Physics, 6th Edition
Chapter 33. Light and Illumination
Chapter 33. Light and Illumination
Light and the Electromagnetic Spectrum
33-1. An infrared spectrophotometer scans the wavelengths from 1 to 16 m. Express these
range in terms of the frequencies of the infrared rays.
f1 
c
1

3 x 108 m/s
c 3 x 108 m/s
13

30.0
x
10
Hz;
f


 1.88 x 1013Hz
2
1 x 10-6 m
2 16 x 10-6 m
Range of frequencies: 1.88 x 1013 Hz to 30 x 1013 Hz
33-2. What is the frequency of violet light of wavelength 410 nm?
f 
c


3 x 108 m/s
;
410 x 10-9 m
f = 7.32 x 1014 Hz
33-3. A microwave radiator used in measuring automobile speeds emits radiation of frequency
1.2 x 109 Hz. What is the wavelength?

c 3 x 108 m/s

;
f 1.2 x 109 Hz
 = 250 mm
33-4. What is the range of frequencies for visible light?
f1 
c
1

( Range of : 700 nm to 400 nm )
3 x 108 m/s
c
3 x 108m/s
14

4.29
x
10
Hz;
f


 7.50 x 1014Hz
2
-9
-9
700 x 10 m
2 400 x 10 m
Range of frequencies: 4.29 x 1014 Hz to 7.50 x 1014 Hz
33-5. If Plank’s constant h is equal to 6.626 x 10-34 J s, what is the energy of light of
wavelength 600 nm?
E  hf 
hc


(6.625 x 10-34 J  s)(3 x 108m/s)
;
600 x 10-9 m
461
E = 3.31 x 10-19 J
Physics, 6th Edition
Chapter 33. Light and Illumination
33-6. What is the frequency of light whose energy is 5 x 10-19 J?
E
5.00 x 10-19 J
f  
;
h (6.625 x 10-34 J  s)
f = 7.55 x 1014 Hz
33-7. The frequency of yellow-green light is 5.41 x 1014 Hz. Express the wavelength of this
light in nanometers and in angstroms?

c
3 x 108 m/s

;
f 5.41 x 1014 Hz
 = 555 x 10-9 m(1 x 1010 A/m);
 = 555 nm
 = 5550 A
33-8. What is the wavelength of light whose energy is 7 x 10-19 J?
E
hc

;

hc (6.625 x 10-34 J  s)(3 x 108m/s)

;
E
7 x 10-19 J
 = 284 nm
The Velocity of Light
33-9. The sun is approximately 93 million miles from the earth. How much time is required for
the light emitted by the sun to reach us on earth?
t
93 x 106 mi
 500 s ;
186,000 mi/s
t = 8.33 min
33-10. If two experimenters in Galileo’s experiment were separated by a distance of 5 km, how
much time would have passed from the instant the lantern was opened until the light was
observed?
t
s
5000 m

;
c 3 x 108 m/s
t = 17.0 x 10-6 s or 17.0 s
462
Physics, 6th Edition
Chapter 33. Light and Illumination
33-11. The light reaching us from the nearest star, Alpha Centauri, requires 4.3 years to reach us.
How far is this in miles? In kilometers?
s = ct = (186,000 mi/s)(3.154 x 107 s/yr)(4.30 yr);
s = ct = (3 x 108 m/s)(3.154 x 107 s/yr)(4.30 yr);
s = 2.53 x 1013 mi
s =4.07 x 1013 km
33-12. A spacecraft circling the moon at a distance of 384,000 km from the earth communicates
by radio with a se on earth. How much time elapses between the sending and receiving of
a signal?
s = (384,000 km)(1000 m/km) = 3.10 x 108 m;
t
4 x 1010 m
;
3 x 108 m/s
t
3.84 x 108 m
= 1.28 s
3 x 108 m/s
t = 133 s
33-13. An spacecraft sends a signal that requires 20 min to reach the earth. How far away is the
spacecraft from earth?
s = ct = (3 x 108 m/s)(20 min)(60 s/min);
s = 3.60 x 1011 m
Light Rays and Shadows
33-14. The shadow formed on a screen 4 m away from a point source of light is 60 cm tall. What
is the height of the object located 1 m from the source and 3 m from the shadow?
Similar trianges:
60 cm
60 cm
h

100 cm 400 cm
1m
h
(400 cm)(60 cm)
;
100 cm
h = 2.40 m
463
3m
h
Physics, 6th Edition
Chapter 33. Light and Illumination
33-15. A point source of light is placed 15 cm from an upright 6-cm ruler. Calculate the length
of the shadow formed by the ruler on a wall 40 cm from the ruler.
h
6 cm

;
55 cm 15 cm
h
6 cm
h = 22.0 cm
15 cm
40 cm
33-16. How far must an 80-mm-diameter plate be placed in front of a point source of light if it is
to form a shadow 400 mm in diameter at a distance of 2 m from the light source?
x
2000 mm

;
80 mm 400 mm
x = 400 mm
80 mm
400 mm
x
2000 mm
33-17. A source of light 40 mm in diameter shines through a pinhole in the tip of a cardboard
box 2 m from the source. What is the diameter of the image formed on the bottom of the
60 mm
box if the height of the box is 60 mm?
y
40 mm

;
60 mm 2000 mm
40 mm
y
y = 1.20 mm
2000 mm
*33-18. A lamp is covered with a box, and a 20-mm-long narrow slit is cut in the box so that light
shines through. An object 30 mm tall blocks the light from the slit at a distance of 500
mm. Calculated the length of the umbra and penumbra formed on a screen located 1.50 m
from the slit. (Similar ’s)
a
500  a

;
20 mm 30 mm
u
20
 ;
1500  a a
500 mm
1500 mm
a
a  1000 mm
30 mm
20 mm
u
20

;
1500  1000 1000
b
c
u = 50 mm
b 500  b
p
20

; b  200 mm; Now,

from which
20
30
1500  b b
464
p = 130 mm
u
p
Physics, 6th Edition
Chapter 33. Light and Illumination
Illumination of Surfaces
33-19. What is the solid angle subtended at the center of a 3.20-m-diameter sphere by a 0.5-m2
area on its surface?
A
0.5 m2
 2 
;
R
(1.6 m)2
 = 0.195 sr
33-20. A solid angle of 0.080 sr is subtended at the center of a 9.00 cm diameter sphere by
surface area A on the sphere. What is this area?
A = R2 = (0.08 sr)(0.09 m)2;
A = 6.48 x 10-4 m2
33-21. An 8½ x 11 cm sheet of metal is illuminated by a source of light located 1.3 m directly
above it. What is the luminous flux falling on the metal if the source has an intensity of
200 cd. What is the total luminous flux emitted by the light source?
A = (0.085 m)(0.11 m);

A = 9.35 x 10-3 m2
A 9.35 x 103m2

 5.53 x 10-3sr
2
2
R
(1.3 m)
F = I = (200 cd)(5.53 x 10-3 sr);
Total Flux = 4I = 4(200 cd);
1.3 m
F = 1.11 lm
Total Flux = 2510 lm
33-22. A 40-W monochromatic source of yellow-green light (555 nm) illuminates a 0.5-m2
surface at a distance of 1.0 m. What is the luminous intensity of the source and how
many lumens fall on the surface?
F = (680 lm/W)(40 W) = 27,200 lm;
I
FR 2 (27, 200 lm)(1 m) 2

;
A
0.5 m 2
465
I
F
A
;  2

R
I = 54,400 cd
Physics, 6th Edition
Chapter 33. Light and Illumination
33-23. What is the illumination produced by a 200-cd source on a small surface 4.0 m away?
E
I
200 cd

;
2
R
(4 m) 2
E = 12.5 lx
33-24. A lamp 2 m from a small surface produces an illumination of 100 lx on the surface.
What is the intensity of the source?
I = ER2 = (100 lx)(2 m)2;
I = 400 cd
33-25. A table top 1 m wide and 2 m long is located 4.0 m from a lamp. If 40 lm of flux fall on
this surface, what is the illumination E of the surface?
E
F
40 lm

;
A (1 m)(2 m)
E = 20 lx
33-26. What should be the location of the lamp in Problem 33-25 in order to produce twice the
illumination? ( Illumination varies inversely with square of distance.)
R2 
E1R12 = E2R22 = (2E1)R22;
R12
(4m) 2

;
2
2
R2 = 2.83 m
*33-27. A point source of light is placed at the center of a sphere 70 mm in diameter. A hole is
cut in the surface of the sphere, allowing the flux to pass through a solid angle of 0.12 sr.
What is the diameter of the opening?
[ R = D/2 = 35 mm ]
A = R2 = (0.12 sr)(35 mm)2;
A
 D2
4
;
D
4A


A = 147 mm2
4(147 mm2 )

466
;
D = 13.7 mm
Physics, 6th Edition
Chapter 33. Light and Illumination
Challenge Problems
33-28. When light of wavelength 550 nm passes from air into a thin glass plate and out again
into the air, the frequency remains constant, but the speed through the glass is reduced to
2 x 208 m/s. What is the wavelength inside the glass? (f is same for both)
f 
vair
air

vglass
glass
glass 
;
vglass air
vair

(2 x 108 m/s)(550 nm)
;
3 x 108 m/s
glass= 367 nm
33-29. A 30-cd standard light source is compared with a lamp of unknown intensity using a
grease-spot photometer (refer to Fig. 33-21). The two light sources are placed 1 m apart,
and the grease spot is moved toward the standard light. When the grease spot is 25 cm
from the standard light source, the illumination is equal on both sides. Compute the
unknown intensity? [ The illumination E is the same for each. ]
I x Is
 ;
rx2 rs2
Ix 
I s rx2 (30 cd)(75 cm)2

;
rs2
(25 cm)2
Ix
30 cd
75 cm
25 cm
Ix = 270 cd
33-30. Where should the grease spot in Problem 33-29 be placed for the illumination by the
unknown light source to be exactly twice the illumination of the standard source?
E x  2 Es ;
I x 2I s
 2 ;
rx2
rs
4.5
1
 2;
2
(1- x)
x
270 cd 2(30 cd)

;
(1- x) 2
x2
4.5 x 2  (1  x) 2 ;
3.12 x  1;
x
1m
30 cd
2.12 x  1  x;
1
 0.320 m;
3.12
467
x
Ix
270 cd
1m–x
x = 32.0 cm from standard
Physics, 6th Edition
Chapter 33. Light and Illumination
33-31. The illumination of a given surface is 80 lx when it is 3 m away from the light source. At
what distance will the illumination be 20 lx? ( Recall that I = ER is constant )
E1R12  E2 R22 ;
R2 
E1 R12
(80 lx)(3 m) 2

;
E2
(20 lx)
R2 = 600 m
33-32. A light is suspended 9 m above a street and provides an illumination of 35 lx at a point
directly below it. Determine the luminous intensity of the light.
E
I
; I  ER 2  (36 lx)(9 m) 2 ;
2
R
I = 2920 cd
*33-33. A 60-W monochromatic source of yellow-green light (555 nm) illuminates a 0.6 m2
surface at a distance of 1.0 m. What is the solid angle subtended at the source? What is
the luminous intensity of the source?
F = (680 lm/W)(60 W) = 40,800 lm;
I
F (40,800 lm)

;

0.60 sr

A 0.6 m2
= 0.600 sr

R 2 (1 m)2
I = 68,000 cd
*33-34. At what distance from a wall will a 35-cd lamp provide the same illumination as an 80cd lamp located 4.0 m from the wall? [ E1 = E2 ]
I1 I 2
 ;
r12 r22
r2 
I 2 r12
(35 cd)(4 m) 2
;

I1
80 cd
r2 = 2.65 m
*33-35. How much must a small lamp be lowered to double the illumination on an object that is
80 cm directly under it?
R2 
R12
(80 cm) 2

;
2
2
E2 = 2E1 and E1R12 = E2R22 so that: E1R12 = (2E1)R22
R2 = 56.6 cm;
468
y = 80 cm – 56.6 cm = 23.4 cm
Physics, 6th Edition
Chapter 33. Light and Illumination
*33-36. Compute the illumination of a given surface 140 cm from a 74-cd light source if the
normal to the surface makes an angle of 380 with the flux.
I cos (74 cd)(cos380 )
E

;
R2
(1.40 m)2
E = 29.8 lx
*33-37. A circular table top is located 4 m below and 3 m to the left of a lamp that emits 1800
lm. What illumination is provided on the surface of the table? What is the area of the
table top if 3 lm of flux falls on its surface?
tan  
3m
4m
 = 36.90
I
F = 4I
F 1800 lm

;
4
4
I cos (143 lm/sr) cos36.90
E

;
R2
(5 m)2
R=5 m
I  143 lm/sr
E = 4.58 lx;

4m
3m
A
F
= 0.655 m2
E
*33-38. What angle  between the flux and a line drawn normal to a surface will cause the
illumination of that surface to be reduced by one-half when the distance to the surface has
not changed?
E1 
I1
;
R12
E2 
Substitution yields:
I cos
;
R22
E1  2 E2 ;
2I cos = I
I1  I 2 and
and cos  = 0.5
469
or
R1 = R2
 = 600
Physics, 6th Edition
Chapter 33. Light and Illumination
*33-39. In Michelson’s measurements of the speed of light, as shown in Fig. 33-11, he obtained
a value of 2.997 x 108 m/s. If the total light path was 35 km, what was the rotational
frequency of the eight-sided mirror?
The time for light to reappear from edge 1 to edge 2 is:
t
s 35, 000 m

 1.167 x 10-4s
8
c 3 x 10 m/s
The time for one revolution is 8t: T = 8(1.167 x 10-4 s)
f 
1
1

;
T 8(1.167 x 10-4s)
f = 1071 Hz
*33-40. All of the light from a spotlight is collected and focused on a screen of area 0.30 m2.
What must be the luminous intensity of the light in order that an illumination of 500 lx
be achieved?
E
I
;
A
I  EA  (500 lx)(0.30 m 2 )  150 cd
I = 150 cd
*33-41. A 300-cd light is suspended 5 m above the left edge of a table. Find the illumination of
a small piece of paper located a horizontal distance of 2.5 m from the edge of the table?
R  (2.5 m) 2  (5 m) 2  5.59 m
R 
2.5 m
tan  
;
5m
E
  26.6
0
2.5 m
I cos (300 cd)(cos 26.60 )

;
R2
(5.59 m)2
470
E = 8.59 lx
5m
Physics, 6th Edition
Chapter 33. Light and Illumination
Critical Thinking Problems
33-42. A certain radio station broadcasts at a frequency of 1150 kHz; A red beam of light has a
frequency of 4.70 x 1014 Hz ; and an ultraviolet ray has a frequency of 2.4 x 1016 Hz.
Which has the highest wavelength? Which has the greatest energy? What are the
wavelengths of each electromagnetic wave? [ Recall that h = 6.625 x 10-34 J.]

c
3 x 108 m/s

;
f 1.150 x 106 Hz
 = 261 m
E = hf = 7.62 x 10-28 J

c
3 x 108m/s

;
f 4.70 x 1014 Hz
 = 639 nm
E = hf = 3.11 x 10-19 J

c
3 x 108m/s

;
f 2.40 x 1016 Hz
 = 12.5 nm
E = hf = 1.59 x 10-17 J
The radio wave has highest wavelength; The ultraviolet ray has highest energy.
*33-43. An unknown light source A located 80 cm from a screen produces the same illumination
as a standard 30-cd light source at point B located 30 cm from the screen. What is the
luminous intensity of the unknown light source?
I x Is
 ;
rx2 rs2
Ix 
I s rx2 (30 cd)(80 cm)2

;
rs2
(30 cm)2
Ix = 213 cd
*33-44. The illumination of a surface 3.40 m directly below a light source is 20 lx. Find the
intensity of the light source? At what distance below the light source will the
illumination be doubled? Is the luminous flux also doubled at this location?
I = ER2 = (20 lx)(3.40 m)2 ;
I = 231 cd
R2
E2 = 2E1 and E1R12 = E2R22 = (2E1)R22
R1
R2 
2
1
2
R
(3.40 m)

;
2
2
R2 = 2.40 m
471
F =0
Physics, 6th Edition
Chapter 33. Light and Illumination
*33-45. The illumination of an isotropic source is EA at a point A on a table 30 cm directly below
the source. At what horizontal distance from A on the table top will the illumination be
reduced by one half?
EA 
I
;
RA2
EB 
I
2 I cos

;
2
RA
RB2
I cos
;
RB2
E A  2 EB ;
R 2A
1

;
2
R B 2cos
So that: (cos ) 2 
1
;
2 cos
tan  
but
I A  IB
RA
 cos
RB
1
(cos  )3  ;
2
x
;
30 cm
RB

RA
30 cm
x
cos   3 0.5  0.794; and  = 37.50
x  (30 cm) tan 37.50 ;
x = 23.0 cm
*33-46. In Fizeau’s experiment to calculate the speed of light, the plane mirror was located at a
distance of 8630 m. He used a wheel containing 720 teeth (and voids). Every time the
rotational speed of the wheel was increased by 24.2 rev/s, the light came through to his
eye. What value did he obtain for the speed of light?
An increase of f = 24.2 Hz is needed for light to reappear from Edge 1 to Edge 2.
Thus, the time for one revolution of entire wheel is:
T
1
1

 0.0413 s
f 24.2 Hz
Now, the time for one tooth and one void between 1 and 2 is:
t
0.0413 s
 5.74 x105 s;
720 teeth
c
2(8630 m)
;
5.74 x 10-5s
Time for one round trip.
c = 3.01 x 108 m/s
472