambatan cahaya dalam gentian optikPropagation of light in an optical fibre


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F

f

F

f

Sinar cahaya yang selari dengan paksi utama

dibiaskan ke dalam dan menumpu di titik fokus.Light rays parallel to principal axis are refracted inwards and converge to the focal point.

Sinar cahaya yang selari dengan paksi utama

dibiaskan keluar dan mencapah dari titik fokus.Light rays parallel to the principal axis are refracted outwards and diverge from the focal point.

Kanta cembung = kanta penumpu = kanta positifConvex lens = converging lens = positive lens

Kanta cekung = kanta pencapah = kanta negatifConcave lens = diverging lens = negative lens

•Kantacembungadalahlebihtebal di tengah-tengah daripada di sisinya.

Convex lenses are thicker in the centre than at the edge.

•Kantacekungadalahlebihnipisdi tengah-tengah daripada di sisinya.

Concave lenses are thinner in the centre than at the edge.

•Kantacembungmembiaskansinar tuju, yang selari dengan paksi utama, oleh itu, ia tertumpu kepada suatu titik di atas paksi utama.

Convex lenses refract incident rays of light, which are parallel to the principal axis, so that they are converged, to a point on the principal axis.

•Kantacekungmembiaskansinar tuju yang selari dengan paksi utama dan seolah-olah sinar yang terbias itu mencapah dari titik fokus yang terletak di sebelah yang sama dengan sinar tuju di paksi utama.

Concave lenses refract incident rays of light, which are parallel to the principal axis, so that they appear to diverge from a point located on the incident ray of the principal axis.

Kanta cembungConvex lens

Kanta cekungConcave lens

F

f

F

f

Titik fokus, F/Focal point, FTitik penumpuan di mana semua sinar yang selari dengan paksi utama bertumpu

kepadanya (kanta cembung) atau menyimpang daripadanya (kanta cekung) selepas

melalui kanta.

The concentration point where all rays parallel to the principal axis converge to it

(convex lens) or diverge from it (concave lens) after passing through the lens

Panjang fokus, f /Focal length, f Jarak di antara titik fokus dan pusat optik kanta

The distance between the focal point and optical centre

of the lens

MEMAHAMI KANTAUNDERSTANDING LENSES5.4


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Istilah untuk kantaCommon terminology of lenses

Pusat optik, COptical centre, C

Pusat geometri kanta. Sinar yang merambat melalui pusat optik kanta dalam garis lurus.The geometric centre of the lens. Rays travelling through the optical centre pass through the lens in a straight line.


Garis lurus yang melalui pusat optik pada sudut tegak (90°) dengan satah kanta.Straight line which passes through the optical centre at right angle (90°) to the plane of the lens.

Titik fokus, FFocal point, F

Titik pada paksi utama yang mana sinar tuju yang selari dengan paksi utama:Is the point on the principal axis to which incident rays of light travelling parallel to the principal axis:a. menumpu daripada titik itu selepas

pembiasan melalui kanta cembung converge after refraction through a convex

lens.b. mencapah daripada titik itu selepas

pembiasan melalui kanta cekung. diverge after refraction through a concave

lens.


Jarak antara pusat optik, O, dengan objek.Distance between the optical centre, O, to the object.


Jarak antara pusat optik, O, dengan imej.Distance between the optical centre, O, to the image.

Peraturan untuk rajah sinarRules for ray diagrams

Imej terhasil melalui kanta Images formed by lenses

ObjekObject

ObjekObject

F F

FF

Rajah sinar untuk kanta cembungRay diagrams for convex lenses

(a) u > 2f

(b) u = 2f

F

f

vu

2F F ImejImage

ObjekObject

ImejImage

F2F F

f

vu

Kanta cekungConcave lenses

Kanta cembungConvex lenses

(a)

(b)

(c)

(a)

(b)

(c)

ObjekObject

ObjekObject

F F

FF


Songsang/Inverted

Nyata/Real

Mengecil/Diminished


Songsang/Inverted

Nyata/Real

Sama saiz dengan objekSame size as the object


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(c) f < u < 2f

(d) u < f

F2F F 2F

ImejImage

f

vu

F2F F

ImejImage

f

v u


Songsang/Inverted

Nyata/Real



Tegak/Upright

Maya/Virtual


Rajah sinar untuk kanta cekungRay diagrams for concave lenses

F2F

ObjekObject

ImejImage F F2F

ObjekObject

ImejImage

F

Ciri-ciri imej bagi kanta cekung adalah sama dengan ciri-ciri cermin cembungThe characteristics of the image for concave lens are the same as the characteristics of convex mirrors

DUVD = Mengecil DiminishedU = Tegak UprightV = Maya Virtual


Diperkecilkan/Diminished Tegak/Upright Maya/Virtual

Pembesaran Magnification

Pembesaran, mMagnification, m =

Saiz imej/Size of image

Saiz objek/Size of object =

hI

hO

Pembesaran, mMagnification, m =

Jarak imej/Image distance

Jarak objek/Object distance =

v

u

m = hI

hO =

v

u


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Hubungan antara u, v, f Relationship between u, v and f

1

f =

1

u + 1

v

Jenis kantaType of lens

Panjang fokus, fFocal length, f



Kanta cembungConvex lens

Nilai positifPositive value


Nilai positif atau negatifPositive value or negative value

Kanta cekungConcave lens

Nilai negatifNegative value


Nilai negatifNegative value

Satu objek dengan tinggi 8 cm diletakkan pada jarak 20 cm dari kanta cekung. Panjang fokus ialah10 cm. Berapakah(i) jarak imej dan (ii) saiz imej?An object of height 8 cm is placed at a distance of 20 cm from a concave lens. Its focal length is 10 cm. What is (i) the image distance and (ii) size of the image?

Penyelesaian/Solution:(i) ho = 8 cm u = + 20 cm f = –10 cm

1f =

1u +

1v

1

–10 cm =

120 cm +

1v

v = –6.7 cm

(ii) hI

ho =

vu

hI = 6.7 cm20 cm × 8 cm

= 2.68 cm

Rajah menunjukkan kedudukan imej, I, bagi objek, O yang dibentuk oleh kanta cembung. Ketinggian objek, ho ialah 6 cm.The diagram shows the position of an image, I, of an object, O formed by a convex lens. The height of the object, ho is 6 cm.

O

I

70 cm

100 cm

Berapakah tinggi imej, hI?What is the height of the image, hI?

Penyelesaian/Solution:Jarak imejImage distance, v = 70 cmJarak objek/Object distance, u = 100 cm – 70 cm = 30 cmho = 6 cm

m = hIhO

= vu

hI6 cm =

70 cm30 cm

hI = 70 cm30 cm × 6 cm

= 14 cm

Latihan/Exercises

Latihan/Exercises


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Kegunaan kanta dalam alat optik The uses of lenses in optical devices

1 Kanta pembesarMagnifying glass

FF

I

O

MataEye

Jarak objek, u: u < f

Ciri-ciri imej: Diperbesarkan, tegak, maya

Object Distance, u: u < f

The characteristics of the image: Magnified, upright, virtual

2 Mikroskop majmukCompound microscope

Kanta objektifObjective lens

MataEye

ObjekObject

Kanta mataEyepiece

F0Fe I1

I2

FeF0

Ciri-ciri imej pertama, I1 Characteristics of first image, I1

Songsang, nyata, diperbesarkan

Inverted, real, magnified Ciri-ciri imej terakhir, I2 Characteristics of final image, I2

Songsang, maya, diperbesarkan Inverted, virtual, magnified

3 Teleskop astronomi (digunakan dalam pelarasan normal)Astronomical telescope (used in normal adjustment)

Kanta objektifObjective lens

Imej, I2, di in�nitiImage, I2, is formed

at in�nity

Kanta mataEyepiece lens

Fo Fe

MataEye

Cahaya selari dariobjek jauhParallel raysfrom distantobject

I1

F0 = titik fokus kanta objektif

focal point of objective lens

Fe = titik fokus kanta mata

focal point of eyepiece


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Perbandingan antara mikroskop majmuk dan teleskop astronomi Comparison between a compound microscope and an astronomical telescope

Persamaan /Similarities

1 Terdiri daripada dua kanta cembung

Consists of two convex lenses

2 Imejyangpertamaadalah nyata dan songsang dan bertindak sebagai objek bagi kanta mata

The first image is real and inverted and acts as the object for the eyepiece

3 Kanta mata bertindak sebagai kanta pembesar

The eyepiece acts as a magnifying lens

4 Imejakhiradalah maya , songsang dan diperbesarkan

The final image is virtual , inverted and magnified

Perbezaan/Differences

Mikroskop majmukCompound microscope

AspekAspects

Teleskop astronomiAstronomical telescope

Dua kanta cembung berkuasa tinggiTwo high-powered convex lenses

Jenis kanta dankuasa kanta

Types of lenses andpower of lenses

Kanta cembung berkuasa rendah dan kanta cembung berkuasa tinggiA low powered convex lens and a high poweredconvex lens

fo < fe Panjang fokusFocal length

fo > fe

ImejpertamaadalahbesarFirst image is magnified

ImejpertamaFirst image

ImejpertamaadalahkecilFirst image is diminished

Kedudukan dekat dengan mata pemerhatiAt the near point of the observer’s eye

Kedudukan imej terakhir

Position of final image

Pada kedudukan infinitiAt infinity

Lebih besar/Greater than ( f o + fe)D > ( f o + fe)

Jarak antara dua kantaDistance between lenses

(D)

Sama/Equal to ( f o + fe)D = ( f o + fe)

m = mo × mePembesaran linear, mLinear magnification, m

m = fofe


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1 Rajah menunjukkan alur cahaya bergerak dari minyak ke udara.

The diagram shows a light ray travels from the oil into the air.

Apakah nilai k?/What is the value of k? [Indeksbiasanminyak/Refractive index of oil = 1.4] A 44.4º D 55.4º B 45.6º E 58.9º C 54.5º

2 Laju cahaya di dalam udara ialah 3 × 108 m s-1. Apakah laju cahaya di dalam blok plastik?/The speed of light in the air is 3 × 108 m s-1. What is the speed of light in a plastic block?

[Indeksbiasanplastik/Refractive index of plastic = 1.2]

A 1.0 × 108 m s-1 B 1.5 × 108 m s-1 C 2.0 × 108 m s-1

D 2.5 × 108 m s-1

E 3.0 × 108 m s-1

3 Kanta cekung dengan panjang fokus 20.0 cm membentuk imej yang merupakan 2 kali ganda saiz objek. Apakah jarak objek tersebut?

A convex lens with a focal length of 20.0 cm forms an image which is 2 times the size of the object. What is the object distance?

A 10.0 cm B 15.0 cm C 30.0 cm D 40.0 cm

Penyelesaian / SolutionsJawapan B / Answer B

Penerangan / ExplanationMenggunakan rumus / Using formula

1

1.4 = sin 30ºsin r

sin r = 1.4 sin 30º = 0.7 r = 44.4º,Oleh itu / Therefore, k = 90º – 44.4ºk = 45.6º

Penyelesaian / SolutionsJawapan D / Answer D


n = cv

1.2 = 3 × 108 m s–1

v

v = 3 × 108 m s–1

1.2

= 2.5 × 108 m s-1

Penyelesaian / SolutionsJawapan C / Answer C


1f =

1u +

1v and / dan m =

vu

2 = vu

2u = vGantikan ke dalam persamaan pertama Substitute into the first equation

1

20 cm = 1u +

12u

1

20 cm = 2 + 1

2u

1

20 cm = 32u

2u = 60 cm u = 30 cm

k

60º

Water/Air

Minyak/Oil

n = sin i

sin r

n =

Laju cahaya dalam vakum (udara)speed of light in vacuum (air)

Laju cahaya dalam suatu mediumspeed of light in medium

Latihan/Exercises


5

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1 Rajah 1 menunjukkan sinar cahaya yang merambat dari air ke udara.Diagram 1 shows light ray travels from water to air.

sr

pq

UdaraAirAir

Water

Rajah 1 / Diagram 1

IndeksbiasanbagiairituialahRefraction index of water is

A sin r——–sin q

B sin p——–sin s

C sin s——–sin p

D sin p——–sin r

2 Imejmanakahyangdihasilkanolehkantapenumpupada skrin?Which image is obtained by a convex lens on the screen?A Songsang dan sahih. Inverted and real.B Songsang dan maya. Inverted and virtual.C Tegak dan sahih. Upright and real.D Tegak dan maya. Upright and virtual.

3 Rajah 3 menunjukkan sinar X ditujukan ke dalam blok kaca. Sudut genting kaca ialah 42°. Diagram 3 shows a ray, X, directed into a glass block. The critical angle of the glass is 42°.

48˚

B

A

X

Y

KacaGlass

CD

Rajah 4 / Diagram 4

Ke arah manakah sinar merambat dari titik Y?In which direction does the light move from point Y?

4 Antara berikut, yang manakah adalah ciri-ciri imej yang terbentuk di atas cermin satah?Which of the following are the characteristics of the image formed on a plane mirror?A Songsang, sama saiz dengan objek dan nyata. Inverted, same size as the object and real.B Tegak, sama saiz dengan objek dan nyata. Upright, same size as the object and real.C Tegak, diperbesarkan dan songsang sisi. Upright,magnifiedandlaterallyinverted.D Songsang sisi, sama saiz dengan objek dan maya. Laterally inverted, same size as the object and virtual.

5 Rajah 5 menunjukkan suatu objek diletakkan di hadapan satu kanta cembung.Diagram 5 shows an object placed in front of a convex lens.

F2F 2FF

ObjekObject

Rajah 5 / Diagram 5

Apakah ciri-ciri imej yang terbentuk?What are the characteristics of the image formed?A Nyata, songsang, diperkecilkan Real, inverted, diminishedB Nyata, songsang, sama saiz Real, inverted, same sizeC Maya, tegak, diperbesarkan Virtual,upright,magnifiedD Nyata, songsang, diperbesarkan Real,inverted,magnified

6 Rajah 6 menunjukkan suatu sinar cahaya merambat dalam satu blok semi bulatan yang lutsinar.Diagram 6 shows a ray of light propagating in a transparent semi-circular block.

35°

Rajah 6 / Diagram 6

Apakah indeks biasan bagi blok lutsinar itu?What is the refractive index of the transparent block?

A 1.00 C 1.74B 1.35 D 2.80

Latihan Pengukuhan/Enrichment Exercises


5

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7 Sebuah kanta cembung yang berpanjang fokus 15.0 cm digunakan untuk menghasilkan imej yang sama saiz dengan objek. Jarak objek ialahA convex lens of focal length 15.0 cm is used to produce an image of the same size as the object. The object distance isA 0.8 cm C 15.0 cmB 3.0 cm D 30.0 cm

8 Suatu objek diletakkan 20.0 cm dari sebuah cermin cembung. Panjang fokus cermin cembung ini adalah 10.0 cm. Apakah jarak imej?An object is 20.0 cm in front of a convex mirror. The focal length of the mirror is 10.0 cm. What is the image distance?A 5.0 cm C 10.0 cmB 6.7 cm D 20.0 cm

9 Sebuah kanta cekung akan sentiasa menghasilkan imej dengan ciri-ciri yang berikut.A concave lens will always produce an image with the following characteristics.A Maya, tegak, diperbesarkan Virtual,upright,magnifiedB Maya, tegak, diperkecilkan Virtual, upright, diminishedC Nyata, songsang, diperbesarkan Real,inverted,magnifiedD Nyata, songsang, diperkecilkan

Real, inverted, diminished

1 Dalam satu eksperimen dalam makmal, pelajar mengkaji hubungan antara jarak objek, u dan jarak imej, v bagi kanta cembung. Daripada data yang diperoleh, graf diplot seperti di bawah.In a laboratory experiment, students investigate the relationship between the object distance, u and the image distance, v for a convex lens. From the data obtained, a graph is plotted as shown below.

0

0.4

0.8

1.2

(cm–1)

1.61.41.21.00.80.60.40.2 (cm–1)

Rajah 1 / Diagram 1

(a) (i) Nyatakan nilai pintasan di 1—v paksi apabila

1—u = 0 dan oleh itu, cari nilai panjang fokus kanta.

State the value of the intercept at the 1—v axis when 1—u =0andtherefore,findthevalueofthefocallengthofthelens.

(ii) Namakan kuantiti yang diperoleh dari pintasan di paksi 1—u apabila

1—v = 0.

Name the quantity obtained from the reciprocal of the intercept at the 1—u axis when 1—v = 0.

Jarak fokus / Focal length

(b) Kirakan kecerunan graf. / Calculate the gradient of the graph.

Soalan Struktur/Structure Questions

1.4 cm–1, f = 1———–

1.4 cm–1 = 0.714 cm

m = – 1.4—–1.4 , m = –1.0


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UNIT



(c) Dapatkan ciri-ciri imej yang dihasilkan apabila / Obtain the characteristics of the image produced when (i) objek u pada jarak 0.5 cm. / the object distance u is 0.5 cm.

Maya, menegak, diperbesarkan / Virtual,upright,magnified

(ii) objek u pada jarak lebih daripada 1.0 cm. / the object distance u is more than 1.0 cm.

Sahih, songsang, diperbesarkan / Real,inverted,magnified

2 Rajah2.1menunjukkanlintasansinarcahayayangtidaklengkapmemasukisuatuprismakacaPyrex.Indeksbiasankaca Pyrex ialah 1.47.Diagram 2.1 shows an incomplete light ray path entering a Pyrex glass prism. The refractive index of Pyrex glass is 1.47.

45˚

0Prisma kaca PyrexPyrex glass prism

Rajah 2.1 / Diagram 2.1

(a) (i) HitungsudutgentingprismakacaPyrexitu./Calculate the critical angle of the Pyrex glass prism.

(ii) Pada Rajah 2.1, lengkapkan lintasan sinar cahaya dari titik O. On Diagram 2.1, complete the light ray path from point O.

(iii) Berdasarkan jawapan di 2(a)(ii), namakan fenomena cahaya yang terlibat. Based on the answer in 2(a)(ii), name the light phenomenon involved.

Pantulan dalam penuh / Totalinternalreflection

(b) Rajah 2.2 menunjukkan binokular berprisma. Kedudukan bagi dua prisma pada satu sisi binokular adalah seperti yang ditunjukkan. / Diagram 2.2 shows prism binoculars. The position of two prisms on one side of the binoculars are as shown.

Rajah 2.2 / Diagram 2.2 Rajah 2.3 / Diagram 2.3

Rajah 2.3 adalah rajah skema bagi Rajah 2.2. Pada Rajah 2.3, lukis lintasan sinar cahaya yang memasuki kedua-dua prisma itu. Dalam lukisan anda, tunjukkan arah lintasan sinar cahaya itu.

Diagram 2.3 is a schematic diagram of Diagram 2.2. On Diagram 2.3, draw the ray path entering both prisms. In your drawing, indicate the direction of the ray path.

n = 1——–sin C

sin C = 1—n

= 1——1.47

= 0.6803 Sudut genting / Critical angle = 42.86°


LIGHT - yuhuaphysics.com Fizik Tg4 B5 2015(FSY4p) n.pdf · • Melukis gambar rajah sinar untuk...

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