Light
PHYSICS – UNIT TWO
Light
What is light?
Light exhibits properties of both waves and particles.
Light as a wave and a particle https://www.youtube.com/watch?v=J1yIApZtLos
Properties of light
https://www.youtube.com/watch?v=bP4i1KUfB3o
Waves
Light as a wave
Light behaves in a way that is consistent with our conceptual and
mathematical understanding of waves, as it……
• Reflects like a wave
• Refracts like a wave
• Diffracts like a wave
• Undergoes interference like a wave
Now we will take a closer look at waves to understand what all this means
Waves
What are waves?
• A disturbance that travels through a medium from one place to another.
• They originate from a vibration that initially disturbs the medium.
• They transport energy as they travel through the medium
Lets view an example of energy transfer via waves.
Guitar video The string is plucked, a wave transfers this energy to sound.
You can see that different sided waves produce different sounds
https://mrsamsingyr11physics.wordpress.com/
Waves
Transverse vs Longitudinal Waves
Longitudinal Waves are waves in which the particles of the medium are
In Transverse Waves, the particles of the medium are displaced in a direction
perpendicular (right angled) to the direction of the energy transport
https://www.youtube.com/watch?v=7cDAYFTXq3E
displaced in a direction parallel to the energy transport.
Waves
Transverse vs Longitudinal Waves
Longitudinal Waves are waves in which the particles of the medium are
In Transverse Waves, the particles of the medium are displaced in a direction
perpendicular (right angled) to the direction of the energy transport
https://www.youtube.com/watch?v=7cDAYFTXq3E
displaced in a direction parallel to the energy transport.
Waves
Longitudinal Waves
Longitudinal Waves are waves in which the particles of the medium are
displaced in a direction parallel to the direction of the energy transport.
Direction of the vibration through the medium
The wave moves in this direction
Waves
Transverse Waves
Transverse Waves are waves in which the particles of the medium are
displaced in a direction perpendicular (right angled) to the direction of the
energy transport
The wave moves in this direction
Direction of the
vibration through the
medium
Transverse
Waves
Pulsed vs Continuous Wave
Pulsed Wave
• When a single disturbance is passed through
a medium
• The wave carries the energy which gradually
oscillates away once the energy is used
eg. An Explosion or a Sudden Impact
Continuous Wave
• When a continuous disturbance is
passed through a medium
generated by repetitive motion
• Energy is carried away in the form of
a continuous wave
eg. A vibration of sound waves
from a speaker
NOW DO
Problems from the alternative text
Ex1.1A Q 1, 2, 3, 4, 5, 6, 7
Transverse
Waves
Amplitude
• Is a measure of how much energy the wave has.
• Is measured from the midpoint to the peak of the crest (or bottom of the trough).
• Can be defined as the maximum displacement from the average position.
Amplitude
Crest
Trough
Transverse
Waves
Wavelength (𝜆)
• Is the distance between two peaks (or the distance between two troughs)
• Can be defined as ‘the distance the wave has travelled during one complete cycle’.
• Given by the symbol
𝜆
( pronounced ‘lam-da’ )
• Measured in metres (as it’s a length)
Wavelength ( λ )
Crest
Trough
Transverse
Waves
Frequency (𝑓)
• The Frequency of a wave is defined as ‘the number of complete cycles in one second’.
• Units: Hertz (Hz)
• 1 Hz = 1 cycle per second
One complete cycle
Transverse
Waves
Period (𝑇)
• The Period of a wave is defined as ‘the time taken for one complete wave cycle’.
• Measured in seconds
𝑃𝑒𝑟𝑖𝑜𝑑 =
1
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
Which can be rearranged to give:
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 =
Period - One complete cycle
1
𝑝𝑒𝑟𝑖𝑜𝑑
Transverse
Waves
Velocity of a wave
In a constant medium, the velocity of the wave is given by
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 × 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ
→
𝑣 = 𝑓λ
Which can be rearranged to give:
𝑓=
𝑣
λ
or
λ=
𝑣
𝑓
Or as usual, if we know a distance
travelled and a time taken to travel
that distance, we can use
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑣=
𝑡𝑖𝑚𝑒
Transverse
Waves
Examples
eg1. A wave undergoes 42 vibrations in 60 seconds.
a) Find the frequency of the wave.
b) Find the period of a wave.
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝐶𝑦𝑐𝑙𝑒𝑠 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑
42
=
60
= 0.7 𝐻𝑧
𝑃𝑒𝑟𝑖𝑜𝑑 = 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑜𝑛𝑒 𝑐𝑦𝑐𝑙𝑒
1
=
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
1
0.7
=
= 1.43 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
Transverse
Waves
Examples
eg2. On a guitar string, the trough of the wave travels 10cm in 2 seconds.
Find the speed of the wave in metres per second..
𝑆𝑝𝑒𝑒𝑑 =
=
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
0.10 𝑚
2𝑠
= 0.05 𝑚/𝑠
Transverse
Waves
Examples
eg3. A bird lands on a lake, causing vibrations resulting in a ripple of 30 waves
in 5 seconds.
a) Find the frequency of the waves
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝐶𝑦𝑐𝑙𝑒𝑠 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑
30
=
5
= 6 𝐻𝑧
b) Find the period of each wave
𝑃𝑒𝑟𝑖𝑜𝑑 = 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑜𝑛𝑒 𝑐𝑦𝑐𝑙𝑒
1
=
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
1
6
=
= 0.167 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
Transverse
Waves
Examples
eg4. If waves on a lake vibrate at a frequency of 30 Hz and the wave crests are
1cm apart, what is the speed of the waves?
𝑣 = 𝑓λ
= 30 × 0.01
= 0.30 𝑚/𝑠
NOW DO
Problems from the alternative text
Q1.2A Q1 – 7
Q1.3A Q1 – 4
NOW DO
Problems from the alternative text
Worksheet – Waves equations
Q1.2A Q9 – 10
Behaviour of light
Light can be absorbed, reflected or transmitted
We will look closer at these behaviours, but
firstly we need to look at another model for describing light.
The Ray model
We can understand many phenomena involving light without having to use
sophisticated models such as the wave model or the particle model.
Instead, we describe light and it’s interactions with objects more simply according to
the path it takes.
To model and draw light, we can think of light travelling in a straight line path, called
light rays. This is called the ray model.
The Ray model
Light rays
transmitted
from a light
source
Transmitted light rays from the
sun hit a tree – the shadow
formed shows than light does
not pass through the tree –
where does it go?
Some light is reflected –
enabling us to see the tree
Some light is absorbed –
providing energy to the tree.
Straight rays of
light transmit
through a
curved medium
– the light
refracts and
converges
Rays of light reflect off
the fish in the medium
(water) and transmit
through to a different
medium (air) causing the
rays to diverge due to
the refraction.
Some Light Terminology
Luminous – Objects that we see as they
give off their own light. eg. sun, lamp.
Non-Luminous – Objects that we see as
light reflects off them. eg. people, book.
Incandescent – A luminous object that
produces light due to being heated.
Brightness and colour changes
depending on how hot the object is.
eg. Gas heater/cooker – blue at source = hot!
Yellow flame = cooler
Some Light Terminology
Opaque – Opaque objects reflects some
light and absorbs the rest. No light passes
through. eg. A brick wall, tiled floor.
Transparent – Objects that allow a
significant amount of light to pass through.
Some light may also be absorbed and/or
reflected. eg. Clear windows, cling wrap.
Translucent – Objects that allow a some light
to pass through, but no clear image can be
seen through the material.
eg. Frosted bathroom windows, tissue paper.
Behaviour of Light Rays
Ex 1.1B Q1 – 6
Have a think about Question 2,
attempt it and then we can do it together
Reflection
off perfectly flat surfaces
Light rays travel off a surface into the direction of the mirror
and reflect off into the direction of the eye.
Plane Mirror Reflection
A normal household mirror is constructed with 3 separate layers:
• Transparent Glass
• Thin coating of Aluminium or Silver deposited on the glass to reflect light
• A backing layer of protective paint
When a beam of light strikes the surface of the mirror;
~4% of the light is reflected off the glass,
~96% of the light is reflected of the metal.
Because this is near perfect reflection, we use
the plane mirror as our model to display reflection
Plane Mirror Reflection
The light ray that approaches the mirror is called the incident ray.
The ray leaving the mirror is called the reflected ray.
The normal is perpendicular (right angle) to the mirror.
The angle of incidence (𝑖) is the angle between the
incident ray of the normal.
The angle of reflection 𝑟 is the angle between the
reflected ray and the normal.
𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒 𝑖 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 (𝑟)
Behaviour of Light Rays –
Reflection off irregular surfaces
When a surface is not perfectly flat like a plane mirror, the reflection is known as
diffuse reflection.
Each light ray still reflects off the surface with
𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒 𝑖 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 (𝑟),
however because the surface is irregular, the
reflected rays emerge in different directions.
Behaviour of Light Rays –
Forming images with a plane mirror
• Images are a likeness of an object.
• When viewing an object with a mirror, we see a virtual image.
• Rays of light from the cats ear reflect off the mirror and enter the viewers eye and
appear to come from a region behind the mirror.
• The real image and the virtual image are located at equal distances from the
mirror.
“Real
Image”
“Virtual
Image”
Behaviour of Light Rays – Drawing ray diagrams
Summary, Drawing ray diagrams for plane mirror reflection
https://www.youtube.com/watch?v=OHqYB42WfZk
Drawing ray diagrams – Plane Mirrors
“Virtual
Image”
“Real
Image”
Drawing ray diagrams – Plane Mirrors
Plane Mirror Reflection – The
Image you see is:
UPRIGHT,
LATERALLY INVERTED
Behaviour of Light Rays - Reflection
Ex 1.2B Q1, 2, 4, 6, 7
Refraction
• Light travels in straight line paths while travelling in a uniform medium.
• When it passes through another medium, the light ray changes direction, or its
path bends.
• Refraction occurs as the light changes speed in different media, due to the
change in density of different substances.
Behaviour of Light Rays - Refraction
• In 1621, Dutch physicist, Willebrand Snell investigated refraction of light
• He found that the ratio of the sines of the angles of incidence and angles of
refraction was constant for all angles of incidence for a given pair of media.
𝑆𝑛𝑒𝑙𝑙 ′ 𝑠 𝐿𝑎𝑤:
𝑠𝑖𝑛 𝜃𝑖
𝑠𝑖𝑛 𝜃𝑟
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
This ‘constant’ value (n)
changes depending on
the pair of media
Behaviour of Light Rays - Refraction
• Snell repeated the experiment & found the ratio changed with different substances.
• This showed that different substances bend light by different amounts.
• The value of the ratio (of any medium compared to a vacuum) is called the
refractive index, given by the symbol
n.
• We can rewrite 𝑆𝑛𝑒𝑙𝑙 ′ 𝑠 𝐿𝑎𝑤 to solve angles
and n values for any refracted ray:
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
Behaviour of Light Rays - Refraction
eg1. A ray of light travelling through air (n≃1), strikes a glass block with a refractive
index of 1.8 at an angle of incidence of 40°. What is the angle of refraction?
𝑛1 = 1
𝑛2 = 1.8
𝜃1 = 40°
𝜃2 = ?
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
1 sin 40 = 1.8 sin 𝜃2
𝜃2 =
sin−1 (
sin(40)
)
1.8
𝜃2 = 20. 92°
Behaviour of Light Rays - Refraction
eg2. A ray of light travelling through air (n≃1), strikes a plastic block at an angle of
incidence of 35°, refracting at an angle of 15°. What is the refractive index of the
plastic block?
𝑛1 = 1
𝑛2 =?
𝜃1 = 35°
𝜃2 = 15°
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
1 sin 35 = 𝑛 sin(15)
sin( 35 )
𝑛=
sin( 15 )
𝑛 = 2. 22
Behaviour of Light Rays - Refraction
Table:
Some common refractive index values
These were derived by:
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑎 𝑣𝑎𝑐𝑢𝑢𝑚
𝑛=
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑒𝑑𝑖𝑢𝑚
Behaviour of Light Rays - Refraction
𝜃𝑖 < 𝜃𝑟
𝑛𝑚𝑒𝑑 1 > 𝑛𝑚𝑒𝑑 𝐴
𝑣𝑚𝑒𝑑 1 < 𝑣𝑚𝑒𝑑 𝐴
𝜃𝑖 > 𝜃𝑟
𝑛𝑚𝑒𝑑 1 < 𝑛𝑚𝑒𝑑 𝐵
𝑣𝑚𝑒𝑑 1 > 𝑣𝑚𝑒𝑑 𝐴
𝜃𝑖 = 𝜃𝑟
𝑛𝑚𝑒𝑑 1 = 𝑛𝑚𝑒𝑑 𝐶
𝑣𝑚𝑒𝑑 1 = 𝑣𝑚𝑒𝑑 𝐴
Behaviour of Light Rays - Refraction
As the speed of light in a vacuum (c) = 𝟑 × 𝟏𝟎𝟖 𝒎/𝒔, we can use this to
calculate the speed of light in different media:
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑣𝑎𝑐𝑢𝑢𝑚
𝑐
3 × 108
=
=
=
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑒𝑑𝑖𝑢𝑚
𝑣𝑚𝑒𝑑𝑖𝑢𝑚
𝑣𝑚𝑒𝑑𝑖𝑢𝑚
∴ 𝑣𝑚𝑒𝑑𝑖𝑢𝑚 =
𝑐
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
sin 𝜃𝑖 sin 𝜃1 𝑣1 𝑛2
∴
=
=
=
sin 𝜃𝑟 sin 𝜃2 𝑣2 𝑛1
eg1. A ray of light travels through a medium at a speed of 1.2 × 108 𝑚/𝑠.
a) What is the refractive index of the medium?
𝑛 =?
b) The light enters a different medium with a
refractive index of 1.7. What speed is the light
travelling at in this medium?
𝑣 = 1.2 × 108 𝑚/𝑠
𝑛𝑚𝑒𝑑𝑖𝑢𝑚 =
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
𝑐
𝑣𝑚𝑒𝑑𝑖𝑢𝑚
𝑣1 𝑛2
=
𝑣2 𝑛1
3 × 108
=
𝑣𝑚𝑒𝑑𝑖𝑢𝑚
𝑛2
𝑣1 = 𝑣2 ×
𝑛1
8
108 ×
3 × 10
=
1.2 × 108
= 2.5
𝑣1 = 1.2 ×
2.5
1.7
𝑣 ≅ 1.7647 × 108 𝑚/𝑠
eg2. A ray of light travels through a medium that has a refractive index of 𝑛 = 1.96
a) At what speed does the light
travel through the medium?
𝑛 = 1.96
𝑣 =?
𝑣𝑚𝑒𝑑𝑖𝑢𝑚 =
𝑐
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
𝑣𝑚𝑒𝑑𝑖𝑢𝑚
3 × 108
=
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
𝑣𝑚𝑒𝑑𝑖𝑢𝑚
3 × 108
=
1.96
= 1.53 × 108 𝑚/𝑠
b) The ray leaves the medium at an angle of
incidence of 25°, into air.
What angle is the light at once entering air?
sin 𝜃1 𝑛2
=
sin 𝜃2 𝑛1
sin 𝜃𝑎𝑖𝑟
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
=
sin 𝜃𝑚𝑒𝑑𝑖𝑢𝑚
𝑛𝑎𝑖𝑟
sin 𝜃𝑎𝑖𝑟 1.96
=
sin 25
1
𝜃𝑎𝑖𝑟 = sin−1 (1.96 sin 25)
𝜃𝑎𝑖𝑟 = 55.9°
Angle of Deviation
The term angle of deviation refers to the difference between the
Angle of incidence and the Angle of refraction
eg. If the 𝜃𝑖 = 40° and 𝜃𝑟 = 25°,
find the angle of deviation.
Angle of deviation = 40° − 25°
= 15°
Apparent Depth
• Water is more dense than air so it has a larger refractive index than air.
• This causes light to travel slower in water than it does in air.
• When viewed from above, objects appear to be shallower than they actually are.
• Where the object is called ‘Real Depth’
• Where the object appears to be is ‘Apparent Depth’
We can calculate these using:
𝑅𝑒𝑎𝑙 𝐷𝑒𝑝𝑡ℎ
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝐷𝑒𝑝𝑡ℎ
= 𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝐼𝑛𝑑𝑒𝑥
Apparent Depth
𝑅𝑒𝑎𝑙 𝐷𝑒𝑝𝑡ℎ
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝐷𝑒𝑝𝑡ℎ
= 𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝐼𝑛𝑑𝑒𝑥
eg1. Water has a refractive index ≅ 1.33. A fish appears to be 1.2 metres
beneath the surface, but how far is it in reality?
𝑅𝑒𝑎𝑙 𝐷𝑒𝑝𝑡ℎ
1.2
= 1.33
𝑅𝑒𝑎𝑙 𝐷𝑒𝑝𝑡ℎ = 1.33 × 1.2
𝑅𝑒𝑎𝑙 𝐷𝑒𝑝𝑡ℎ = 1.596 𝑚
Apparent Depth
𝑅𝑒𝑎𝑙 𝐷𝑒𝑝𝑡ℎ
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝐷𝑒𝑝𝑡ℎ
= 𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝐼𝑛𝑑𝑒𝑥
eg2. Water has a refractive index ≅ 1.33. If the fish is 4 metres beneath the
surface, how deep does it appear to be?
4
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝐷𝑒𝑝𝑡ℎ
= 1.33
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝐷𝑒𝑝𝑡ℎ =
4
1.33
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝐷𝑒𝑝𝑡ℎ = 3.01 𝑚
Behaviour of Light Rays - Refraction
Ex2.2B Q1-10
Internal Reflection
Depending on:
• The angle of incidence, and
• The medium that light is travelling from/to;
a single light ray may be:
* Refracted only
* Both Reflected and refracted
* Reflected only
Internal Reflection
• Internal reflection involves light moving to a less dense medium
(ie. Lower refractive index ‘n’ )
• As the angle of incidence of a light ray increases, more of the light ray is
reflected and less refracted, as seen below.
• As the angle of incidence increases, so does the angle of refraction
Internal Reflection – Critical Angle
• As we increase the angle of incidence, eventually this results in a reflected angle
of almost 90° - the transmitted ray runs along the surface of the water
• When this occurs, the angle of incidence is called the critical angle 𝜃𝑐
Total Internal Reflection
• If the angle of incidence increases beyond the ‘critical angle’, no light is
refracted – it is all reflected
• This is called Total Internal Reflection
Critical Angle
The critical angle can be calculated using Snell’s law.
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
For these problems always use angle of refraction (𝜃2 ) = 90°
eg. If a light ray travels from water to air, calculate the critical angle.
𝑛1 = 1.3
𝜃1 =?
1.3 𝑠𝑖𝑛 𝜃1 = 1 𝑠𝑖𝑛 90°
sin 90°
−1
sin (
1.3
𝑛2 = 1
𝜃1 =
𝜃2 = 90°
𝜃1 = 50.28°
)
Behaviour of Light Rays
Practical Activity
Behaviour of Light Rays
Ex 2.3B Q1 – Q3 & Q5 – 6
Research Questions
Ex 2.3B Q7 – 8
(all answers can be found within the text I sent to you)
Dispersion
The separation of white light into it’s different colours is called dispersion.
White light is made up of a series of colours, which separate in order:
Red Orange Yellow Green Blue Indigo Violet
Dispersion
The colours separate into this order as each colour:
• Has a different frequency
• Travels at slightly different speeds
to each other in the same medium
Refractive Index of colour
This means that the refractive index of a given medium varies for different
colours of light travelling through the medium.
Red travelling through a medium refracts the least, Violet refracts the greatest.
ROYGBIV
Refractive index increasing
Refractive Index of colour
ROYGBIV
Refractive index increasing
In glass, the refraction of colour can be expressed in terms of Snell’s Law by:
𝑛𝑎𝑖𝑟 sin 𝜃𝑖 = 𝑛𝑔𝑙𝑎𝑠𝑠(𝑟𝑒𝑑) sin 𝜃𝑟𝑒𝑑 = 𝑛𝑔𝑙𝑎𝑠𝑠(𝑣𝑖𝑜𝑙𝑒𝑡) sin 𝜃𝑣𝑖𝑜𝑙𝑒𝑡
Which is still the same Snell’s Law that
we saw last term…recall:
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
eg. A beam of white light enters a glass prism with an angle of incidence of
40°. In air, white light travels at 3 × 108 𝑚/𝑠. In the prism the different
colours of light are slowed to varying degrees. The refractive index for red
light travelling through the prism is 1.50 and for violet light is 1.53. Find:
a) The angle of refraction for red light
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
𝑠𝑖𝑛 40 = 1.5 𝑠𝑖𝑛 𝜃𝑟
sin 40
−1
𝜃𝑟 = sin (
)
1.5
= 25.4 °
b) The speed of the red light in the prism
𝑣𝑚𝑒𝑑𝑖𝑢𝑚 =
=
𝑐
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
3×108
1.50
= 2 × 108 𝑚/𝑠
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
eg. A beam of white light enters a glass prism with an angle of incidence of
40°. In air, white light travels at 3 × 108 𝑚/𝑠. In the prism the different
colours of light are slowed to varying degrees. The refractive index for red
light travelling through the prism is 1.50 and for violet light is 1.53. Find:
c) The angle of refraction for violet light
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
𝑠𝑖𝑛 40 = 1.53 𝑠𝑖𝑛 𝜃𝑟
𝜃𝑟 =
sin−1 (
sin 40
)
1.53
= 24.8 °
d) The speed of the violet light in the prism
𝑣𝑚𝑒𝑑𝑖𝑢𝑚 =
=
𝑐
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
3×108
1.53
= 1.96 × 108 𝑚/𝑠
e) The angle through which the spectrum is dispersed
25.4° − 24.8° = 0.6°
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
Eg2. A beam of white light enters a different glass prism with an angle of
incidence of 30°. The refractive index for red light travelling through the
prism is 1.65 and for violet light is 1.68. Find:
a) The angle of refraction for red light
b) The speed of the red light in the prism
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
𝑠𝑖𝑛 30 = 1.65 𝑠𝑖𝑛 𝜃𝑟
sin 30
−1
𝜃𝑟 = sin (
)
1.65
= 17.64 °
𝑣𝑚𝑒𝑑𝑖𝑢𝑚 =
=
𝑐
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
3×108
1.65
= 1.81 × 108 𝑚/𝑠
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
Eg2. A beam of white light enters a different glass prism with an angle of
incidence of 30°. The refractive index for red light travelling through the
prism is 1.65 and for violet light is 1.68. Find:
c) The angle of refraction for violet light
𝑛1 𝑠𝑖𝑛 𝜃1 = 𝑛2 𝑠𝑖𝑛 𝜃2
𝑠𝑖𝑛 30 = 1.68 𝑠𝑖𝑛 𝜃𝑟
𝜃𝑟 =
sin−1 (
sin 30
)
1.68
= 17.31 °
d) The speed of the violet light in the prism
𝑣𝑚𝑒𝑑𝑖𝑢𝑚 =
=
𝑐
𝑛𝑚𝑒𝑑𝑖𝑢𝑚
3×108
1.68
= 1.79 × 108 𝑚/𝑠
e) The angle through which the spectrum is dispersed
17.64° − 17.31° = 0.33°
Polarisation
• The transverse wave model of electromagnetic radiation, proposes that
light and other electromagnetic waves travel in many planes.
• The electromagnetic model of light has oscillating electric and magnetic
fields at right angles to each other.
Polarisation
• The blocking of transverse waves except for those travelling in a single
plane is called Polarisation.
Longitudinal waves
cannot be polarised
Polarisation
• Because of these observations of polarisation, light can be described as a
transverse wave, rather than a longitudinal wave.
• The polarisation of light is observed when it passes through some materials.
• These materials which allow light waves in one plane to pass through, while
blocking light in other planes are called Polarising Filters.
Dispersion
Ex 2.5A Q1, Q3, Q5, Q6, Q7, Q9
Polarisation: Read pages 3 – 4 from your 2.5A Booklet, then do
Ex 2.5A Q2, Q4, Q10
Seeing Colour
When we look at an object and perceive a colour, you are not necessarily
seeing a single frequency of light.
For example – You are looking at a shirt which appears yellow to your eye.
In this instance, there may be several frequencies of light striking your eye
and varying intensities.
Your eye-brain system interprets the frequencies that strike your eye and
perceive the colour to be yellow.
The colour of the object IS NOT in the object, but rather in the light which
reflects off or transmits through the object.
Primary Colours of Light
• White is not a colour, but rather the presence of all frequencies of visible
light – we have learnt this to be ROYGBIV
• However this is not the only way of producing white light – it can be
produced by combining only 3 distinct frequencies of light, providing
they are widely separated on the visible light spectrum and combined at
the correct intensity.
• The most common set of 3 colours, which we refer to as the primary
colours of light, are:
Red, Green and Blue
Primary Colours of Light
Adding 2 combinations of these colours together produces other colours.
Red + Green = Yellow
Blue + Red = Magenta
Green + Blue = Cyan
Red + Green + Blue = White
Secondary Colours of Light
• We can refer to Yellow, Magenta, Cyan as the secondary
colours of light, since they are produced by the addition of
equal intensities of 2 primary colours of light.
• Varying the intensity of the
combinations will produce
countless other colours
• Play around with the widget to
produce other colours
www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=827957100468fbe97741fbd144f66
199&title=Color%20Addition&theme=red&i0=0.81&i1=0.66&i2=0.14&podSelect=&includepo
did=ColorSwatch%3AColorData&includepodid=ColorSchemaConversions%3AColorData
Secondary Colours of Light
• We can refer to Yellow, Magenta, Cyan as the secondary
colours of light, since they are produced by the addition of
equal intensities of 2 primary colours of light.
• Varying the intensity of the
combinations will produce
countless other colours
• Play around with the widget to
produce other colours
www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=827957100468fbe97741fbd144f66
199&title=Color%20Addition&theme=red&i0=0.81&i1=0.66&i2=0.14&podSelect=&includepo
did=ColorSwatch%3AColorData&includepodid=ColorSchemaConversions%3AColorData
Complementary Colours of Light
• Any 2 colours of light that when mixed together form white light are called
complementary colours of each other.
Red and Cyan are complementary colours of each other
Which other combinations do you think may be complimentary?
Blue and Yellow
Green and Magenta
Colour Subtraction
• This relates to colours being absorbed and reflected rather than added together
• Materials contain atoms which are capable of selectively absorbing one or more
frequencies of light.
• We can say that colours which are absorbed are ‘subtracted’ in order to find out
what colour is seen by the eye.
eg. White light shines on a ball - ie. Red, Blue, Green are shining onto the ball.
If Blue is absorbed and Red and Green are reflected, what colour does the ball
appear to be?
Colour Subtraction
eg. Cyan light is shining on the same material (ie. A material that is capable of
absorbing Blue and reflects Red and Green).
What colour light is reflected from the ball?
Cyan light is produced by Blue and Green light.
Blue light absorbed.
Green light reflected
C – B = (G + B) – B = G
eg. Magenta light shines on a material that absorbs blue.
What colour does the material appear to be?
Magenta light is produced by Red and Blue light.
Blue light absorbed. Red light reflected.
Material Appears RED.
M – B = (R + B) – B = R
eg. Red light shines on a material that absorbs blue.
What colour does the material appear to be?
No blue light. Red light reflected.
Material Appears RED.
eg. Blue light shines on a material that absorbs blue.
What colour does the material appear to be?
Blue light is a primary colour.
Blue light absorbed. No colour is left to be reflected so,
in the absence of reflected light,
Material Appears BLACK
Now Try Ex2.1B
Q5, Q6, Q7, Q9
Further Reading: Read Ch 2.1B and attempt Q1, 2, 3