Physics Module

Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure

Guru Cemerlang Physics MSAB Date:

3.5.3 Application of Archimedes’ Principle

3.5.3.1 Submarine

3.5.3.2 Hot-air Balloon

12

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Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure

Guru Cemerlang Physics MSAB Date:

3.6 UNDERSTANDING BERDOULLI’S PRINCIPLE

1 Bernoulli's Principle is the principle that allows
wings to produce lift and planes and helicopters to
fly. There are many factors that can affect the lift
produced under this principle, but in order to fully
understand how and why things can effect flight one
must understand how Bernoulli's principle works.

2 Bernoulli's principle works on the idea that as a wing passes through the air and its shape make
the air travel more over the top of the wing than beneath it. This creates a higher pressure are
beneath the wing than above it. The pressure difference cause the wing to push upwards and lift
is created.

3 Bernoulli’s principle states that when the speed of a fluid increases, the pressure in the fluid
decreases, and when the speed of the fluid decreases, the pressure in the fluid increases.

4 Bernoulli's principle is valid for incompressible flows (e.g. most liquid flows).
5 If a fluid is flowing horizontally and along a section of a streamline, where the speed increases it

can only be because the fluid on that section has moved from a region of higher pressure to a
region of lower pressure.

6 If its speed decreases, it can only be because it has moved from a
region of lower pressure to a region of higher pressure.
7 Consequently, within a fluid flowing horizontally, the highest speed
occurs where the pressure is lowest, and the lowest speed occurs where
the pressure is highest.

3.6.1 Applications of Bernoulli’s Principle 13
1 The shape of the Aerofoil

 The air travels faster over the curved upper surface
than it does over the flatter lower surface.

 As a result, the pressure above the aerofoil is lower than that below it.
 The difference in pressures lifts the aerofoil upwards.
2 Lifting paper
 When the air is blown up in the surface of a piece of paper, it’s

observed that the paper moves up.
 This happened because the air moved at a very high velocity.
 According to Bernoulli’s Principle, the pressure of the moving

air decreases as the speed of the air increases.
 The higher atmospheric pressure which acts at the bottom of

the paper pushes up the paper.

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Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:
3 “Banana Ball” in sport

 In a sport such as baseball, cricket, soccer or table tennis, the athlete deliberately spins the

ball to make it curve in a particular direction.

 The air comes into contact with the spinning ball. The air moving in direction of the spin is

speed up and the air in the opposite direction of the spin is slowed down.

 Because of this the air pressure on one side is higher than that on the other side.

 The high air pressure tries to compensate for the low air pressure and pushes the ball

resulting in the curve.

4 Ping-Pong ball closer to each other when air blown between.
 When the air is blown harder through the straw, the two Ping-Pong balls
will move closely to each other.
 The air moved at a very high velocity between the balls.
 According to Bernoulli’s Principle, the pressure of the moving air
decreases as the speed of the air increases.
 The higher atmospheric pressure caused the Ping-Pong balls closer to
each other.

5 Insect Piston Spray
 When the piston is pushed, air is forced out through the jet of gas at a high speed.
 According to Bernoulli’s Principle, the pressure of the moving air decreases as the speed of
the air increases.
 The higher atmospheric pressure in the insect poison container will push the insect poison
liquid up through the narrow metallic tube

14

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Mr Ng Han Guan Form 4 Chapter 3: Force and Pressure
Guru Cemerlang Physics MSAB
Date:
6 Bunsen burner

 When the jet of gas flows out from the nozzle with high

velocity, the pressure in the Bunsen burner becomes low.

 A higher external atmospheric pressure will be sucked into

the air hole and be mixed with the gas.

 The mixture of gas and air allows more complete combustion

of the gas

15

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Mr Ng Han Guan Form 4 Chapter 4: Heat Date:
Guru Cemerlang Physics MSAB
CHAPTER 4: HEAT

4.1 UNDERSTANDING THERMAL EQUILIBRIUM

4.1.1 What is Thermal Contact?
1. Two objects are in thermal contact if energy (heat) can flow between them.
2. It does not depend on mass, size, shape and types of material.
3. Energy is transferred at a faster rate from the hotter object to the colder object.
4. Energy is also transferred from the colder object to the hotter object, but at a slower rate.
5. There is a net flow of energy from the hotter object to the colder object

and is known as the heat transferred.
6. Therefore, the hotter object cools down (releases heat) while the

colder object warms up (absorbs heat).
7. After some time, energy is transferred at the same rate (no net heat

transfer) between the objects.
8. Now the objects are said to be in thermal equilibrium.

4.1.2 Thermal Equilibrium
1. Two objects are same temperature.
2. The rates of heat transfer between two objects are equal.
3. There is no net flow of heat between two objects.

4.1.3 Calibration of Liquid-in-glass Thermometer

1. Temperature is a measure of the degree of hotness of an object.

2. Temperature is measured using a liquid-in-glass thermometer. When

liquid in liquid-in-glass thermometer is in thermal equilibrium with an

object, they have equal temperature.

3. The bulb contains a fixed mass of liquid. The volume of the liquid

increases when it absorbs heat. The liquid expands and rises in the capillary tube. Therefore

the length of the liquid column in the capillary tube indicates the magnitude of the temperature.

4. The traditional method of putting a scale on a liquid-in-

glass or liquid-in-metal thermometer was in three stages:

i. Immerse the sensing portion in a stirred mixture of

pure ice and water at 1 Standard atmosphere

(101.325 kPa; 760.0 mmHg) and mark the point
indicated when it had come to thermal equilibrium.  lo for 0 ℃

ii. Immerse the sensing portion in a steam bath at 1 Standard atmosphere (101.325 kPa; 760.0
mmHg) and again mark the point indicated.  l100 for 100 ℃

iii. Divide the distance between these marks into equal portions according to the temperature

scale being used. 1

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Mr Ng Han Guan Form 4 Chapter 4: Heat
Guru Cemerlang Physics MSAB
Date:

5. The liquid commonly used in the liquid-in-glass thermometer is mercury because:
i. It is a good conductor of heat  faster react to heat and reach thermal equilibrium.
ii. It has a high boiling point (357 ℃)  can measure high temperature.

iii. It expands uniformly when heated  suitable for measuring temperatures.
iv. It is opaque a can be seen easily
6. Mercury freezes at temperature of – 39 ℃ and is not suitable for measuring temperatures below

this temperature.
7. So, liquid-in-glass thermometers filled with alcohol, which freezes at – 115 ℃, can measure lower

temperatures.

Example 1 Example 2
The lengths of the mercury column in a The lengths of the mercury thread in a
thermometer at the ice point and the steam point thermometer are 4.0 cm and 24.0 cm
are 12 cm and 20 cm respectively. When the respectively at 0 oC and 100 oC. What is the
thermometer is placed in a liquid, the length of length of the thread when the thermometer is
the mercury column is 15 cm. What is the placed in a substance at 65oC?
temperature of the liquid? Solution
Solution

2

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Mr Ng Han Guan Form 4 Chapter 4: Heat
Guru Cemerlang Physics MSAB
Date:

4.2 UNDERSTANDING SPECIFIC HEAT CAPACITY

4.2.1 Concept of Heat Capacity

1. The heat capacity of a body is amount of heat energy that supplied to the body to increase its

temperature by 1 oC.

2. The unit of heat capacity is J oC-1.

3. Each object has its own heat capacity.

4. Example 1:

 The figure shows two objects A and B are both made of
aluminium, but different amounts of heat are required to raise
their temperature by 1 oC.

 Object B has higher heat capacity than object A.
 Heat capacity of A is 1800 J oC-1.
 Heat capacity of B is 3600 J oC-1.
 They have different heat capacity because the masses are

different.

5. Example 2:

 The figure shows two objects A and C have equal masses;
but different amounts of heat are required to raise their
temperature by 1 oC.

 Object A has higher heat capacity than object C.
 Heat capacity of A is 900 J oC-1.
 Heat capacity of C is 130 J oC-1.
 They have different heat capacity because they are made of

different materials.

6. Note that although A and C have equal masses, different amount of heat are required to raise

the temperature by 1 oC because they are made of different materials.

7. Different materials are said to have different specific heat capacities, c.

4.2.2 Concept of Specific Heat Capacity, c

1. The specific heat capacity, c, of a substance is the amount of heat that must be supplied to

increase the temperature by 1 oC or 1 K for a mass of 1 kg of the substance.

2. The unit of specific heat capacity is J kg-1 oC-1 or J kg-1 K-1.

3. Different materials have different specific heat capacities, c.
4. The word “specific” refers to per unit mass (per kg). This means that specific heat capacity, c, is

the heat capacity for 1 kg of a material.

5. Heat capacity only relates to a particular object whereas specific heat capacity, c, relates to a

material.

6. Specific heat capacity, c =  Q = mc∆θ
∆

(Q = quantity of heat absorbed or released; ∆θ = change of temperature; m = mass of a substance)

3

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Mr Ng Han Guan Form 4 Chapter 4: Heat
Guru Cemerlang Physics MSAB
Date:

4.2.3 Application of Specific Heat Capacity

1. Cooking instruments

 Cooking instruments such as frying pans, pots, kettles, electric iron

and so on made of substances with low specific heat capacities

and good heat conductor for the body.

 This is because they can quickly heated up when there is only

small heat absorption.

 The handle of the cooking instruments are made by the substances with high specific heat

capacities and poor heat conductor.

 This is because these materials will not become too hot even if heat is absorbed.

2. The cooling system of a car engine

 Water is very useful as a cooling agent because

 water has a very high specific heat

capacity can storing more energy.
 water is cheap, safe and readily available.

 In the cooling system, heat is removed from the engine and released to the surroundings by

use of water.
 A water pump circulates the water. Heat generated from the combustion of the petrol-air

mixtures is absorbed by the water that flows along the space in the engine walls.

 The hot water flows to the radiator where heat is lost to the cooler that flows through the

cooling fins.

 The cooling fins are made from the materials with low specific heat capacities so that these

materials will cool down faster after absorbed heat from water.

3. Sea breeze and Land breeze
 In daytime the sun warms the land to higher temperature than the

sea.

 It is because land has a lower specific heat capacity than sea-

water.
 The air above the land is heated and rises, and its place is taken

by cooler air above the sea moving inland (convection currents)

 Air higher in the atmosphere completes the circulation, and hence

a sea-breeze is obtained.
 During night time, heat is lost from the sea and the land.
 Land is cools off faster and makes it cooler than sea because land has lower specific heat

capacity than sea.

 Cooler air from the land moves towards the sea and hot air above the sea rises, and hence

a land breeze is obtained.

4

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Mr Ng Han Guan Form 4 Chapter 4: Heat
Guru Cemerlang Physics MSAB
Date:

4.3 UNDERSTANDING SPECIFIC LATENT HEAT

4.3.1 Concept of Latent Heat
1. Latent heat is the heat released or absorbed by a substance or system during a change of

state that occurs without a change in temperature, meaning a phase transition such as the
melting of ice or the boiling of water.
2. When a substance changes phase, that is it goes from either a solid to a liquid or liquid to gas,
the energy, it requires energy to do so.
3. During change of phase, the temperature remains constant even though there is transfer of
heat because energy is needed to overcome the molecular forces of attraction between particles
and does not cause a change in the kinetic energy of the molecules.
4. If we measure the temperature of the substance which is initially solid as we heat it we produce a
graph:

Graph temperature change with time. Phase changes are
indicated by flat regions where heat energy used to
overcome attractive forces between molecules.

5. Latent Heat of Fusion
 The heat absorbed by a melting solid.
 During melting, the heat absorbed is used to break
up the bonds between the particles.
 Latent heat of fusion is absorbed.

6. Latent Heat of Vaporisation
 The heat absorbed during boiling.
 When liquid boils, the heat absorbed is used to completely break the bonds between the
particles and also to do work against atmospheric pressure when the gaseous vapour
expands into the atmosphere.
 Latent heat of vaporisation is absorbed.

5

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Mr Ng Han Guan Form 4 Chapter 4: Heat
Guru Cemerlang Physics MSAB
Date:
4.3.2 Specific Latent Heat

1. The specific latent heat of a substance, , is the amount of heat required to change the phase of

1 kg of the substance at a constant temperature.

 =


 Q is latent heat absorbed or released by the substance; m is mass of the substance

2. The unit of specific latent heat is J kg-1

3. Different materials have different specific latent heat, .

4. The latent heat absorbed or released when a substance of mass, m changes from one phase to

another is given by  =

5. Specific latent heat of fusion

 The amount of heat required to change 1 kg of the substance from solid to liquid phase

without a change in temperature.

6. Specific latent heat of vaporisation

 The amount of heat required to change 1 kg of the substance from liquid to gaseous

phase without a change in temperature.

4.3.3 Application of Specific Latent Heat
1. Steam and boiling water

 Steam will cause a serious burn compare to the boiling water.
 This is because steam has a large specific latent heat of

vaporization.
 When steam condenses on the skin of your arm, the very large

amount of latent heat released and large heat capacity of
boiling water (after steam condenses) are absorbed.
 If only the boiling water pour to the arm, the skin of arm will
absorbed only heat capacity of boiling water.
2. Cooking by steaming
 Water has a large specific latent heat of vaporization.
 This property enables steam to be used for cooking by the
method of steaming.
 When steam condenses on the food, the latent heat is
released directly onto the food enables the food to be
cooked at a faster rate.

6

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Mr Ng Han Guan Form 4 Chapter 4: Heat
Guru Cemerlang Physics MSAB
= = Date:
Principle of Conservation of Energy
=
= ∆ =

All are define energy

Problems Solving for Specific Heat Capacity and Specific Latent Heat
1. A heat transfer of 7.2 x 105 J is required to convert a block of ice at – 8.0 oC to water at 8.0 oC.

What is the mass of the block of ice?
[Specific heat capacity of ice = 2.0 x 103 J kg-1 oC-1;
Specific heat capacity of water = 4.2 x 103 J kg-1 oC-1 and
Specific latent heat of fusion of water = 3.36 x 105 J kg-1]

2. Calculate the heat required to convert 4 kg of ice at – 15 oC into steam at 100 oC.
[Specific heat capacity of ice = 2.0 x 103 J kg-1 oC-1;
Specific heat capacity of water = 4.2 x 103 J kg-1 oC-1;
Specific latent heat of fusion of ice = 3.36 x 105 J kg-1 oC-1 and
Specific latent heat of vaporisation of water = 2.26 x 106 J kg-1]

7

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Mr Ng Han Guan Form 4 Chapter 4: Heat
Guru Cemerlang Physics MSAB
Date:

4.4 UNDERSTANDING THE GAS LAWS

4.4.1 Properties of Gases
1. Gases are easily expandable and compressible unlike solids and liquids.
2. Gases have a measurement of pressure.
3. Pressure is defined as force per unit area of surface. It can be measured in several units such

as kilopascals (kPa), atmospheres (atm), and millimeters of Mercury (mmHg).
4. Gas has a low density because its molecules are spread apart over a large volume.
5. A gas will fill whatever container that it is in. An example of this is a bottle of ammonia being

opened in a room and the smell traveling throughout the room.

4.4.2 Kinetic Molecular Theory
The Kinetic Molecular Theory is the basis of the many properties of gases. The postulates to the
Kinetic Theory are as follows:
 Gases are composed of molecules whose size is negligible compared to the average distance
between them.
 Molecules move randomly in straight lines in all directions and at various speeds.
 The forces of attraction or repulsion between two molecules in a gas are very weak or
negligible, except when they collide.
 When molecules collide with one another, the collisions are elastic; no kinetic energy is lost.
 When a molecule collides with the wall of the container and bounces back, there is a change
in momentum and a force is exerted on the wall.
 The average kinetic energy of a molecule is proportional to the absolute temperature.

4.4.3 Boyle's Law – Relationship between Pressure and Volume
1. Boyle's Law states that for a fixed mass of gas, the volume of the

gas is inversely proportional to its pressure provided the

temperature remains constant.

2. Mathematically Boyle's law can be expressed as
 ∝ 1 , that is PV = constant or P1V1 = P2V2
 V1 is the original volume
 V2 is the new volume
 P1 is original pressure
 P2 is the new pressure

8

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Mr Ng Han Guan Form 4 Chapter 4: Heat

Guru Cemerlang Physics MSAB Date:

4.4.4 Charles' Law – Relationship between Volume and

Temperature

1. Charles' Law states that for a fixed mass of gas, the volume of

the gas is directly proportional to its absolute temperature

provided the pressure is remains constant.

2. Mathematically Charles’ law can be expressed as

 ∝ , that is = constant or 1 = 2
1 2

 V1 is the original volume

 V2 is the new volume

 T1 is original temperature in Kelvin scale

 T2 is the new temperature in Kelvin scale

3. The temperature at which the volume of the gas is expected to

become zero can be obtained by extrapolating the graph of volume

against temperature.
4. It is found that at – 273 oC the volume is expected to become zero.
5. The temperature – 273 oC is the lowest possible temperature and is known as the absolute zero

of temperature.

6. The SI unit for temperature is Kelvin (K). Temperature measured in the Kelvin scale is known as

absolute temperature. Absolute zero = 0 K.
 ℃ = ( + 273)

4.4.5 Pressure Law – Relationship between Pressure and Temperature

1. Pressure Law states that for a fixed mass of gas, the pressure of the

gas is directly proportional to its absolute temperature provided the

volume is remains constant.

2. Mathematically Pressure law can be expressed as

 ∝ , that is = constant or 1 = 2
1 2

 P1 is the original pressure

 P2 is the new pressures

 T1 is original temperature in Kelvin scale

 T2 is the new temperature in Kelvin scale

3. When the temperature of the gas is reduced to 0 oC, the molecules still have kinetic energy which

is less than that at room temperature. The gas exerts a lower pressure.

4. The temperature at which the pressure of the gas is expected to become zero can be obtained by

extrapolating the graph of pressure against temperature.
5. It is found that at – 273 oC the pressure is expected to become zero.

9

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Mr Ng Han Guan Form 4 Chapter 4: Heat Fo
Guru Cemerlang Physics MSAB
Relationship
Gas Law

Boyle’s law For a fixed mass of gas, the volume of the gas
is inversely proportional to its pressure if the PV =
temperature remains constant.
P1V

(T is



For a fixed mass of gas, the volume of the gas =
is directly proportional to its absolute
temperature if the pressure is remains
Charles’ law constant. 1
1

(P is



For a fixed mass of gas, the pressure of the gas = c
Pressure law is directly proportional to its absolute

temperature if the volume is remains constant. 1
1

(V is

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Date: Graph

ormula 10

∝ 1


= constant

V1 = P2V2

constant)

∝

constant

1 = 2
1 2

constant)

∝

constant

1 = 2
1 2

constant)

Mr Ng Han Guan Form 4 Chapter 4: Heat
Guru Cemerlang Physics MSAB
1. The popping of a Balloon
Gas Law Eksperiment When we try to squeeze a ba
the gas inside, which increas
Boyle’s law the added pressure it bursts.
2. Increase in size of bubbl
Charles’ law We've all seen movies show
growing as they rise to the su
Pressure law nature. We know that the de
as the bubbles rise to the su

1. Ping Pong Balls
Little children come up with i
removing the dent from a pin
without being punctured the
Since the air inside the ball t
volume increases as a result
inside keep same as outside
a proportional increase in vo
2. Inflated football deflates
Try inflating a football indoor
seems deflated. This is beca
When the ball was brought o
inside the ball dropped too, m
1. Popping party balloons
During outdoor parties a com
When the air inside the ballo
pressure.
2. Exploding beer/soda can
Beer or soda cans and bottle
place”. The reason being tha
them. When exposed to dire
However since the volume is
pressure. The temperature in

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Date:

Applications

n
alloon we are actually trying to reduce the volume of
ses the pressure. Since the balloon cannot withstand
.
les as they rise to the surface
wing scuba divers breathing under water and bubbles
urface. This is a good example of Boyle's Law in
eeper we go in the sea, more the pressure. Therefore
urface the pressure decreases and accordingly the volume increases.

ingenious ways of mending their toys. One of them is
ng pong ball. When a ping pong ball gets dented
best solution is to dip it for a while in warm water.
tries to match the temperature of the water outside,
t popping the dented part back into place (pressure
e). This shows how an increase in temperature caused
olume according to Charles’ law.
in winter
rs on a chilly winter day. When playing outside it will be noticed that the football
ause of the change in temperature from the warm indoors to the chilly outdoors.
outside the temperature dropped and proving Charles’ law, the pressure of the air
making the ball seem deflated.

mmon nuisance has to replace popped party balloons.
oon heats up due to the sun they pop due to increased

ns
es have a label on them stating “Store in a cool, dry
at these cans have a lot of artificial pressure stored in
ect sunlight/heat, the pressure inside the cans rise.
s constant the pressure increases to a limit where they burst, letting out all the
ncrease in the can resulted in the increase in pressure resulting in the explosion.

11

Mr Ng Han Guan Form 4 Chapter 5: Light Date:
Guru Cemerlang Physics MSAB
CHAPTER 5: LIGHT

5.1 UNDERSTANDING REFLECTION OF LIGHT

5.1.1 The Law of Reflection of Light
1. The angle of incidence equals the angle of reflection.
2. The incident ray, the reflected ray and normal line are

all lie at the same plane.

5.1.2 Reflection of Plane Mirror Normal Reflected angle
Incident angle

Incident ray Reflected ray

ir

Plane mirror

1. The characteristics of an image formed by a plane mirror:
i. same size as an object,
ii. the image distance from the mirror same as the object distance
iii. upright
iv. virtual
v. laterally inverted (left and right are interchanged)

2. Virtual image is that image which cannot be projected (focused) onto a screen.

5.1.2.1 Ray Diagram

1. The following steps show how to draw the ray diagram for the formation of an image:

i. Consider an object O placed in front of a plane mirror. Measure object distance.

ii. Measure the same distance at the other side of the mirror (difference side with object) and

mark the position as the image, I. A B Image C
iii. Draw two diverging rays from a point on object

the image to the corner of the eye. i1
iv. Use dotted line (virtual) for ray from the r1

image to the mirror.

v. Finally, draw two diverging rays from the

Eye

object to the mirror to meet the diverging

rays from the image.

2. Complete the image of L in the diagram of

1

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Mr Ng Han Guan Form 4 Chapter 5: Light Date:
Guru Cemerlang Physics MSAB
Mirror
reflection below.

i
r

i. If the plane mirror above move towards the object L by 1 m, the image of L will move towards
how far? Answer: 2 m

3. Determine the angle of incidence in the diagram of reflection below. Answer: 30o

i. Angle between the incident ray and reflected ray is 60o.
ii. If the incident ray remain the same but the mirror rotate clockwise direction by 10o, determine

the angle between incident ray and reflected ray. Answer: 40o

5.1.3 Reflection of Curved Mirrors

1. Figure shows the curvature of a curved mirror.

2. There are to types of curved mirrors:
i. Convex mirror – surface curved outwards
ii. Concave mirror – surface curved inwards

3. The centre of curvature, C

- The centre of sphere of the mirror

4. The radius of curvature, r 6. The focal point, F

- The distance between C and surface of - The point inside the mirror where the rays

the mirror parallel to the principal axis converge or

- r = 2f diverge

5. The focal length, f 7. The principle axis

- Distance FP // ½ CP // ½ r - The connecting line from the centre of

curvature, C to point P

2

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Mr Ng Han Guan Form 4 Chapter 5: Light
Guru Cemerlang Physics MSAB
Date:

5.1.3.1 Characteristics of a Concave Mirror’s Image

It depends on the object distance:

- u<f - 2f > u > f - u > 2f
- u at infinity
- u=f - u = 2f
- upright
5.1.3.2 Characteristics of a Convex Mirror’s Image

It no depends on the object distance, and its image always:

- Virtual - Smaller than the object

5.1.3.3 Principle of Drawing Ray Diagrams for Curve Mirror:
1. Rays parallel to the principal axis are reflected through the principal focus, F.

CF P P FC

Concave mirror Convex mirror

2. Rays passing through the principal focus are reflected parallel to the principal axis.

CF P P FC

Concave mirror Convex mirror

3. Rays passing through the centre of curvature are reflected directly back.

CF P P FC

Concave mirror Convex mirror

3

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Mr Ng Han Guan Form 4 Chapter 5: Light
Guru Cemerlang Physics MSAB
Date:
5.1.3.4 Image Formed by Concave Mirror
r = radius of curvature
u = object distance; v = image distance; f = focal length;

Case 1: u at infinity

CF F

image Diminished ii) Real iii) Inverted
Characteristics of image formed: i) iii) Inverted

Case 2: u > 2f iii) Inverted

object F iii) Inverted

C F 4
image

Characteristics of image formed: i) Diminished ii) Real

Case 3: u = 2f or u = r

object F F

C
image

Characteristics of image formed: i) Same size ii) Real

Case 4: f < u < 2f

object F F

C

image

Characteristics of image formed: i) Magnified ii) Real

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Guru Cemerlang Physics MSAB
F
Case 5: u = f

object

CF

Characteristics of image formed: i) Image at infinity

Case 6: u < f

object image

CF F

Characteristics of image formed: i) Magnified ii) Virtual iii) Upright

5.1.3.5 Image formed by convex mirror:
u = object distance ; v = image distance ; f = focal length ; r = radius of curvature

object F image

C F

Characteristics of image formed: i) Diminished ii) Virtual iii) Upright

5.1.4 Application of Reflection of Light – Plane Mirror
1. Mirror in a meter (voltmeter or ammeter)

- The mirror is used to help our eyes to see at the correct position
and prevent parallax error in our reading.

- Eyes are perpendicularly (normal) to the pointer when the image of
the pointer in mirror cannot be seen.

2. Periscope
- Used to see over the top of high obstacles such as a wall. It is also
used inside a submarine.

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5.1.5 Application of Reflection of Light – Curved Mirror

Newton’s Telescope:

Plane mirror

Lens
Eye

Concave mirror

Car head lamp Curved mirror

lamp

OFF ON Where the lamp should be placed to achieve the
above result? Answer: At the principal focus

5.2 UNDERSTANDING REFRACTION OF LIGHT

1. Light travels in a straight line through transparent materials.
2. Refraction of light is a phenomenon where the direction of

light is changed when it crosses the boundary between two
materials of different optical densities.
3. When light travels from a less dense medium to a denser
medium, the ray is refracted toward the normal at point of
incidence.
4. When light travels from a denser medium to a less dense
medium, the ray is refracted away from the normal at point
of incidence.

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5.2.1 The Law of Refraction of Light

1. The incident ray, the refracted ray and normal line at the point of incidence all lie in the same

plane.

2. The ration of sin i is a constant, where i is the angle of incidence and r is the angle of refraction.
sin r

- This is also known as Snell’s Law

- sin i (vacuum)  n , n is the refractive index for that medium.
sin r (medium)

 Value of n for all transparent materials must be larger than 1 (n > 1) except for air or

vacuum (n =1).

 n  sin i (air or vacuum)
sin r (medium)

 n  c (speed of light in air or vacuum, 3 x 108 m s-1)
v (speed of light in medium)

 n  H (real depth)
h (apparentdepth)

3. Solve all the questions below:
i)

Refractive index of liquid X is 1.45 Refractive index of medium Y is 1.40

Angle of r is 22o
Angle of i is 35o

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4. Steps to construct the refraction of light of an

object in the water.

i) Draw the two normal lines at the boundary

of liquid

ii) Draw and show the two rays refracted at

the air (further from normal line)

iii) Draw an eye at the correct position

iv) From the eye, followed the refracted rays,

extrapolated dotted lines backward and

meet above the original object. Draw a

dotted image.

5. i) Complete the diagram below to shows the
image of the coin in the water.

ii) If the coin is at an actual depth of 4 m and
the refractive index of water is 1.33, what is
the apparent depth of the image?
Answer: 3 m

6. The speed of light in liquid X is 1.5 x 108 m s-1.
Calculate the refractive index of the liquid X.

7. Draw the path of a ray of light with an angle of incidence 45o refracted through water and then

through a glass block, as shown in figure. Use a protractor to mark the angle of incidence and the

angle of refraction to actual scale. Given the refractive index of water is 1.33 and the refractive

index of glass is 1.50.

45°

32° 32°
Water
Glass block 28°
28°

45° 8
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5.3 UNDERSTANDING TOTAL INTERNAL REFLECTION OF LIGHT

1. Diagrams below show a light ray travels through a semicircle glass block. If the angle of incident,

i, is small, most of the light is refracted but some is reflected.

2. The angle of refracted, r, increases as the angle of incident, i, increases.

3. When the angle of refracted, r, increases until maximum at 90o, now the angle of incident, i, is
known as critical angle, c.

4. The total internal reflection occurs when the incident angle, i, is greater than the critical angle, c.

5. The total internal reflection only occurs when,
i) Light travels from a optically denser medium to a optically less dense medium;

ii) The incident angle, i, is greater than the critical angle, c.
6. The critical angle, c, of the medium can related to the refractive index of the medium:

n  sin 90o  1  c  sin1 1 
sin c sin c n

7. Figure shows a ray of light travelling from air to medium and to the air again.

i) Determine the reflective index of the medium.

n sin 90o  1  1.35
sin (90o  42o ) sin 48o

ii) Find the angle of y.

n  sin i
sin r

1.35  sin(90o  y)
s in42o

y  25o

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5.3.1 The Phenomenon Involving Total Internal Reflection of Lights

1. Mirage

 On a hot day, a distant road will appear to have pools of water lying on the surface. This
phenomenon calls as mirage.

 It’s occurs is due to the different temperature layers of air, hot air on the ground is less dense
compare with cold air at higher level on hot day.

 The sun ray will travel from cool air layer to hot air layer (means from denser to less dense).
The sun ray will always refract away from the normal.

 The total internal reflection will occurs when the angle of incident ray is larger than critical
angle and the light ray is reflected upwards.

 Our eye sight sees straight as though there are pools of water (virtual image of reflection of
sky) on the ground.

2. Rainbow

 When sunlight shines on millions of water droplets in the air
after rain, we see a colourful natural phenomenon called
rainbow. It caused by refraction, dispersion* and total internal
reflection of light within water droplets.

 When white light from the sun enters the raindrops (less dense to denser), it is refracted and
dispersed into its various colour components inside the raindrops.

 When the dispersed light hits the back of the raindrop, it undergoes total internal reflection
(denser to less dense, incident angle greater than critical angle).

 It is then refracted again as it leaves the drop (incident angle less than critical angle).

*Dispersion is the separation of white light into its component colours:
ROYGBIV (Red, Orange, Yellow, Green, Blue, Indigo and Violet)

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3. Sparking of Diamond

 It is caused by multiple total internal reflections and dispersion of white light.

 Density of diamond is very high and makes its critical angle

become very small (about 24.4o). So that the light will be very

easily to occurs total internal reflection.

 Any ray which strikes the surface on the side at an angle greater

than 24.4o will not escape from the diamond.
 Reflected light gives the diamond its “sparkle”.

5.3.2 Applications of Total Internal Reflection glass
1. Glass prism

 The critical angle of the glass is about 42o.
 Optical instruments make use of total internal reflection within

prisms rather than reflection by mirrors.
 Images produced by total internal reflection are brighter than

those produced by mirrors.
 An image produced is upright and not laterally inverted.

i) Prism Periscope

ii) Prism Binoculars

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2. Optical Fibres

 An optical fibre consists of an inner core of high refractive index glass and surrounded by an
outer cladding/protective material of lower refractive index.

 When light is introduced into the inner core at one end, it will propagate along the fibre.
 This is because of light in the inner core will undergo a series of total internal reflection.
 It uses in many important applications like

fibre optic diagnostic tools in medicine and
fibre optic cables in telecommunications.

5.4 UNDERSTANDING LENSES ii) Concave lens
There are two types of lenses:

i) Convex lens

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Mr Ng Han Guan Form 4 Chapter 5: Light
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Date:

5.4.1 Important Terms In Ray Diagram of Lenses:

i) Focal point, F: A common point on the principal axis where all rays parallel to the

axis converge to it after passing through a convex lens, or

appear to diverge from it after passing through a concave lens.

ii) Focal length, f: The distance between the focal point and the optic centre.

iii) Object distance, u: The distance between the object and optic centre

iv) Image distance, v: The distance between the image and the optic centre

5.4.2 Image Formed by a Convex Lens

It depends on the object distance:

- u<f - 2f > u > f - u > 2f
- u at infinity
- u=f - u = 2f

5.4.3 Image Formed by a Concave Lens

It no depends on the object distance, and its image always:

- Virtual - Upright - Diminished

5.4.4 Principle of Drawing Ray Diagrams for Lenses:
1. Ray parallel to the principal axis is refracted through the principal focal point, F.

2. Ray passing through the principal focal point, F is refracted parallel to the principal axis.

3. Ray passing through the optical centre, C travels straight without bending.

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Mr Ng Han Guan Form 4 Chapter 5: Light
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5.4.5 Image Formed by a Convex Lens
iii) Diminished
Case 1: u at infinity iii) Diminished
iii) Same size
Objec F iii) Magnified
t ii) Inverted
14
2F F

Characteristics of image: i.) Real
Case 2: u > 2f

Objec F
t ii) Inverted

2F F F
ii) Inverted
Characteristics of image: i) Real
Case 3: u = 2f

Object F
2F

Characteristics of image: i) Real
Case 4: 2f > u > f

Object F F
2F ii) Inverted

Characteristics of image: i) Real
Case 5: u = f

Object F
2F F

Characteristics of image: i) Image at infinity

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Case 6: u < f

2F F F
Object ii) Upright

Characteristics of image: i) Virtual iii) Magnified

Note:
All real images are inverted (u>f).
All virtual images are upright (u<f).

5.4.6 Image Formed by a Concave Lens

Case 1: u > 2f

Object F
2F F ii) Upright

Characteristics of image: i) Virtual iii) Diminished
Case 2: 2f > u > f

Object F

2F F

Characteristics of image : i) Virtual ii) Upright iii) Diminished

Note:
Image is always upright, diminished, and virtual.
Position of image is on the same side of the lens as the object.

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Mr Ng Han Guan Form 4 Chapter 5: Light
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5.4.7 Equations for Lenses

1. The relationship between object distance, image distance and focal length of the lens

 1 11
f uv

 For convex lens, the focal length, f is positive.
 For concave lens, the focal length, f is negative.
 If v is positive, means that image is real image and formed behind the lens.
 If v is negative, means that image is virtual image and formed in front of the lens.

Note:
Characteristics of image formed by lenses can be determined
by this equation instant of ray diagram. Can you do it?

2. Magnification

 Linear magnification, m  Height (size) of image  hi  v
Height (size) of object ho u

 For telescope at normal adjustment, m  fo
fe

- fo = focal length of objective lens
- fe = focal length of eyepiece lens
 For microscope, M  mo  me

- mo = magnification by the objective lens = mo  First image, h1
Object, h0

- me = magnification by the objective lens = me  Final image, h2
First image, h1

3. The power of a lens

 Power, p  1  100
f (m) f (cm)

 Measured in unit dioptres, D
 Power of convex lens is positive
 Power of concave lens is negative.
 High power lenses are ticker and have short focal lengths.

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5.4.8 The Use of Lenses in Optical Devices

1. Magnifying glass (simple microscope)
 A magnifying glass consists of a converging lens.
 The object must be placed at a distance less than f in order for the lens to act as a
magnifying glass.
 The characteristics of the image formed by a magnifying glass are virtual, upright,
magnified and on the same side as the object.
 Greater magnification can be obtained by using a lens which has short focal length (high
power lens).

2. Astronomical telescope

 It used to view very distant objects like the planets and the stars.

 The astronomical telescope consists of 2 convex lenses as the

objective lens and the eyepiece lens.

 The objective lens has long focal length, fo (low power lens).
 The eye lens has short focal length, fe (high power lens).
 The distance between the objective lens and the eye lens is equal to the sum of their

individual focal lengths: l = fo + fe  normal adjustment or l < fo + fe

 Magnification of telescope at normal adjustment, m  fo
fe

 Image formed by objective lens (first image) is real, inverted and diminished.

 Image formed by eyepiece lens (final image) is virtual, upright and magnified (compare with

first image that act as an object at eyepiece lens).

 Image formed by eyepiece lens (final image) is virtual, inverted and magnified (compare

with an object at objective lens).

 Ray diagram of telescope in normal adjustment:

Objective Eye
lens lens

Fo Fe

 Ray diagram of telescope to see distant objects: 17

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Objective lens Eye lens

Fe Fo

3. Compound Microscope
 It used to view very small objects like micro organisms.
 It is consists of 2 powerful convex lenses as the

objective lens and the eyepiece lens.
 The objective lens has shorter focal length, fo (higher

power lens).
 The eye lens has long focal length, fe (lower power

lens).
 The distance between the objective lens and the eye

lens is greater than the sum of their individual focal lengths: l > fo + fe
 Magnification of telescope at normal adjustment, M  mo  me
 Image formed by objective lens (first image) is real, inverted and magnified.
 Image formed by eyepiece lens (final image) is virtual, upright and magnified (compare with

first image that act as an object at eyepiece lens).
 Image formed by eyepiece lens (final image) is virtual, inverted and magnified (compare

with an object at objective lens).
 Ray diagram of microscope:

Objective lens Eye lens

Fo Fe

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4. Simple camera

5. Slide projector

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