CHAPTER 9. PHOTOMETRY AND RADIOMETRY
Photometry is the study of the measurement of light in terms of the visual response
it produces. Photometric measurement involves the visual response of the observer.
9.1. Visual Sensitivity
The eye is not equally sensitive to the different colors of light in the visible
spectrum. The visual sensitivity varies with wavelength. The sensitivity curve, sometimes
referred to as the luminosity curve of an equal energy spectrum, give the relative
brightness, as assessed by the average eye, of the different colors of the spectrum when the
incident energies at each wavelength have been reduced to the same mechanical value.
For the average light adapted eye at moderate intensities (photopic vision) the maximum
visual effect is obtained with light of wavelength 555nm (yellow-green).
Fig.9.1. Luminosity curve. The
shift of the visual sensitivity
curve from a maximum at
555nm in photopic vision to
500nm in scotopic vision (for
average dark adapted eye) is
known as the Purkinje effect.
9.2. Solid Angle
area
r2
The total solid angle surround a point is 4 steradians.
solid angle  
r

Fig.9.2. Solid angle
1
(9.1)
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.
(Ans: 0.625 sr.)
9.3. Standard Source and Candela
A standard must give a constant luminous output for an indefinite period of time.
This restriction excludes all gas lamps, electric filament lamps, and discharge tubes from
being used for standardization. The present primary standard source of light is based on
the concept of a black body radiator. A black body is a device which absorbs all the
radiation incident on it at all temperatures. When at a constant temperature a black body
will be radiating the same energy that it receives, the total quantity depending on its
temperature. Thus a black body is also a perfect emitter of radiation, and will emit more
energy than any other body at the same temperature. The radiation from a black body is
independent of the nature and material of the body.
Advanced Study:
The radiation field () of a blackbody has the characteristic spectral distribution,
8h 3
1
   
3
h kT
c
e
1
where h is the Planck constant, k is the Boltzman constant, c is the speed of light in vacuum,  is the
frequency of light and T is the temperature of the blackbody.
In basic terms, a small opening in a hollow vessel with a blackened interior will
absorb all radiations incident on it by virtue of repeated internal reflections and
absorptions. The small baffle is to avoid a single normal reflection escaping form the
opposite site of the enclosure. When such a device is heated the small hole acts as a black
body radiator.
Fig.9.3. A black body is a perfect
absorber and radiator. A hollow
vessel with a blackened interior and
single small hole provides the basis
for a standard source.
baffle
Since 1948 the primary international standard of light has been adopted in terms of
the visible radiation from a small hole in the end of a cylinder made of pure fused thorium
oxide as shown in Fig.9.4.
The unit of measurement of the luminous intensity of a source is candela. 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.
2
Fig.9.4. The international black body
standard. The thorium oxide tube is
maintained at the temperature of
1773C by immersion in a mixture of
molten and solid pure platinum. The
central thorium oxide tube is about
40mm long and the radiation outlet has
a diameter of 1.5mm. Platinum is used
because it can be obtained in a pure
state, has a high freezing point, and
does not oxidize in air. Thorium oxide
has a melting point higher than
platinum’s and it does not react with or
dissolve in the platinum.
Advanced Study:
Since 1979 the candela has been realized by a cryogenic absolute radiometer. This device
equates the heating effect of optical radiation with that of an electric current. The candela is defined
as the luminous intensity (in a given direction) of a source which emits monochromatic radiation of
frequency 5.41014 Hz and that has a radiant intensity in that direction of 1/683 watt per steradian.
This corresponds to a wavelength of 555nm, for which the photopic sensitivity of the eye is a
maximum.
9.4. Working Standards
Electric filament lamps are much easier to set up and more convenient to use than
the primary black body standard source. Consequently, electric lamps are used as
everyday working standards for luminous intensity, and are sometimes called substandards. They are calibrated periodically with the primary standard in case there is any
deterioration in their light output.
9.5. Luminous Flux
The transfer of light from a source is expressed in terms of luminous flux . It is
defined as the rate at which light energy flows. The unit of luminous flux is lumen. One
lumen is the luminous flux emitted into a unit solid angle (1 steradian) from a point source
of intensity 1 candela. Hence the total luminous flux emitted in all directions by a point
source of intensity 1 candela is 4 lumens.
or,
Suppose I the intensity of a source,  luminous flux and  the solid angle. Then,
=I
(9.2)
Φ
I
(9.3)

For a specified direction, we use infinitesimal small solid angle to substitute into the above,
dΦ
I
(9.4)
d
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9.6. Luminous Intensity of a Source
The luminous intensity I of a source in a specified direction is defined by the
equation (9.4). Practical sources of light do not radiate equally in all directions, and are
referred to as non-uniform sources. In commercial use it may be necessary to refer to the
average value of the luminous intensity in all directions. This is referred to as the mean
spherical intensity.
Φ
mean spherical int ensity 
cd
(9.5)
4
Fig.9.5. The knowledge how the intensity of a lamp varies with
direction is best represented by a light distribution curve
called polar curve. Because of the metal cap and the shape
of the filament, light does not radiate uniformly in all
directions. By means of a suitable photometer, the intensity
of the lamp may be determined in various directions in a
particular plane. The polar curve is drawn by taking the
source as the origin, the luminous intensity in any particular
direction is proportional to the radius in that direction. The
figure shows a polar curve typical of a 40W domestic light
bulb.
Example: A source of light has a mean spherical intensity of 20 cd. How much total flux
does it emit?
(Ans: 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?
(Ans: 250 lm)
9.7. Illuminance
The illuminance E at a point on a surface is the amount of luminous flux falling on
unit area of the surface.
Φ lm
E
(9.6)
A m2
The basic unit of measurement of illuminance is lux, which is abbreviated as lx. For a
specific direction,
dΦ
E
(9.7)
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. If the illuminance of a surface is due to two
or more sources, then the illuminance is equal to the sum of the illuminances due to each
source separately. It is interesting to note that the eye functions quite well, and can adapt
quickly, over a wide range of illuminances corresponding to about six or seven orders of
magnitude. As the variation between maximum and minimum opening of the pupil
represents a factor of only about 10 in area, there are obviously other factors involved in
the adaptation process.
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9.8. Two Fundamental laws of Photometry
From Eq.(9.4), the flux flowing in a given direction is dΦ  Id , substituting into
d
dA cos 
Eq.(9.7), we have E  I
. From geometry, d 
, therefore,
dA
r2
I
E  2 cos 
(9.8)
r
Fig.9.6. The figure shows the geometrical
relation that the projected area is a spherical
one that subtends a solid angle d to the
point source. Therefore, dA cos   r 2 d
Two fundamental laws of photometry can be immediately derived from Eq.(9.8):
The first is that 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 .
Advanced Study:
N
Ik
cos  k
2
k 1 rk
cos 
For extended sources, the illuminance is E   dE   2 dI
r
For multiple point sources, the illuminance is E  
S (I=100 cd)
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
4m
below the source, (ii) at 6 m from this point?
(Ans: (i) 6.25 lx, (ii) 1.07 lx)
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?
(Ans: 55.77)
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9.9. Reflectance, Transmittance, and Optical Density
The reflectance or reflection factor  of a surface is the ratio,
luminous flux per unit area reflected by a surface

luminous flux per unit area incident on the surface
(9.9)
A mirror, or other reflecting surface, placed near to a light source can increase the
illuminance on any surface which receives the luminous flux. Consider a light source S in
Fig.9.7. The reflected flux from the mirror will be , and the illuminance on the surface
at B due to the reflected light is,
Φ I I A2 I
EB 



(9.10)
A2
A2
A2 d 2 d 2
Thus, 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. The total illuminance at B will be due
to the reflected light from the mirror together with the light reaching B directly from the
source. A similar procedure may be adopted for curved mirrors.
Fig.9.7. A light source S, of intensity I cd, is
placed a distance a in front of a plane
mirror M, having a reflectance . The
luminous flux  contained in the solid
angle  is incident on the area A1 of the
mirror, and is then reflected to form an
illuminated patch, of area A2 on the
surface at B.
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.
(Ans: 116.1 lx)
The transmittance  of a transparent body is given by the ratio,
luminous flux per unit area transmitted by the body

(9.11)
luminous flux per unit area incident on the surface
The value of  depends on the nature and thickness of the substance. It may also depend
on the wavelength of the light used. Some optical materials display an almost constant
transmittance across the whole visible spectrum. Such materials are called neutral
substances.
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
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its original value?
(Ans: 32 lx; 14.4 lx; 1.68 m)
The transmittance of a transparent body can be further divided into the
transmittance at the 1st (T1) and 2nd (T2) surface and the transmittance of the material (Tm)
with a certain thickness, that is,
  T1TmT2
(9.12)
If the material’s thickness has increased by a factor of n, then the transmittance of the
material is changed to (Tm)n. The surface transmittance T1 and T2 can be calculated as,
4 nn 
T1  T2 
(9.13)
n  n2
where n and n are the refractive indices of the air and the material, respectively.
The optical density D of a substance is defined as,
1
D  log

(9.14)
Then we have,
1
1
1
 log
 log
 D1  Dm  D2
(9.15)
T1
Tm
T2
It can be seen that 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.
D  log
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.
(Ans: 0.24; 0.17)
9.10. Luminance
The luminance L of any surface (self-luminous, transmitting, or reflecting) in a
specified direction is defined as the luminous intensity per unit projected area in the
direction concerned.
I
L
(9.16)
A cos 
Fig.9.8. Luminance of a surface area A in a given
direction . In the  direction, the intensity is I cd
and the projected area of A is Acos.
It is important to note that I is not a constant but is -dependent. The unit of luminance is
cd/m2.
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It is extremely important to distinguish between the terms illuminance and
luminance. The former is concerned with the luminous flux incident on a surface and this
does not depend on the nature of the surface. The latter term concerns the flux which is
emitted (or transmitted, or reflected) in a given direction and this will be dependent on the
nature of the surface.
A surface which obeys the Lambert’s Law of Emission below is referred to as a
uniformly diffusing surface.
I  cos
(9.17)
The luminance L of the surface is thus the same in all directions.
Let’s consider, as shown in Fig.9.9, a converging lens forming an image of area a
of an object of area a. Let the area of the lens be A, with  and  representing the solid
angles of the incident and refracted cones of light rays. Let the luminance of the object
area a in the direction of the lens be L. Then,
I
L
a
2
Since   A l , the flux incident on the lens from the area a is,
Φ  I  aL lumen
Fig.9.9.
The illuminance of an
image.
If we now assume no absorption or reflection losses occur in the system, the entire
light incident on the lens will be incident on the image. The illuminance of image is then,
Φ a L
E 
lux
a
a
Ll 2 AL
a l2

Since
, we have E  2  2 . If the lens diameter is d, we can write,
l
l
a l  2
 d2
E L 2
(9.18)
4 l
It can be seen that the illuminance of the image of a surface is proportional to the
luminance of the object surface and the area of the lens aperture. In a camera, for distant
objects, where l is equal to the focal length f, the illuminance of the image on the film is,
E
 d
2
L 
4  f  
(9.19)
8
The quantity d/f is called the aperture ratio of the lens and is the reciprocal of the fnumber of the lens.
In Fig.9.9, the luminous flux  from the area a is,
Φ  I   La  La
Since a  a , we have
L  L
(9.20)
That is, the luminance of an image is equal to the luminance of the object.
9.11. Photometers
Photometers may be classed as visual or non-visual. In the former group of
instruments the basic principle applicable to their use is that 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 (Fig.10).
Fig.9.10. The principle of the visual
photometer is to compare the
intensities of two lamps if the colors
of their lights is similar. The distance
of one source from one diffusing
surface is adjusted until the two
surfaces appear equally bright.


2
As shown in Fig.9.10, the illuminance on the left hand screen is E L  I1 d1 cos 
2
and the illuminance on the right hand screen is ER  I 2 d 2 cos . When balanced,
EL  ER . Therefore,

d 
I1  I 2  1 
 d2 

2
(9.21)
The eye cannot give any quantitative comparison between different illuminances. It
can only judge with a fair degree of accuracy when two adjacent surfaces appear to be
equally bright, provided they appear the same or nearly the same color. The ability of the
eye to judge equality of brightness of two surfaces is expressed by the Weber-Fechner
Law: If L is a value for the prevailing luminance of a surface and dL is the minimum
noticeable increment, then
dL
 constant
(9.22)
L
9
dL/L is referred to as the Fechner fraction and has a value of about 0.01 to 0.03 over a
wide range of luminances. At very low luminances there is a rapid fall off in the ability of
the eye to discriminate between the luminances of two surfaces.
Advanced Study:
There are several kinds of visual photometers, such as, the grease-spot photometer, the wax block
photometer, the shadow photometer, the Lummer-Brodhun photometer, the flicker photometer, and the
integrating photometer. A brief description of these photometers can be found from the reference.
The non-visual (physical) photometers directly measure the illuminance falling on
them. The most commonly used non-visual photometers are photovoltaic cells (Fig.9.11).
The cell does require a battery for its operation. The response time is quick and the output
current depends on the illuminance on its surface.
Fig.9.11. Schematic of a photovoltaic
cell. 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 Se-layer.
9.12. Luminous Efficacy of a Source
To describe the effectiveness of a source, two concepts should be distinguished:
luminous flux emitted  lumens 
luminous efficiency 


(9.23)
total flux radiated
 lumens 
luminous flux emitted by a lamp  lumens 
luminous efficacy 

 (9.24)
power consumed
 watts 
The luminous efficiency indicates the fraction of total flux radiated that is actually visible.
The 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.
Advanced Study:
At a wavelength of 555nm which is the one most sensitive to the eye, the theoretical maximum
luminous efficacy is 649.35 lm/W. In practical, light sources have much lower values of efficacies
since they do not radiate all the light at 555nm and much energy is lost as heat.
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/m 2 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.
(Ans: 55)
10
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?
(Ans: 4000 lx; 4000 lx)
Advanced Study:
Radiometry is the science to measure all radiant energy while photometry is applied only to the
visible part of the whole optical spectrum. Another distinction between radiometry and photometry is
that radiometry involves only physical measurement while photometry involves the response of human
eye. Nevertheless, the mathematical treatments on photometry can be similarly extended to radiometry.
The radiant flux e is defined as the radiate energy flowing per second through a surface. Its unit is
watt. The radiant intensity Ie (in watts per steradian) of a point source in a specific direction is,
dΦe
Ie 
d
The irradiance Ee (in watts per square meter) is the radiant flux received on a unit surface area,
dΦe I e
Ee 
 cos 
dA r 2
The radiance Le (in watts per square meter per steradian) is the flux per steradian emitted in a
specified direction,
I  
Le  e
A cos 
The radiant intensity of a black body (or a perfectly diffuse scattering surface) follows the
Lambert’s law, that is,
I e    I e 0 cos 
The radiance of such a surface is constant as view from different angles.
Exercises:
9.1 A 75 W lamp is rated as producing 15 lumens per watt. Determine the intensity of
the lamp in candela, and its rating in watts per candela.
9.2 Calculate the intensity of a light source which emits 6500 lumens of flux in
directions below the horizontal and no flux in any direction above the horizontal.
9.3 The luminance flux incident on the condenser lens of a projector is 12000 lumens
and the average illuminance on the screen 5 m square is 50 lumen/m2. Determine
the fraction of the incident light transmitted to the screen by the optical system.
9.4 Calculate the illuminance due to a small source of intensity 100 cd on a screen 2 m
away (a) for normal incidence; (b) for an angle of incidence of 30; and (c) for an
angle of incidence of 60.
9.5 Two lamps each of 500 cd are suspended 8 m above a road 6 m wide. The lamps
are placed above the centre line of the road 30 m apart. Find the illuminance at a
point halfway between them (a) in the center of the road and (b) at the side of the
road.
9.6 A photometer bench is 2m long. At one end is placed a source of intensity 16 cd
whilst at the other end is placed a source of intensity 25 cd. Determine the position
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of a screen between the two sources so that each side is equally illuminated.
9.7 Two sources of light are arranged to produce equal illuminances on opposite sides
of a photometer screen. One source at 50 cm distance has to be moved 5 cm nearer
to the screen to restore the balance when a sheet of glass is interposed between it
and the screen. What is the percentage of light transmitted by the glass? How much
nearer would this source have to be moved if a second similar sheet of glass were
introduced?
9.8 A small lamp of intensity 40 cd is placed 60 cm from a screen. A plane mirror
which reflects 70% of the light incident on it is placed 15 cm behind the lamp,
parallel to the screen. Find the illuminance of the screen.
9.9 At 2 mm thickness, a certain tinted glass has a transmittance for a specified
wavelength of 0.44. If the index of glass is 1.530, determine the transmittance at
1mm and 3mm thickness.
9.10 The luminance of a flat uniformly diffusing surface of area 2.5mm2 is 20cd/mm2.
Determine the luminous intensity of the surface along the normal and along the
direction inclined at 50 to the normal.
9.11 A point source of intensity 48 cd is placed at the principal focus of a concave mirror
which subtends 1 steradian at the source. If the reflectance of the mirror is 90%,
find the amount of luminous flux in the reflected pencil.
9.12 A parallel pencil of light is obtained by placing a small source of light of intensity
20 cd at the first principal focus of a converging lens of focal length 20 cm and
diameter 10 cm. The light falls obliquely on a screen at an angle of incidence of 30.
Find the illuminance of the screen if the lens transmits 80% of the light incident
upon it.
Reference:
A.H.Tunnacliffe and J.G.Hirst, Optics, The Association of British Dispensing Opticians,
1996.
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