In place of the term intensity, the text distinguishes three quantities with units of power
per unit area. Measured at the surface of the emitter it is called exitance, at the surface of
a detector it is called irradiance. In either case, the generic name is flux density.
Problem 14
20W
W
I
 1.59 2
2
m
4 1.00m 
Problem 16
Intensity (irradiance) of sunlight at top of earth's atmosphere =
I
P
3.9 1026 W
W

 1.38 103 2
2
2
11
4 r
m
4 1.5 10 m 
I   0c E 2
E
2 1/ 2

E2 
I
 0c

W
2
5 V
m2

5.19

10
m2
C2 
8 m
 3 10 
N  m2  
s
1.38 103

12
 8.854 10

V2
V
 5.19 10 2  721
m
m
5
Problem 19
E0 
2I

 0c
2 1020
2
W
m2
 2.74 1011
V
m
C
m
 3 108
2
Nm
s
Notice the difference here between the amplitude (maximum value of the electric field)
and the rms amplitude of the previous problem. While the difference between the
definition of the two is a matter of 2 , the polychromatic radiation of the previous
problem implies that only average electric fields are meaningful while here the
monochromatic nature of the radiation implies that it is sensible to speak of a welldefined electric field.
It is interesting to compare this with the electric field holding an electron to a proton in a
hydrogen atom. This is
Nm 2
9  109 2  1.6  1019 C
ke
V
C

 1011
2
2
r
m
1010 m 
8.854 1012
so this laser is powerful enough to ionize matter!
Problem 21
v
With the insects flying to the right, the number of
insects leaving the shaded cross-sectional region
shown, of area 1 m2, in a time of 1s is the number
of insects contained in the box-like volume shown,
with dimension
m


1m 1m   0.1 1s   0.1m3
s


There are 100 insects per m 3 ,so in a volume of
0.1m3 there will be
insects
 0.1m3  10 insects
3
m
Ten insects per second cross the shaded region.
100
Problem 23
Frequency of 550nm radiation =
c


3 108
m
s
5.50 107 m
In the past we've used the letter "f" for frequency.
 5.45 1014 Hz  
Energy of one photon = h  6.626 1034 Js  5.45 1014 s 1  3.611 1019 J
J
J
1 photon
photons
100W  100  100 
 2.77 1020
19
s
s 3.61110 J
s
Problem 25
100W
W
I
 7.95 2
2
m
4 1.00m 
E0 
2I

 0c
2  7.95
8.854 1012
W
m2
2
C
m
 3 108
2
Nm
s
V
E
m  2.58 107 T
B0  0 
c 3 108 m
s
77.4
 77.4
V
m
Problem 38
E  E0 cos  t  kx 
rad

  (or f ) =
 5.00 1014 Hz
s
2
 1015 s 1
k
 1.61 107 m 1
m
0.65  3 108
s
2
2


 3.90  107 m  390nm
7
1
k 1.61 10 m
m
3 108
c
s  600nm
The vacuum wavelength would be 
 5.00 1014 s 1

600nm
n  vacuum 
 1.54
glass 390nm
   1015
Problem 44
glass 
vacuum
nglass

500nm
 312.5nm
1.60
1 cm
102 m
# complete waves 

 3.20  104  32, 000
9
312.5nm 312.5 10 m