The Hong Kong Polytechnic University
Optics II (AP229)
Dr. Haitao Huang
Department of Applied Physics, Hong Kong PolyU
Tel: 27665694; Office: CD613
Reference:
Optics, A.H.Tunnacliffe and J.G.Hirst, Association of British Dispensing
Opticians 1996
Optics II----by Dr.H.Huang, Department of Applied Physics
1
The Hong Kong Polytechnic University
Reference Books:
Introduction to Optics, 2nd Edition, F.L.Pedrotti and L.S.Pedrotti (Prentice Hall
1993)
Optics & Vision, L.S.Pedrotti and F.L.Pedrotti, (Prentice-Hall 1998)
Optics, 4th Edition, E.Hecht (Addison Wesley )
Optics, 3rd Edition, A.Ghatah (McGraw Hill 2005)
Waves—Berkeley Physics Course V.3, F.S.Crawford, Jr., (McGraw Hill 1968)
Fundamentals of Physics, 7th Edition, D.Halliday, R.Resnick, and J.Walker (John
Wiley & Sons 2005)
University Physics, 11th Edition, H.D.Young and R.A.Freedman (Pearson 2004)
Physics for Scientists and Engineers, R.D.Knight (Pearson 2004)
Physics for Scientists and Engineers, 3rd Edition, D.C.Giancoli, (Prentice Hall
2000)
College Physics, A.Giambattista, B.Mc.Richardson, and R.C.Richardson,
(McGraw Hill 2004)
Advanced Studies:
The Feynman Lectures on Physics, Definitive Edition, R.P.Feynman,
R.B.Leighton, and M.Sands (Addison-Wesley 2006)
Principles of Optics, 7th Edition Expanded, M.Born et al.
Optics II----by Dr.H.Huang, Department of Applied Physics
2
The Hong Kong Polytechnic University
Assessment Weighting:
Final Exam: 60% (A minimum grade D is required)
The final exam will be in the form of open-book test. You are allowed to take at most
one book plus the lecture notes.
Mid-term test: 15% (1.5 hour on 17/03/2006)
Course work: 40%
Lab reports: 20%
Homework: 5%
Features:
• Many physics concepts will be learned. Calculus is unavoidable but will be kept to
the least extent.
• Plenty of exercises and examples are provided in each chapter.
• Some contents labeled Advanced Studies in the lecture notes are only for those
students who are interested in the related topics. They will not be examined.
• The PowerPoint of each chapter can be downloaded from my website.
Optics II----by Dr.H.Huang, Department of Applied Physics
3
The Hong Kong Polytechnic University
Photometry
Photometry=photo+metry=visible light + measurement
Visual Sensitivity:
The eye is not equally sensitive to the
different colors of light in the visible
spectrum.
daylight
night
Luminosity Curve:
For the average light adapted eye at
moderate intensities (photopic vision) the
maximum visual effect is obtained with
light of wavelength 555nm (yellow-green).
For average dark adapted eye, the
maximum is around 500 nm.
Optics II----by Dr.H.Huang, Department of Applied Physics
4
The Hong Kong Polytechnic University
Photometry
Solid Angle:
solid angle  
area
r2
r

The total solid angle around a point is 4 (steradians).
Example:
Calculate the solid angle subtended at the center of a sphere, radius 2 m, by an
area of 2.5 m2 on the surface of the sphere.

area 2.5
 2  0.625 steradian
2
r
2
Optics II----by Dr.H.Huang, Department of Applied Physics
5
The Hong Kong Polytechnic University
Photometry
Standard Source and Candela:
A standard requires a constant luminous output for all the
time.
The present primary standard source of light is based on
the concept of a black body radiator. The radiation solely
depends on its temperature and does change with time.
baffle
International black body standard
One candela is equal to one sixtieth (1/60) of
the luminous intensity per square centimeter of
a black body at the temperature of solidification
of platinum (1773°C).
Electric lamps are used as everyday working standards for luminous intensity, and
are sometimes called sub-standards. They are calibrated periodically with the
primary standard in case there is any deterioration in their light output.
Optics II----by Dr.H.Huang, Department of Applied Physics
6
The Hong Kong Polytechnic University
Photometry
Luminous Flux and Intensity:
Luminous flux : the rate at which light energy flows. Unit: lumen
One lumen is the luminous flux emitted into a unit solid angle (1 steradian) from a
point source of intensity 1 candela. Φ  I
Luminous intensity of a source: I 
For a specified direction: I 
Φ

dΦ
d
mean spherical int ensity 
Example:
Φ
4
cd
A source of light has a mean spherical intensity of 20 cd. How much total flux does
it emit?
Φ  I  4  20  251.3 lm
Example:
A source has an intensity of 250 cd in a particular direction. How much flux is
emitted per unit solid angle in that direction?
Φ  I  250 1  250 lm
Optics II----by Dr.H.Huang, Department of Applied Physics
7
The Hong Kong Polytechnic University
Photometry
Illuminance:
Φ lm
dΦ
E

A m2
dA
The illuminance at a point on a surface does not depend on the nature of the surface
since it is only concerned with incident light.
d
E

I
dΦ  Id
dA
dA cos 
I
d 
E

cos 
2
2
r
r
E
Two laws:
The illuminance at a point on a surface is inversely proportional to the square of the
distance between the point and the source. This law applies strictly only in the case
of point sources.
The second is that if the normal to an illuminated surface is at an angle  to the
direction of the incident light, the illuminance is proportional to the cosine of .
Optics II----by Dr.H.Huang, Department of Applied Physics
8
The Hong Kong Polytechnic University
Photometry
Example:
A point source of light S, of intensity 100 cd, is suspended 4 m above a horizontal
surface. What is the illuminance on the surface (i) at the point vertically below the
source, (ii) at 6 m from this point?
I  100 cd; rA  4 m;  A  0
EA 
I  100 cd; rB  7.211 m; cos  B 
4
rB
I
cos  A  6.25 lux
2
rA
S (I=100 cd)
4m
I
EB  2 cos  B  1.07 lux
rB
A
6m
B
Example:
Light falls normally on a surface at 4 m from a point source of light. If the surface is
moved to a distance 3 m from the source, at what angle must the surface be inclined
in order that the illuminance is the same value?
For =0, E  I cos   I
A
rA2
Afer movement : E A 
A
16
I
I
cos


cos 
2
rA
9
  55.77 
Optics II----by Dr.H.Huang, Department of Applied Physics
9
The Hong Kong Polytechnic University
Photometry
Reflectance:

luminous flux per unit area reflected by a surface
luminous flux per unit area incident on the surface
Reflecting Mirror:
The illuminance on the surface at B due to the
reflected light is,
Φ I I A2 I
EB 


 2
2
A2
A2
A2 d
d
The illuminance at B, due to the reflected light, is as if from a source of intensity I
situated at the position of the image in the mirror.
Optics II----by Dr.H.Huang, Department of Applied Physics
10
The Hong Kong Polytechnic University
Photometry
Example:
A small 50 cd source which may be assumed to
radiate uniformly in all directions, is placed 75 cm
above a horizontal table, and a plane mirror is fixed
horizontally 25 cm above the source. If the mirror
reflects 85% of the incident light, calculate the
illuminance on the table at the point vertically below
the source.
By direct light : EB 
By reflection : EB 
I
50
cos


 88.9 lux
2
2
r1
0.75
I
I
cos


 27.2 lux
2
2
r2
0.75  0.25  0.25
Total illuminanc e  EB  EB  116.1 lux
Optics II----by Dr.H.Huang, Department of Applied Physics
11
The Hong Kong Polytechnic University
Photometry
Transmittance:

luminous flux per unit area transmitted by the body
luminous flux per unit area incident on the surface
It depends on the nature and thickness of the substance. It may also depend on the
wavelength of the light used.
Neutral substances: materials that display an almost constant transmittance across
the whole visible spectrum.
Optics II----by Dr.H.Huang, Department of Applied Physics
12
The Hong Kong Polytechnic University
Photometry
Example:
A point light source of intensity 200 cd is 2.5 m from a screen. Calculate the
illuminance on the screen for normal incidence. If a neutral filter of transmittance
45% is placed between the source and the screen, what is the new value of the
illuminance? Where must the source be placed such that, with the above
mentioned filter in place, and with normal incidence, the illuminance on the screen
is restored to its original value?
Without filter :
I
200
E1  2 cos  
 32 lux
2
d1
2.5
With filter :
I
200
E2  2 cos   2 lux
d2
d2
 E1  E2
 d2  1.68 m
Optics II----by Dr.H.Huang, Department of Applied Physics
13
The Hong Kong Polytechnic University
Photometry
Optical Density:
For a transparent plate:   T1TmT2
If the material’s thickness has increased by a factor of N,
then the transmittance of the material is changed to (Tm)N.
Transmittance at two surfaces: T1  T2 
4 nn 
n  n2
Optical density D of a substance is defined as, D  log
1
1
1
D  log  log
 log
 D1  Dm  D2
T1
Tm
T2
N slices
1

The optical density of the sample is simply the sum of the optical densities of the
surfaces and the stated material.
To find new optical density we can simply multiply the optical density Dm of the
material by whatever factor that gives the new thickness.
Optics II----by Dr.H.Huang, Department of Applied Physics
14
The Hong Kong Polytechnic University
Photometry
Example:
At 2 mm thickness, a certain tinted glass has a transmittance for a specified
wavelength of 0.47. Determine the transmittance at 4 mm and 5 mm thickness.
ng=1.5.
T1  T2 
4nn
4  1  1. 5

 0.96
2
2
n  n  1.5  1
At 2mm thickness : Tm 

T1T2

0.47
 0.51
2
0.96
At 4mm thickness :   T1 Tm  T2  0.96  0.512  0.96  0.24
42
At 5mm thickness :   T1 Tm  T2  0.96  0.512.5  0.96  0.17
52
Optics II----by Dr.H.Huang, Department of Applied Physics
15
The Hong Kong Polytechnic University
Photometry
Luminance:
The luminance L of any surface in a specified direction is
defined as the luminous intensity per unit projected area
in the direction concerned.
L
I  
A cos 
A uniformly diffusing surface obeys the Lambert’s Law of Emission.
I  cos
Illuminance is concerned with the luminous flux incident on a surface and this does
not depend on the nature of the surface.
Luminance is concerned the flux which is emitted (or transmitted, or reflected) in a
given direction and this will be dependent on the nature of the surface.
Optics II----by Dr.H.Huang, Department of Applied Physics
16
The Hong Kong Polytechnic University
Photometry
Luminance of a Image:
Luminance of the object area a in the
direction of the lens
I
L
a
The flux incident on the lens is
Φ  I  aL lumen
Φ a L

lux
a
a
a l2
Ll 2 AL
 2 , we have E  2  2
Since
l
l
a l 
 d 2 Illuminance of the image of a surface is
If the lens diameter is d : E  L 2 proportional to the luminance of the object surface
4 l
and the area of the lens aperture.
2
 d
In camera, E  L  d/f is the aperture ratio of the lens and is the reciprocal of the
4  f   f-number of the lens.
The illuminance of image is then E 
Luminous flux  from the area a is
Φ  I   La  La
a  a
L  L
Optics II----by Dr.H.Huang, Department of Applied Physics
17
The Hong Kong Polytechnic University
Photometry
Photometer (visual):
Principle: If two adjacent identical white
reflecting surfaces appear to be equally bright
when illuminated with two sources, then the
surfaces will be receiving the same
illuminance and the boundary between the
surfaces will be difficult to see.


2
Illuminance on the left hand screen E L  I 1 d1 cos 


2
Illuminance on the right hand screen ER  I 2 d 2 cos
 d1 
I1  I 2  
 d2 
2
Weber-Fechner Law for human eye to judge the equality of brightness of two
surfaces: If L is a value for the prevailing luminance of a surface and dL is the
minimum noticeable increment, then
dL
 constant
L
Optics II----by Dr.H.Huang, Department of Applied Physics
18
The Hong Kong Polytechnic University
Photometry
Photometer (non-visual):
The non-visual (physical) photometers directly measure the illuminance falling on
them. The most commonly used non-visual photometers are photovoltaic cells.
Due to the photoelectric effect, electrons are emitted from the selenium layer when
light is incident on the cell. A thin transparent gold film is deposited on the top of Selayer.
Optics II----by Dr.H.Huang, Department of Applied Physics
19
The Hong Kong Polytechnic University
Photometry
Luminous Efficacy:
To describe the effectiveness of a source, two concepts are used:
luminous efficiency 
luminous efficacy 
luminous flux emitted
total flux radiated
 lumens 


lumens


luminous flux emitted by a lamp
power consumed
 lumens 


 watts 
Luminous efficiency indicates the fraction of total flux radiated that is actually visible.
Luminous efficacy is the fraction of the total consumed power that is used to emit
the visible light. A large amount of energy can be wasted in the form of heat or
infrared radiation.
Optics II----by Dr.H.Huang, Department of Applied Physics
20
The Hong Kong Polytechnic University
Photometry
Example:
A football pitch 110 m by 87.5 m is illuminated for evening matches by equal banks
of 1000 W lamps supported on 16 towers, which are located around the ground to
provide approximately uniform illuminance of the pitch. Assuming 35% of the total
light emitted reaches the playing area and that an illuminance of 800 lm/m2 is
necessary for TV purposes, calculate the number of lamps on each tower. The
luminous efficacy of each lamp may be taken as 25 lm/W.
Flux emitted by one lamp  1000  25  25000 lm
E
Φ 25000 N  0.35

 800 lm/m 2
A
110  87.5
Each tower will carry
N  880
N
 55 lamps
16
Optics II----by Dr.H.Huang, Department of Applied Physics
21
The Hong Kong Polytechnic University
Photometry
Example:
A point source of intensity 40 cd is placed on the axis and 20 cm from a +10 D lens
of aperture 4 cm. Find the illuminance on a screen placed 10 cm from the lens,
neglecting reflection and absorption losses. What will be the illuminance on the
same screen if the aperture of the lens is reduced to 3 cm?
l  0.2 m; F  10 D;
1 1
   F  l   0.2 m
l l
0.022 
Φ  I  40 
 0.4 lm
2
0.2
d  l   0.1
For similar triangles :

 d   0.02 m
d
l
Φ
0.4
E 
 4000 lx
2
A 0.01 
Homework:
Φ I
 d 2
1
I d
E 
I

  const.
2
2
2 
A A
l
 d  2 l  d  
2
2
9.2; 9.6; 9.9; 9.12; 9.16;
9.18; 9.20
Optics II----by Dr.H.Huang, Department of Applied Physics
22