Download Powerpoint

Light and Colour（光與顏色）
CHENG Kai Ming
Department of Physics
CUHK
Time allocation: 6 hours
1
Content

Reflection of Light（光的反射）





Geometrical Optics（幾何光學 ）
Law of Reflection（反射定律 ）
Images（像）
Plane Mirrors（平面鏡）
Spherical Mirrors（球面鏡）



Concave Mirrors（凹面鏡）
Convex Mirrors（凸面鏡）
Parabolic Mirror（拋物面鏡）
2

Refraction of Light（光的折射）



Law of Refraction（折射定律 ）
Refractive index（折射率）
Total Internal Reflection（全內反射）


Critical Angle（臨界角 ）
Thin Lenses（薄透鏡）





Convex Lenses（凸透鏡）
Concave Mirrors（凹透鏡）
Normal Lenses
Short-sighted（近視）
Long-sighted（遠視）
3

Magnification Equation & Mirror/lens
Equation



Telescope（望遠鏡）and Microscope（顯微鏡）
Fermat’s Principle of Least Time（費爾馬
最短時間原理）
Wave Properties of Light（光的波動特性）



Electromagnetic Waves（電磁波）
Electromagnetic Spectrum（電磁波譜）
Blackbody Radiation（黑體輻射）
4

Colour（顏色）




Dispersion（色散）
Primary Colours（原色）
Complementary Colours（互補色）
Selective Reflection（選擇反射）






Pigments（顏料）
Selective Transmission（選擇透射）
Selective Scattering（選擇散射）
Rainbow（彩虹）
Laser（激光）
Colour Deficiency（色弱）
5
Part 1
Reflection of Light
6
Geometrical Optics

Light travels in straight paths called
rays.
7
Law of Reflection


Incident ray, reflected ray and normal
all lie on the same plane.
i   r
Normal
Incident ray
Reflected ray
i
r
8
Law of Reflection

Regular (specular) /diffuse reflection
Regular (specular) reflection
Diffuse reflection
9
Image




The reflected ray appears to come from a
point behind the mirror.
This point is called the image.
Real image  can be captured by a screen as
a sharp image.
Virtual image  rays of light seems to
emanate from the image.
Real image
Produced by converging beams
Virtual image
Produced by diverging beams
10
Plane Mirrors

1.
2.
3.
4.
5.
mirror
Image of a real object
virtual,
upright,
laterally inverted,
the same size as the object,
and
as far behind the mirror as
the object is in front of it.
11
Plane Mirrors
B
A
C
D
12
Example
Q.
A.
A person is sitting in
front of two mirrors that M1
intersect at an angle of
90. How many images
I1
can he see?
3 images
90
I12 or I21
O
M2
I2
13
M1
I1
I21
O
M2
60
I212 or I121
I2
I12
No. of images n 
360

1
14
Spherical Mirrors

A spherical mirror: a part of a spherical
surface
Concave
Mirror
Convex
Mirror
15
Spherical Mirrors




centre of curvature C = centre of the
sphere
radius of curvature R = radius of the
sphere
focal point (principal focus) F =
midpoint between C and the mirror
focal length f = R/2
16
Spherical Mirrors
f
C
F
17
Ray Tracing





The law of reflection applies just as it does
for a plane mirror.
The normal for the reflection is drawn
between the point of incidence and C.
Principal axis = straight line drawn through C
and the midpoint of the mirror
Paraxial rays = rays that lie close to the
principal axis
Object/image at infinity = parallel rays
18
Ray Tracing (Concave Mirrors)

1.
2.
3.
For paraxial rays:
Rays parallel to the
principal axis will be
reflected passing
through the focal point.
Rays passing through
the focal point F will be
reflected parallel to the
principal axis.
Rays passing through
C will be reflected back
along its own path.
C
F
19
Concave Mirrors
Real
Object
Image
Properties of image
Beyond C
Between C
and F
At C
Real
Inverted
Real
Inverted
Between C
and F
Beyond C
Real
Inverted
At F
At 
Between F
and mirror
Behind
mirror
At C
Virtual
Upright /
Erect
Diminished
/ Reduced
Same Size
Magnified /
Enlarged
Magnified /
Enlarged 20
Concave Mirrors
21
Ray Tracing (Convex Mirrors)

1.
2.
3.
For paraxial rays:
Rays parallel to the principal axis will be reflected in a
way that it appears to be originated from the focal point.
Rays directing towards the focal point F will be reflected
parallel to the principal axis.
Rays directing towards C will be reflected back along its
own path.
C
F
22
Convex Mirrors
The image of a real object is always
1.
2.
3.
Virtual
Erect
Diminished
23
Concave and Convex Mirrors
Converging
F
Diverging
F
24
Think 1
Q.
Tom is observing a concave mirror
and claimed that he found an image
between the focus and the mirror.
What would you say about his finding?
25
Think 1
A.

Tom must be either lying or
performing the experiment
perfunctorily.
The image of a real object for a
concave mirror can be anywhere
(including anywhere behind the mirror)
except between F and the mirror.
26
Principle of Reversibility

If the direction of a light ray is reversed,
the light retraces its original path.
IO
O
I
f
27
Parabolic Mirror
For a parabolic mirror, all rays
parallel to the principal axis
(not necessarily paraxial) will
be reflected passing through
the focal point F as shown.
principal axis
28
Reflector (telescope)
(Mount)
(Aperture)
(Primary mirror)
(Incident light)
(Eyepiece)
(Focal length of primary mirror)
29
Part 2
Refraction of Light
30
Law of Refraction





Incident ray, refracted ray and normal all
lie on the same plane.
n1 sin 1  n2 sin 2
1
Medium 1
Medium 2
Snell’s law
c
2
refractive index ni  vi
c = speed of light in vacuum, defined to
be exactly 299,792,458 m/s (~3108 m/s)
31
Substance
Refractive index / Index of refraction n
Solids at 20C
Diamond
2.419
Glass, crown
1.523
Ice (0C)
1.309
Sodium chloride
1.544
Quartz - Crystalline
1.544
Quartz – Fused
1.458
Liquids at 20C
Benzene
1.501
Carbon disulfide
1.632
Carbon tetrachloride
1.461
Ethyl alcohol
1.362
Water
1.333
Gases at 0C and 1 atm
Air
1.000293
Carbon dioxide
1.00045
Oxygen
1.000271
Hydrogen
1.000139
Example
Q.


A.
A light ray strikes an air/water surface at an
angle of 46 with respect to the normal. The
index of refraction for water is 1.33. Find the
angle of refraction when the direction of the ray
is from air to water.
Medium 1 = medium of incidence, i.e. air
Medium 2 = medium of refraction, i.e. water
n1 sin 1  n2 sin  2
1sin 46  1.33sin  2
sin  2  0.54   2  32.74
33
Use the same example
Q.


A.
Find the angle of refraction when the
direction of the ray is from water to air.
Medium 1 = medium of incidence, i.e. water
Medium 2 = medium of refraction, i.e. air
n1 sin 1  n2 sin  2
1.33sin 46  1sin  2
sin  2  0.96   2  73.08
34
Refraction by a Slab
n1 sin 1  n2 sin 2
n2 sin  2  n3 sin 3
 n1 sin 1  n3 sin 3
n1  n3
 sin 1  sin  3
 1   3


Medium 1
1
2
2
3
Medium 1
Medium 2
The emergent and incident rays are parallel.
Yet is displaced laterally relative to the
incident ray.
35
Total Internal Reflection




Occurs only when n1>n2
Normal incidence means 1 = 0
When 1 , it reaches a certain value,
called the critical angle c, such that 2
= 90.
When 1  further, there is no more
refraction.
c
36
Critical Angle
n1 sin 1  n2 sin  2
n1 sin  c  n2 sin 90
n2
sin  c 
n1
sin  c  1  n2  n1
c
37
Example
Q.
A.
A beam of light is propagating through
diamond (n1 = 2.42) and strikes a
diamond-air interface at an angle of
incidence of 28. Will part of the beam
enter the air (n2 = 1) or will the beam
be totally reflected at the interface?
n2
1
sin  c 

n1
2.42
 c  24.41
38
Example


Since 28 > c, there is no refraction, and the
light is totally reflected back into the diamond.
Similarly, many of the rays of light are
striking the bottom facet of the diamond at 1
> c, they are totally reflected back into the
diamond, eventually exiting the top surface to
give the diamond its sparkle.
39
Thin Lenses


A convex lens is known as a converging lens
because paraxial incident rays will be
converged to the principal axis.
A concave lens is known as a diverging lens
because paraxial incident rays will be
diverged away from the principal axis.
Convex lens
Concave lens
40
Convex Lenses

1.
2.
3.
For paraxial rays:
Rays parallel to the principal axis will be
refracted passing through the focal point.
Rays passing through the focal point will be
refracted parallel to the principal axis.
Rays passing through the centre of the lens will
be passing through straightly without bending.
O
F
F
I
41
Object
Image
Properties of image
Beyond 2F
Between
2F and F
At 2F
Real
Inverted
Real
Inverted
Between
2F and F
Beyond 2F
Real
Inverted
At F
At 
Between F
and lens
Same side
as Object
At 2F
Virtual
Upright /
erect
Diminished
/ Reduced
Same Size
Magnified /
Enlarged
Magnified /
Enlarged
Concave Lenses

1.
2.
3.
For paraxial rays:
Rays parallel to the principal axis will be refracted in a
way that it appears to be originated from the focal
point.
Rays directing towards the focal point will be refracted
parallel to the principal axis.
Rays passing through the centre of the lens will be
passing through straightly without bending.
F
F
43
Concave Lenses

1.
2.
3.
The image of a real object is always
Virtual
Erect
Diminished
44
Example
Q.
A.
An object 2 cm tall is placed 10 cm away from
a convex lens with a focal length of 5 cm. Find
the image position and its size.
Image distance = 10 cm, image size = 2 cm.
5 cm
O
I
10 cm
45
Normal Eyes


Far point at 
Near point at about 25 cm
46
Short-sighted

Image of distant object formed in front
of retina



Far point not at 
Eyeball too long
Focal length too short
47
Short-sighted


Corrective lens: Concave lens
Object at , image at far point of eye
48
Long-sighted

Image of close object formed behind
retina



Near point too far away
Eyeball too short
Focal length too long
49


Corrective lens: Convex lens
Close objects form images at near point
of eye
50
Example
Q.
A student sees the top and the bottom
edges of a pool simultaneously at an
angle of 14 above the horizontal as
shown in the Figure. What is the new
view angle, if he wants to see the top
edge and the bottom center of the pool
(nwater = 1.33 and nair = 1)?
51
2004 IJSO
A.
In order to see the bottom edge of the
n1 sin 1  n2 sin  2
pool,
1.33sin 1  1sin 90  14 
1  46.85
x
 tan 1  1.0667
h
In order to see the bottom centre of the
x/2
pool,
 tan 1 '
h
1 '  28.07 
1.33sin 1 '  1sin  2 '
 2 '  38.75
The new view angle is
90  38.75  51.25
Magnification Equation:
hi
di
M  
ho
do
Where
• ho is the object height and is always +ve.
• hi is the image height and is +ve if the image is an upright image (and therefore, also
virtual) and is -ve if the image is an inverted image (and therefore, also real).
• do is the object distance from the lens/mirror and is always +ve.
• di is the image distance from the lens/mirror. It is +ve if the image is a real image
and located on the opposite(same) side of the lens(mirror) and is -ve if the image is a
virtual image and located on the same(opposite) side of the lens(mirror).
53
Mirror/lens Equation:
1 1 1
 
d o di
f
where
• f is the focal length and is +ve if the lens(mirror) is convex(concave) and is -ve
if the lens(mirror) is concave(convex).
54
Let’s consider the ray diagram of
a convex lens
A
B
f
ho
I
F
O
do
D
hi
di
C
55
Proof of Magnification Equation
 AOD ~  CID
=>
=>
hi di
 
ho d o
hi
di
M  
ho
do
Note: The Magnification Equations for concave lens and mirrors can
be proved similarly by considering appropriate ray diagrams.
56
Proof of lens Equation
 BDF ~  CIF
hi di  f
=> 

ho
f
=>
di di  f di

 1
=>
do
f
f
1 1 1
 
d o di
f
Note: The Lens/Mirror Equations for concave lens and mirrors can
be proved similarly by considering appropriate ray diagrams.
57
Example
Q.
A.
A 2.0-cm diameter coin is placed a distance
of 20.0 cm from a convex mirror which has a
focal length of -12.0 cm. Determine the
image distance and the diameter of the image.
By Mirror Equation, we have
1
1
1
 
d o di
f
1
1
1
15

    di  
 7.5(cm)
20 di
12
2
58
Example

By Magnification Equation, we have
hi
di 7.5
M 


 0.37
ho
do
20

Therefore, a virtual image forms 7.5 cm
behind the mirror and the diameter of the
coin is 0.75 cm.
Check the answers by drawing an appropriate ray diagram
59
Telescope and Microscope
L1(Objective)
do1
L2(Eyepiece)
di1
F1
-di2
To eye
do2
F2
60
Telescope and Microscope



Always converging mirrors or lenses since
diverging mirrors or lenses always give smaller
images
The focal length, F1, of the objective lens is always
longer (shorter) than the focal length,F2, of the
eyepiece in telescope (microscope) – Why?
The magnification, M, is equal to the product of
the magnifications of the individual lenses:
 di1  di 2 
M  M 1M 2   
 

d
d
 o1  o 2 
61
Fermat’s Principle of Least Time


Out of all possible paths that light
might take to get from one point to
another, it takes the path that requires
the shortest time.
The Principle is true for both reflection
and refraction!
62
Part 3
Wave Properties of Light and Colour
63
Waves
Amplitude
Wavelength
64
Electromagnetic Waves
v
E
B
65
Electromagnetic Spectrum
Increase in frequency
66
Image credit: http://imagers.gsfc.nasa.gov/ems/waves3.html
Blackbody Radiation
T4
T3
T4>T3>T2>T1
T2
T1
67
Sodium Lamps, Florescent
Tubes, Laser

Electrons inside atoms jump from outer
orbits to inner orbits and release energy
http://hal.physast.uga.edu/~rls/1020/ch6/emission.swf
68
Colour



From longest to shortest wavelength:
red, orange, yellow, green, blue, indigo,
violet
Light of different wavelengths are
perceived as different colours.
All the colours combine to make white.
69
Dispersion



Due to the difference in refractive index
for different colours
Angle of deviation d
d
Violet deflected most
Crown glass
Colour
Red
Orange
Yellow
Green
Blue
Violet
Wavelength in vacuum (nm) Refractive index n
660
1.520
610
1.522
580
1.523
550
1.526
470
1.531
410
1.538
70
Light in diamond
White light
Violet
Dispersion + Total internal reflection
Red
71
Primary Colours





3 types of cone-shaped receptors in our
eyes perceive colour
Light that stimulates the cones sensitive
to longest wavelengths appears red.
…middle…green
…shortest…blue
Red + Green + Blue = White
72
Complementary Colours



Red+Blue=Magenta
Red+Green=Yellow
Blue+Green=Cyan
Magneta+Green=White
Yellow+Blue=White
Cyan+Red=White
(Magneta,Green), (Yellow,Blue) and
(Cyan, Red) are complementary colours73
Selective Reflection





Most objects reflect rather than emit
light.
Many of them reflect only part of the
light that shines upon them.
If a material absorbs all light except red,
it appears red.
If it reflects all, it appears white.
If it reflects none, it appears black.
74
What do you see?



If white light shines on a red ball, the
red
ball appears ___.
If red light shines on a red ball, the ball
appears red
___.
If green light shines on a red ball, the
black
ball appears ___.
75
Pigments






Pigments are tiny particles that absorb
specific colours.
Magenta = white – green (absorb green)
Yellow = white – blue (absorb blue)
Cyan = white – red (absorb red)
Red, green, blue are additive primaries.
Magenta, yellow, cyan are subtractive
primaries.
76
Pigments
77
Selective Transmission




Colour of a transparent object depends
on the light it transmits.
Pigments in a red glass absorb all
colours except red.
Energy of the absorbed light warms the
glass.
Can we have something “transparent
white”?
78
Which disc is warmer in sunlight?
79
Selective Scattering




Light that incidents on an atom sets the
atom into vibration.
The vibrating atom then re-emit light in
all directions.
Violet light is scattered the most by
nitrogen and oxygen which make up
most of our atmosphere.
But why does the sky appears blue
80
instead of violet?
Why do we have a whitish sky?


When the atmosphere contains a lot of
particles of dust and other particles
larger than oxygen and nitrogen, light
of the longer wavelengths is also
scattered strongly.
After a heavy rainstorm when the
particles have been washed away, the
sky becomes a deeper blue.
81
Why is the setting sun red?


Light that is not scattered is light that is
transmitted. Red, which is scattered the
least, passes through more atmosphere
than any other colour. So the thicker
the atmosphere through which a beam
of sunlight travels, the more time there
is to scatter all the shorter wavelengths.
Why is the rising sun less red?
82
Why are the clouds white?


Different sizes of water molecule
clusters scatter different wavelengths.
The overall result is a white cloud.
Why are the rain clouds dark?
83
Rainbows (double rainbows)
Secondary rainbow
Primary rainbow
84
Primary rainbows
refraction
reflection
violet
refraction
red
85
Secondary rainbows
sunlight
red
violet
86
Laser


Monochromatic = single wavelength / colour
Laser (Light Amplification by Stimulated
Emission of Radiation)
A laser is an instrument that produces a beam of coherent light.
87
Colour Deficiency





ability to distinguish colours and shades is
less than normal
Though “colour blind” is often used, only a
very small number of people are completely
unable to identify any colours.
more common in males than females
usually inherited, but can also result from
certain diseases, trauma or as a side effect of
certain medications
occurs when an individual partially or
completely lacks one or more types of the
three kinds of cones
88
Types of Colour Deficiencies


two different kinds of red-green
deficiency and one blue-yellow
deficiency
red-green deficiencies are by far the
most common
89
Think 2
Q.
If you hold a small source of white light
between you and a piece of red glass,
you’ll see two reflections from the glass:
one from the front surface and one
from the back surface. What colour is
each reflection?
90
Think 2
A.
The reflection from the front surface is
white because the light does not go
far enough into the coloured glass to
allow absorption of non-red light. Only
red light reaches the back surface
because the pigments in the glass
absorb all the other colours, and so
the back reflection is red.
91