PH 0101
UNIT- 3 LECT - 1
1
BASIC DEFINITIONS
2
PRINCIPLES OF PHOTOMETRY
3
LUMMER – BRODHUN PHOTOMETER
4
PROBLEMS
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PHOTOMETRY
Introduction :
The light emitted or reflected by the objects can be
measured in comparison with a standard source.
The branch of optics which deals with the
measurement of light is known as ‘Photometry’.
(‘Photo’
means light and ‘metry’
means
measurement).
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BASIC DEFINITION :
Luminous Flux (F) :
The amount of light energy radiated from a
source or an illuminating object in all directions per
second is known as ‘luminous flux’. It is denoted
by the letter F or and its unit is ‘lumen’.
Definition of 1 Lumen :
1 Lumen =
F
4
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Luminous Intensity or Illuminating Power (I) :
The luminous intensity (or) illuminating power (I)
of a point source in any direction is defined as the
luminous flux emitted or radiated per unit solid angle
in that direction.
If ‘F’ is the luminous flux radiated by a source
within a solid angle ‘’ in any particular direction,
then luminous intensity
I=F/
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‘F’ is measured in lumen and ‘’ in steradian , the
unit for ‘I’
Lumen
I 
or 'Candela'
Steradian
Definition of One Candela :
It is the luminous intensity of one sixtieth
(1/60)th of a square centimetre of platinum at its
fusion point. (1769oC) or one candela is the
luminous intensity of the source if it emits 1 lumen
per steradian.
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Illumination (or) Intensity of Illumination (E) :
The illumination (or) intensity of illumination
at a point on the surface is defined as the luminous
flux (F) received on an unit area (A) of the surface
surrounding the point.
It’s unit is lumen /m2.
E

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F
A
6
Other units of illumination :
Lux : It is equal to 1 lumen per square metre.
It can also be defined as the intensity of illumination
over a surface area of 1m2, placed at a distance of
1m from a point source of 1 Candela.
Phot : It is the intensity of illumination of a
source which receives one lumen per square
centimetre.
1 phot = 104 lux.
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Inverse square law of illumination :
The intensity of illumination (E) of a surface due to
a light source is directly proportional to the luminous
intensity or illuminating power (I) of the source and
inversely proportional to the square of the distance
between the source and the surface (r2).
E  I
E

I
r2

1
r2
(or)
E
E
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
KI
r2
8
1
K 
4
E

I
2
4r
Principle of Photometry :
The principle of photometry is based on the
adjustment of the distances of the two light sources
from the measuring point they produce equal
intensity of illumination at the point.
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E1 and E2 be the intensity of illumination due to
two sources S1 and S2 at a surface ‘S’, r1 and r2
be the distances of the two sources from the
surface. The inverse square law of illumination,
I2
I1
and
E2

E1

2
2
r2
r1
Where I1 and I 2 are the luminous intensities of the
two sources. By adjusting r1 and r2 , E1 and E2
are made equal.
i.e., E1 = E2
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I1
 E 2 2 =
r1
I2
r22
E2
(or)
=
I1
I2
r12
r22
The luminous intensity of one source is known, by
comparison, the luminous intensity of another source
can be determined.
Photometer :
It is an instrument which is used to compare the
luminous intensities of two sources. If the luminous
intensity of one of the source is known, the luminous
intensity of the other one can be measured.
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LUMMER – BRODHUN PHOTOMETER :
This is one of the most accurate photometers used
for comparing the luminous intensities of two sources.
LUMMER – BRODHUN PHOTOMETER
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Screen ‘S’ :
This is a specially prepared
screen made of some white diffusing material of
high reflecting power such as plaster of paris or
magnesium carbonate.
The light rays from the two sources are incident
on the two sides of this screen.
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Totally reflecting prism P1 and P2 :
P1 and P2 are two isosceles right angled prisms
whose angles are 45o, 90o and 45 o.
The light rays scattered from the screen ‘S’, fall
normally on one face of the prism, enter into the
prisms without any change in their direction,
undergo total internal reflection and emerge
normally from the other face of these prisms.
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Lummer Brodhun Cube A and B :
It consists of two right-angled isosceles prisms
A and B in contact with each other.
The edges of prism A are cut in such a way
that it is slightly curved at the outer edge while flat in
the central part.
The prisms are cemented together at the
central part with Canada Balsam, whose refractive
index is the same as that of the material of the
prisms.
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The two prisms are in optical contact with each
other at the central part and enclose an air film
between them at other parts.
Telescope T :
The telescope (T) is used to receive the light
coming from Lummer – Brodhun cube. The whole
apparatus is enclosed in a metallic box blackened
from inside.
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Working :
S1 and S2 are two sources whose luminous
intensities are to be compared.
These sources can be moved along an optical
bench.
They are placed on the opposite side of the screen
‘S’ at equal distances and illuminate the screen
normally.
The light rays reflected diffusively from each side
of screen are incident on the prisms P1 and P2.
These light rays undergo total internal reflection
from the prisms and fall on Lummer Brodhun cube.
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Contd.
 The outer rays from the source S1 undergo total
internal reflection at prism ‘A’, while the middle ray
passes into the prism ‘B’ through the point of contact
of the prisms A and B without any deviation.
 The outer rays from S2 undergo total internal
reflection at ‘B’, while the middle ray passes into the
prism ‘A’ through the point of contact without any
deviation.
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Contd.
In this way, middle ray from source S1 and outer
rays from source S2 enter into the telescope.
 Since the luminous intensities of the two sources
are different, the field of view of the eyepiece of the
telescope consists of a central part illuminated by light
from S1 and an outer part illuminated by light from S2
of different brightness.
 Now the distances of S1 and S2 are adjusted in
such a way that the field of view appears equally
bright.
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Field of view in a Lummer – Brodhun Photometer
When the two sources are of different
intensities
When the sources are having equal intensities.
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Comparison of luminous intensities of two
sources :
When the distances of S1 and S2 are so adjusted
that the illuminations on the two sides of ‘S’ are
equal, the whole field of view appears uniformly bright
and the boundary dividing the two fields disappears.
Let I1 and I 2 be the luminous intensities of S1 and
S2 and r1 and r2 are their respective distances from ‘S’
, then according to inverse square law of illumination,
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E2
(or)
I1
E 2 =
I2
2
1
r
2
2
r
The luminous intensities of the two sources can be
compared.
If ‘I1’ is known, ‘I2’ can be calculated from the relation,
(or) Thus the luminous intensities of the two sources can
be compared.
I2
 r2
 I1 
 r
 1
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



2
22
This photometer is best suited for the
comparison of luminous intensities of two light
sources producing same colour.
If the sources produce different colour the
readings are unreliable, and flicker photometer is
used.
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Solved Problem (1) : Two lamps of 32 candle power and
8 candle power respectively are placed 2 metres apart. A
screen is placed between them. Where should the
screen be placed in order that it may be equally
illuminated by the lamps on both sides ?
Given I1 = 32 candle power
I2 = 8 candle power
r1+ r2 = 2m
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Formula :
I1 I 2
 2
2
r1
r2
Let ‘x’ be the distance from where the screen has to be
placed to get equal illumination on the screen. Then
r1 = x
32
8

2
2
x (2  x)
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32(4+x2-4x)
= 8x2
24x2 – 128x + 128 = 0
3x2 – 16x + 16
=0
3x2 – 4x – 12x + 16 = 0
r1(3x-4)-4(3x-4)
=0
(x-4) (3x-4)
=0
Hence x can be either 4 or 4/3 Since the separation
between the two lamps in 2m, x cannot be 4 m.
Here x = 4/3m. i.e. when the screen is placed at a
distance of 4/3m from the lamp of 32 candle power, the
screen is illuminated equally by the lamps on both sides.
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Solved Problem (2) : Two lamps of 32.c.p. and 8
c.p. respectively are placed 2 meters apart.
Where should a screen be placed in order that it
may be equally illuminated by the lamps ?
There are two possible positions of the
screen.
(i) Both the sources lie on the same side of the
screen. Let the distance of the first lamp (8 c.p.)
from the screen be x as shown. When the screen is
equally illuminated by the two lamps, then
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8
32

2
(x )
(200  x ) 2
or

(200  x ) 2
32

4
2
(x )
8
(200  x )
2
x
or
200  x  2x
 x = 200 cm.
Thus the screen should be placed at a distance 200
cm from the lamp L1(8 c.p.) and 400 cm from the
second lamp L2 (32 c.p.)
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(ii) The screen is between two sources. When the
screen is equally illuminated by the two lamps
..
8
32

(x ) 2
( 2 0 0 x ) 2
(200  x ) 2
32

 4
2
(x )
8
( 200  x )
 2
8
200  3x
x = 66.67 cm.
Thus the screen should be placed at a distance of 66.67 cm
from the first lamp L1 (8 c.p.) and 133.33 cm from the second
lamp L2 (32 c.p.)
or
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Solved Problem (3) : Two identical sources are at a
distance of 60 cm from each other. Where must a screen
be placed in between them so that one side of the screen
may be illuminated four times greater than the other 
Let the screen be placed at a distance xcm from first
source. Now the distance of the second source will be
(60-x) cm. If the illumination of one side is E, then the
illumination of the other side should be 4E. Let both the
sources have illuminating power I. Then
I
E
(x ) 2
and
I
4E 
(64  x ) 2
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x2
4
(60  x) 2
x
or 2  
(60  x)
Case. (i)
Considering +ve sign
120-2x = x or x = 40 cm.
Case. (ii) Considering –ve sign
120 – 2x = - x or x = 120 cm.
which is impossible because it is greater than
60cm. Thus the distance of the screen must be 40
cm from one source and 20cm from the other source.
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Exercise Problem (1) : Two lamps of 20 candle
power and 10 candle power are placed A metre
apart. A screen is placed such that both the
lamps are on the same side of the screen and
line joining the lamps is perpendicular to the
screen. At what distance the screen has to be
placed so that it may be equally illuminated by
the lamps placed on the same side.
At a distance of 9.655m from the lamp of 10
candle power.
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Thank u
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