A small light source with intensity uniform in all

directions is mounted at a height of 10 metres

above a horizontal surface.
Two points A and B both lie on the surface with
point A directly beneath the source. How far is
B from A if the illumination at B
is only 1/10 as great as at A ?
let the intensity of lamp be I and distance
between A and B be x metres
x = Symbol('x')
l = 10.0 #m (vertical distance)
#Illumination at point A
Ea = I/l**2 #lux
#Illumination at point B
Eb = I/(l**2)*(l/(l**2 + x**2)**0.5)**3
I = 10.0 #lm (assumed value as the
equation does not depend on I)
#As Eb = 1/10*Ea
eq = Eq(Eb,Ea/10.0)
x = solve(eq)
x1 = x[1]
Distance between A and B is = 19.08 m.

A small area 7.5 m in diameter is to be

illuminated by a lamp suspended at a height
of 4.5 m over the centre of the area. The lamp
having an efficiency of 20 lm/w is fitted with
a reflector which directs the light output only
over the surface to be illuminated,giving
uniform candle power over this angle.
Utilisation coefficient = 0.40. Find out the
wattage of the lamp. Assume 800 lux of
illumination level from the lamp.'''
dia = 7.5 #m (diameter)
h = 4.5 #m (height)
E = 800.0 #lux (illumination)
eff = 20.0 #lm/w (lamp efficiency)
A = 3.14*(dia**2)/4
#Luminous flux reaching the surface
flux = A*E #lm
#Total flux emmited is
f_out = flux/0.4 #lm
#Lamp in watts
watt = f_out/eff #W
print "lamp wattage =",round(watt,2),"W."
lamp wattage = 4415.63 W.

A drawing office containing a number of

boards and having a total effective area
of 70 m2 is lit by a number of 40 W
incandescent lamps giving 11 lm/W. An
illumination of 80 lux is required on the
drawing boards. Assuming that 60% of the
total light emitted by the lamps is available
for illuminating the drawing boards,
estimate the number of lamps required.'''
A = 70.0 #m^2 (area)
watt = 40.0 #W (each bulb wattage)
eff = 11.0 #lm/W (luminous efficacy)
E = 80.0 #lux (Illumination)
#Output per lamp is
oplamp = watt*eff #lm
#Flux actually used per lamp is
flux = 0.6*oplamp #lm
#Now, Total flux required is Illumination
*Area
flux_tot = E*A #lm
#Therefore number of lamps required are
N = flux_tot/flux
Number of lamps = 21.0

Calculate the brightness (or luminance) of

snow under an illumination of (a) 44,000
lux and (b) 0.22 lux. Assume that snow
behaves like a perfect diffusor having a
reflection factor of 85 per cent.'''
E1 = 44000.0 #lux (illumination 1)
E2 = 0.22 #lux (illumination 2)
rf = 0.85 # (reflection factor)
L1 = (E1*rf/3.14) #cd/m^2 (Brightness 1)
L2 = (E2*rf/3.14) #cd/m^2 (Brightness 2)
Brightness 1 = 1.191083e+04 cd/m^2.
Brightness 2 = 0.06 cd/m^2.

'A 21 cm diameter globe of dense opal glass

encloses a lamp emitting 1000 lumens and
has uniform brightness of 4e+3 lumen/m^2
when viewed in any direction. What would
be the luminous intensity of the globe in
any direction? Find what percentage of the
flux emitted by the lamp is absorbed by the globe.'''
d = 21.0 #cm (diameter)
flux = 1000.0 #lumens (luminous flux)
B = 4e+3 #lm/m^2 (uniform Brightness)
#Surface Area of the globe
sa = 3.14*(d*d)*10e-5 #m^2
#Flux emitted by globe is
fluxe = sa*B #lm
#luminous intensity of globe is
lint = fluxe/(4*3.14) #Cd
print "luminous intensity of globe is =",round(lint),"Cd."
#Flux absorbed by globe is
fluxab = flux - fluxe #lm
#% absorption is
absrp = fluxab/flux*100 #% absorption
luminous intensity of globe is = 44.0 Cd.
percentage absorption = 44.61 %.

A 22cm diameter globe of opal glass encloses a lamp

of uniform luminous intensity 120 C.P. Thirty per cent
of light emitted by the lamp is absorbed by globe.
Determine (a) luminance of globe (b) C.P. of globe in
any direction.'''
d = 0.22 #m (diameter)
I = 120.0 #Cd (luminous intensity)
#surface area is
sa = 3.14*d*d #m^2
#Flux emmited by source is
flux = I*(4*3.14) #lm
#Flux emmited by globe is
reflux = 0.7*flux #lm
#(a)Luminance of globe is
L = reflux/sa #lm/m^2
#(b) C.P of globe is
cp = reflux/(4*3.14) #Cd
Flux emmited by source is = 6942.15 lm/m^2 .
C.P of globe is = 84.0 Cd.

A 0.4 m diameter diffusing sphere of opal glass

(20 percent absorption) encloses an incandescent lamp
with a luminous flux of 4850 lumens. Calculate the
average luminance of the sphere.'''
d = 0.4 #m (diameter)
lflux = 4850.0 #lm (luminous flux)
reflux = 0.8*lflux #lm (flux emmited by globe)
sa = 3.14*d*d #m^2 (surface area)
#Brightness B = flux emmited/surface area . i.e
B = reflux/sa #lm/m^2 (brightness)
print "Average luminance =",round(B,2),"lm/m^2 ."
Average luminance = 7722.93 lm/m^2 .

'If an integrating sphere 0.6 m in diameter whose inner

surface has a reflection coefficient of 0.8 contains a
lamp producing on the portion of the sphere, screened
from direct radiation,a luminance of 1000 cd/m2, what
is the luminous flux yield ofthe source ?'''
from sympy import Eq,solve,Symbol
coef = 0.8 # (reflection coefficient)
L = 1000.0 #cd/m^2 (luminance)
d = 0.6 #m (diameter)
Fl = Symbol('Fl')
E = coef*Fl/(3.14*d*d*(1-coef)) #lm/m^2
L1 = coef*E/3.14 #cd/m^2
#As L is equal to L1
eq = Eq(L,L1)
Fl = solve(eq)
Fl1 = Fl[0] #lumens
Luminous flux yield of the source = 1109.2 lm

A room 8 m * 12 m is lighted by 15 lamps to a fairly

uniform illumination of 100 lm/m^2. Calculate the
utilization coefficient of the room given that the
output of each lamp is 1600 lumens.'''
Area = 8*12 #m^2 (area of room)
num = 15.0 # (number of lamps)
I = 1600.0 #lumens (output of each lamp)
E = 100.0 #lm/m^2 (illumination)
#Lumens emmited by lamp
I_tot = num*I #lumens
#Lumens recieved by working plane
I1 = Area*E #lumens
#Utilization coefficient is
coef = I1/I_tot
Utilization coefficient = 0.4

The illumination in a drawing office 30 m*10 m is to

have a value of 250 lux and is to be provided by a
number of 300 W filament lamps. If the coefficient
of utilization is 0.4 and the depreciation factor 0.9,
determine the number of lamps required. The
luminous efficiency of each lamp is 14 lm/W.'''
A = 30*10.0 #m^2 (area)
E = 250.0 #lm/m^2 (illumination)
coef = 0.4 # (coefficient of utilization)
p = 0.9 # (depriciation factor)
eff = 14.0 #lm/W (luminous efficiency)
watt = 300.0 #W (wattage of eacch lamp)
#Now, flux = E*A/coef*p
flux = E*A/(coef*p) #lm (output in lumens)
#Flux emmited per lamp is
Fl2 = watt*eff #lm
#No. of lamps required are
num = flux/Fl2
Number of lamps required = 50.0

Find the total saving in electrical load and

percentage increase in illumination if instead
of using twelve 150 W tungsten- filament lamps,
we use twelve 80 W fluorescent tubes. It may be
assumed that (i) there is a choke loss of 25
per cent of rated lamp wattage (ii) average
luminous efficiency throughout life for each lamp
is 15 lm/W and for each tube 40 lm/W and
(iii) coefficient of utilization remains the same in
both cases.'''
#Luminous efficiency
eff1 = 15.0 #lm/W
eff2 = 40.0 #lm/W
#Total load in filament-lamps
flamp = 12*150.0 #W
#Total load in fluoroscent tubes
tube = 12*(80 + 0.25*80) #W
#Net saving
load = flamp - tube #W
print "Net saving in load =",round(load,2),"W."
#Let us assume that
#E1 -> illumination with lamps
#E2 -> illumination with tubes
#Now E1/E2 = (O/P in lumens 1)/(O/P in lumens 2)
tube2 = 12*80.0 #W
E1_E2 = flamp*eff1/(tube2*eff2)
#Increase in illumination is given by %increase =
(E2/E1 - 1)*100
increase = (1/E1_E2 - 1)*100.0
Net saving in load = 600.0 W.
Increase in illumination = 42.22 %

A football pitch 120 m * 60 m is to be illuminated

for night play by similar banks of equal 1000 W
lamps supported on twelve towers which are
distributed around the ground to provide
approximately uniform illumination of the pitch.
Assuming that 40% of the total light emitted
reaches the playing pitch and that an illumination
of 1000 lm/m2 is necessary for television purposes,
calculate the number of lamps on each tower.
The overall efficiency of the lamp is to be taken
as 30 lm/W.'''
Area = 120.0*60.0 #m^2 (Area of pitch)
E = 1000.0 #lm/m^2 (Illumination of pitch)
#Flux required is
flux = Area*E #lm
#Since only 40% reaches the ground.Total flux required is
lflux = flux/0.4 #lm
#There are 12 tower banks . Therefore flux by each tower
bank is
flux_each = lflux/12 #lm
#Output of each 1000 W lamp is
I = 30.0*1000 #lm
#Therefore, number of each lamps is
num = flux_each/I
Number of lamps on each tower is = 50.0

'A drawing hall in an engineering college is to be provided

with a lighting installation. The hall is 30 m * 20 m * 8 m
(high).The mounting height is 5 m and the required level
of illumination is 144 lm/m^2. Using metal filament
lamps, estimate the size and number of single lamp
luminaries and also draw their spacing layout. Assume :
Utilization coefficient = 0.6; maintenance factor = 0.75;
space/height ratio=1 lumens/watt for 300-W lamp = 13,
lumens/watt for 500-W lamp = 16.'''
A = 30.0*20 #m^2 (Area)
E = 144.0 #lm/m^2 (Illumination)
coef = 0.6 # (Utilization coefficient)
mfac = 0.75 # (maintenance factor)
#The flux is given by
flux = E*A/(coef*mfac) #lm
#Lumen output for 500-W lamp
I5 = 500.0*16 #lm
#Lumen output for 500-W lamp
I3 = 300.0*13 #lm
#No. of 500 W lamps required is
num5 = flux/I5
print "Number of 500-W lamps required =",round(num5)
#No. of 300 W lamps required is
num3 = flux/I3
print "Number of 300-W lamps required =",round(num3)
Number of 500-W lamps required = 24.0
Number of 300-W lamps required = 49.0

Estimate the number and wattage of lamps which would

be required to illuminate a workshop space 60 *
15 metres by means of lamps mounted 5 metres above
the working plane. The average illumination required is
about 100 lux.Coefficient of utilization=0.4 ; Luminous
efficiency=16 lm/W.Assume a spacing/height ratio of
unity and a candle power depreciation of 20%.'''
A = 60.0*15.0 #m^2
E = 100.0 #lm/m^2
coef = 0.4 # (coefficient of utilization)
lum = 16.0 #lm/W (luminous efficiency)
dep = 1+0.2 # (depriciation factor)
#Total flux is given by
flux = E*A/(coef*1/dep) #lm
#Total wattage required is
watt = flux/lum #W
#Now,space/height ratio is 1.
h = 5.0 #m
#Therefore along breadth , lamps are
num_b = 15.0/h
#Therefore along length , lamps are
num_l = 60.0/h
#Total number of lamps are
num_tot = num_b*num_l
#Wattage of each lamp is
watt_each = watt/num_tot
Total number of lamps = 36.0
Wattage of each lamp is = 500.0 W.

A drawing hall 40 m * 25 m * 6 high is to be

illuminated with metal-filament gas-filled lamps to an
average illumination of 90 lm/m^2 on a working plane 1
metre above the floor.Estimate suitable number, size
and mounting height of lamps. Sketch the Spacing layout.
Assume coefficient of utilization of 0.5, depreciation
factor of 1.2 and spacing/height ratio of 1.2
Size of lamps : 200 W 300 W 500 W
Luminous efficiency (in lm/W) : 16 18 20 '''
A = 40.0*25.0 #m^2 (Area)
E = 90.0 #lm/m^2 (Illumination)
coef = 0.5 # (Coefficient of utilization)
dep = 1.2 # (Depreciation factor)
#Total flux required is
flux = E*A/(coef*1/1.2) #lumens
#Lumen output of each 200-W lamp is
flux_200 = 200.0*16 #lumens
#Lumen output of each 200-W lamp is
flux_300 = 300.0*18 #lumens
#Lumen output of each 200-W lamp is
flux_500 = 500.0*20 #lumens
#Number of 200-W lamps required is
num_200 = flux/flux_200
#Number of 200-W lamps required is
num_300 = flux/flux_300
#Number of 200-W lamps required is
num_500 = flux/flux_500
Number of 200-W lamps required are = 68.0
Number of 300-W lamps required are = 40.0
Number of 500-W lamps required are = 22.0

'A school classroom, 7 m * 10 m * 4 m high

is to be illuminated to 135 lm/m^2 on the
working plane. If the coefficient of utilization
is 0.45 and the sources give 13 lumens per watt,
work out the total wattage required, assuming a
depreciation factor of 0.8 .Sketch roughly the plan
of the room, showing suitable positions for fittings
giving reasons for the positions chosen.'''
A = 7.0*10.0 #m^2 (Area)
E = 135.0 #lm/m^2 (Illumination)
coef = 0.45 # (Coefficient of utilization)
dep = 0.8 # (Depreciation factor)
eff = 13.0 #lm/W (luminous efficiency)
#Total flux is
flux = E*A/(coef*dep) #lumens
#Therefore,total wattage required is
watt = flux/eff #W
Total wattage required is = 2019.23 W.

A hall 30 m long and 12 m wide is to be illuminated

and the illumination required is 50 lm/m2. Calculate
the number,the wattage of each unit and the location
and mounting height of the units, taking a depreciation
factor of 1.3 and utilization factor of 0.5, given that the
outputs of the different types of lamp are as under :
Watts : 100 200 300 500 1000
Lumens : 1615 3650 4700 9950 21500
A = 30.0*12.0 #m^2 (Area)
E = 50.0 #lm/m^2 (Illumination)
coef = 0.5 # (coefficient of utilization)
dep = 1/1.3 # (depreciation factor)
#Total flux required is
flux = E*A/(coef*dep) #lumens
#For 100-W lamps are used ,Number required
num_100 = flux/1615.0
#For 200-W lamps are used ,Number required
num_200 = flux/3650.0
#For 300-W lamps are used ,Number required
num_300 = flux/4700.0
#For 500-W lamps are used ,Number required
num_500 = flux/9950.0
#For 1000-W lamps are used ,Number required
num_1000 = flux/21500.0
Number of 100-W lamps required = 29.0
Number of 200-W lamps required = 13.0
Number of 300-W lamps required = 10.0
Number of 500-W lamps required = 5.0
Number of 1000-W lamps required = 2.0
If we take the mounting height of 5 m, then 300 W
lamps would be suitable. The No.of lamps required
would be 10, arranged in two rows, each row having
5 lamps thus giving space/height ratio of 6/5

'It is desired to floodlight the front of a building 42 m

wide and 16 m high.Projectors of 30° beam spread and
1000-W lamps giving 20 lumen/watt are available. If the
desired level of illumination is 75 lm/m2 and if the
projectors are to be located at ground level 17 m away,
Design and show a suitable scheme.Assume the following :
Coefficient of utilization = 0.4 ; Depreciation factor = 1.3
; Waste-light factor = 1.2. '''
A = 42.0*16.0 #m^2 (Area)
E = 75.0 #lm/m^2 (Illumination)
W = 1.2 # (Waste-light factor)
coef = 0.4 # (Coefficient of utilization)
dep = 1/1.3 # (Depreciation factor)
eff = 20.0 #lm/W (luminous efficiency)
#Total flux is
flux = E*A*W/(coef*dep) #lm/m^2
#Lumen output of each 1000-W lamp is
flux_each = 1000.0*eff
#Number of lamps required are
num = flux/flux_each
Number of lamps required are = 10.0

Estimate the number of 1000-W floodlight projectors

required to illuminate the up per 75 m of one face of a 96 m
tower of width 13 m if approximate initial average luminance
is to be 6.85 cd/m2. The projectors are mounted at ground
level 51m from base of the tower.Utilization factor is = 0.2;
reflection factor of wall = 25% and efficiency of each lamp
= 18 lm/W.'''
A = 13.0*75.0 #m^2 (Area to be flood-lighted)
B = 6.85 #cd/m^2 (Average luminance)
watt = 1000.0 #W (Wattage floodlight projectors)
coef = 0.2 # (Utilization factor)
ref = 0.25 # (Reflection factor)
#Illumination E = pi*B/reflection factor
E = 3.14*B/ref #lm/m^2
#Therefore, total flux required is
flux = E*A #lm
#Flux to be emmited by lamp is
lflux = flux/coef #lm
#Flux from each lamp is
flux_each = 18.0*watt #lm
#The number of floodlight projectors required are
num = lflux/flux_each
The number of floodlight projectors required = 24.0

If the filament of a 32 candela, 100-V lamp has a

length l and diameter d,calculate the length and
diameter of the filament of a 16 candela 200-V lamp,
assuming that the two lamps run at the same
intrinsic brilliance.'''
#As l*d is directly propotional luminous intensity .
#Let a = l1*d1 & b = l2*d2 . Then, l1*d1/l2*d2 = 32/16
a_b = 32.0/16 # (= a/b = l1*d1/l2*d2)
#As luminous intensity is directly propotional to power
output
# 32 o< 100*I1 & 16 o< 200*I2
I1_I2 = 32*200.0/(16*100.0) #( = I1/I2)
#Also , I o< d^3/2
d1_d2 = (I1_I2)**(2.0/3) #( = d1/d2)
#As , d1_d2 = d1/d2
l1_l2 = a_b/d1_d2 #( = l1/l2)
Diameter d2 = 0.4 *d1.
Length l2 = 1.26 *l1.

An incandescent lamp has a filament of 0.005 cm

diameter and one metre length. It is required to
construct another lamp of similar type to work at
double the supply voltage and give half the candle
power. Assuming that the new lamp operates at
the same brilliancy,determine suitable dimensions
for its filament.'''
d1 = 0.005 #cm (diameter)
l1 = 100.0 #cm (length)
#As l*d is directly propotional luminous intensity .
#Let a = l1*d1 & b = l2*d2 . Then, l1*d1/l2*d2 = 2.0/1.0
a_b = 2.0/1 # (= a/b = l1*d1/l2*d2)
#As luminous intensity is directly propotional to power output
# I1 o< V1*i1 & I2 o< V2*i2
I1_I2 = (2.0/1)*(2.0/1) #( = I1/I2)
#Also , I o< d^3/2
d1_d2 = (I1_I2)**(2.0/3) #( = d1/d2)
d2 = d1/d1_d2 #cm
#As , d1_d2 = d1/d2
l1_l2 = a_b/d1_d2 #( = l1/l2)
l2 = l1/l1_l2 #cm
Diameter d2 = 0.00198 cm.
Length l2 = 126.0 cm.