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Physics 1C
Week 3 Questions
Fall 1998
1. The radius of the sun is 6.96 x 108 m and the radiation intensity at its surface
is 64 MW/m2.
(a) Find the total power output of the sun.
(b) The earth has a radius of 6380 km and is 1.49 x 1011m from the sun. Calculate the
total power falling on the earth.
(c) It is suggested that spacecraft can be accelerated due to radiation pressure from
the sun. Find the force exerted on a 100 metre square reflective surface at the
earth orbit.
(d) The sun delivers about 1400 W/m2 to the earth's upper atmosphere. Find the
average magnitude of the electric field vector.
2. In order of decreasing frequency, the entire electromagnetic spectrum is made up of
(a) radiowaves, microwaves, IR, light, UV, and -rays (b) -rays, X-rays, UV, IR,
microwaves, and radiowaves (c) radiowaves, microwaves, IR, light, X-rays, and rays (d) light, IR, and UV (e) none of these.
3. The irradiance 1 m from a candle flame is just about 1.5 x 10-3 W/m2. How much
energy will arrive in 2.00 s on a disk having a 1.00-cm2 area held as close to
perpendicular as possible 1.00 m from the flame?
4. In 1982 workers at Bell Labs produced optical pulses lasting 30 femtoseconds. How
many wavelengths of 620 nm red light correspond to one of these little wavetrains?
Physics 1C
Week 3 Solutions
Fall 1998
1. (a) Find the total power output of the sun:
P = I (area) = IS (4 rS2 ) = 64.3 x 106 (4) (7 x 108)2 = 4 x 1026 W
(b) The area of the earth as seen from the sun (see below) is a flat circular disc of
radius RE with AE =  R E2 .
1
Rs 2
Intensity  2 and intensity at E is IE = IS 2 where RS = radius sun, r = distance
r
r
from earth to sun.
Then total power at the earth is:
PE = IEAE = IS
Rs 2
( R E2 ),
2
r
2
8
6
 RS R E 
7  (6.96  10 )( 6.38  10 ) 
11
PE = IS 
 =  (6.4 x 10 ) 
 = 1.79 x 10 watts.
11
1.49  10
 r 


2
Note: Figure is not drawn to scale! The
"area of the earth" means the projected area
of earth as seen from the sun, i.e. the area
of the shadow earth would make if you put
it on a big screen. This tells you how much
of the sunlight the entire disk of the earth will
intercept. It is not 4 R E2 the area of a sphere.
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r
sun
radius RS
earth
radius RE
Physics 1C
Week 3 Solutions
Fall 1998
~
(c) The Poynting vector S describes the magnitude and direction of energy flow of
electromagnetic energy. The magnitude of this is the irradiance I where
1
~
S = I  c o Eo2 .
2
Photons carry energy and move at the speed of light, so they also have
momentum. If they are absorbed or reflected they impart a force on the absorber
or mirror which results in a pressure. (Remember Ft = (mv), and pressure =
force/unit area) The momentum flow/unit area is given by S/c or I/c. The
radiation pressure, if the em wave is absorbed = I/c. The radiation pressure, if the
em wave is reflected = 2I/c. The force on an area A of reflective surface is
F = 2IA/c.
Hence: A = 100m  100m
R 
I = IE = IS  S 
 r 
2
2
 6.96  10 8  (100) 2
2I A

 0.094 N
F = E  2(6.4  10 7 )
11 
8
c
 1.49  10  3  10
Note that this is a very tiny force. You would need a lot of patience for
"interplanetary sailing".
(d) Find the average magnitude of the electric field vector in 1400 W/m2 sunlight:
As I 
1
c o Eo2 ,
2
Eo 
2I

c o
2(1400)
 1030 V/m
(3  10 )(8.85  10 12 )
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1. (e) See page 921 in text book. *Note that (b) is almost right, but it omits "light".
2. Irradiance is energy per unit time,
I  E / At
E = IAt = (1.5  10 W/m2)(10-4m2)(2.00 s) = 3  10 7 J .
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3. A 30 fs pulse has a length of 1 = ct = (3 x 108)(30 x 10-15) = 9 x 10-6 meters =
9000nm. So 9000/620 = 14.5 wavelengths could fit in one such wavetrain. The
advantage to having such short pulses is that information can be conveyed through a
series of pulses more rapidly if the pulses are shorter in duration. Being able to
transmit a lot of information quickly is critical to many hi-tech operations. Have you
tried "surfing the net" when the system is very busy and the transfer rate is slow?
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