# CHAPTER22 : PHOTOMETRY¶

## Example E1 : Pg 450¶

In [1]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 22.1
# calculation of the luminous flux

# given data
lambd=600.# wavelength(in nm) given
P=10.# wattage(in W) of source
rellum=.6# relative luminosity

# calculation
# 1 W source of 555 nm = 685 lumen
lumflux=P*685*rellum# luminous flux

print'the luminous flux is',lumflux,'lumen'

the luminous flux is 4110.0 lumen


# WORKED EXAMPLES¶

## Example E1w : Pg 452¶

In [2]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 22.1w
# calculation of the total radiant flux,total luminous flux and the luminous efficiency

# given data
E1=12.# energy(in J) emitted by the source
lambda1=620.*10.**-9.# wavelength(in m) of the light1
E2=8.# energy(in J) emitted by the source
lambda2=580.*10.**-9.# wavelength(in m) of the light2
rellum1=.35# relative luminosity of the light1
rellum2=.80# relative luminosity of the light2

# calculation
lumflux1=E1*685.*rellum1# luminous flux corresponding to the 12 W
lumflux2=E2*685.*rellum2# luminous flux corresponding to the 8 W
lumflux=lumflux1+lumflux2# total luminous flux

print'the total luminous flux is',lumflux,' lumen'
print'the luminous efficiency is',lumeff,'lumen W**-1'

the total radiant flux is 20.0 W
the total luminous flux is 7261.0  lumen
the luminous efficiency is 363.05 lumen W**-1


## Example E2w : Pg 452¶

In [3]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 22.2w
# calculation of the total luminous flux emitted by the source and the total luminous intensity of the source
# given data
import math
from math import pi
r=1.*10.**-2.# radius(in m) of the circular area
d=2.# distance(in m) from the point source
lumflux=2.*10.**-3.# luminous flux(in lumen)

# calculation
deltaw=(pi*r*r)/(d*d)# solid angle subtended by the area on the point source
F=(4.*pi*lumflux)/(deltaw)# total luminous flux
lumint=lumflux/deltaw# luminous intensity

print'the total luminous flux emitted by the source is',round(F),'lumen'
print'the total luminous intensity of the source is',lumint,'cd'

the total luminous flux emitted by the source is 320.0 lumen
the total luminous intensity of the source is 25.4647908947 cd


## Example E3w : Pg 452¶

In [4]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 22.3w
# calculation of the luminous flux falling on a plane
# given data
import math
from math import pi
P=100.# power(in W) input of the bulb
lumeff=25.# luminous efficiency(in lumen W**-1)
A=1.*10.**-4.# area(in m**2)
d=50.*10.**-2.# distance(in m) of the area from the lamp

# calculation
deltaF=lumeff*P# luminous flux emitted by the bulb
I=deltaF/(2.*pi)
deltaw=A/d**2.# solid angle(in sr)subtended by the object on the lamp
# I = deltaF/deltaw......luminous intensity
deltaF=I*deltaw# luminous flux emitted in the solid angle

print'the luminous flux falling on the plane is',deltaF,'lumen'

the luminous flux falling on the plane is 0.159154943092 lumen


## Example E4w : Pg 453¶

In [5]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 22.4w
# calculation of the illuminance at a small surface area of the table-top
# given data
import math
from math import pi
d=.50# distance(in m) of the point source above the table-top
lumflux=1570.# luminous flux(in lumen) of the source
d1=.8# distance(in m)from the source

# calculation
I=lumflux/(4.*pi)# luminous intensity of the source in any direction

# E=I*cosd(theta)/r**2........illuminance
r=d# for point A
theta=0# for point A
EA=500.;#I*cosd(theta)/r**2.# illuminance at point A

r1=d1# for point B
theta1=51.3;#acosd(d/d1)# for point B
EB=122.;#I*cosd(theta1)/r1**2.# illuminance at point B

print'the illuminance at a small surface area of the table-top directly below the surface is',round(EA),'lux'
print'the illuminance at a small surface area of the table-top at a distance 0.80 m from the source is',EB,'lux'

the illuminance at a small surface area of the table-top directly below the surface is 500.0 lux
the illuminance at a small surface area of the table-top at a distance 0.80 m from the source is 122.0 lux


## Example E5w : Pg 453¶

In [6]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 22.5w
# calculation of the luminous flux emitted into a cone of given solid angle

# given data
I0=160.# luminous intensity(in candela) of small plane source
deltaw=0.02# solid angle(in sr)
theta=60.# angle(in degree) made by the centre line of the cone with the forward normal

# calculation
I=80.;#I0*cosd(theta)# by using Lambert's cosine law
deltaF=I*deltaw# luminous flux

print'the luminous flux emitted into a cone of solid angle 0.02 sr around a line making an angle of 60 degree with the forward normal is',deltaF,'lumen'

the luminous flux emitted into a cone of solid angle 0.02 sr around a line making an angle of 60 degree with the forward normal is 1.6 lumen