Problem1.
A radar system produces pulses consisting of 100 full cycles of a sinusoidal 70-GHz
electromagnetic wave. The average power while the transmitter is on is 45 MW, and the
waves are confined to a beam 20 cm in diameter. Find (a) the peak electric field, (b) the
wavelength, (c) the total energy in pulse, and (d) a total momentum in a pulse. (e) If the
transmitter produces 1000 pulses per second, what is the averaged power output?
Solution:
(a) The average intensity in a pulse is the average power during the pulse divided by
the beam area S  P A (1 point) and the peak electric field is
(2 points) E p  2 0cS  1.04 MV m .
(b) (1 point)
  c f  4.29 103 m .

(c) (2 points)
Duration of 100 full cycles is t 100 f 1.43 10
2
9
s , so the total
energy in a pulse is U  Pt  6.43 10 J .

(d) (2 points) Using the formula p  U c we obtain

p  2.14 1010 kgm s .

9
(e) (2 points)
Since
duration
of
one
pulse is t  1.43 10 s , the duration of 1000

pulses will
3

9
be t1 10 1.43 10 s 1.43 10
transmitter with 1000 pulses is

Pt
P  1  64.3W
1s

6
s . The averaged power output of the

Problem 2.
Two homogeneous isotropic dielectrics have a boundary plane z = 0. For z > 0 the
 dielectric constant (relative permittivity) is K E1  4 and for z < 0 K E2  3. A uniform
electric field E1  5iˆ  2 ˆj  3kˆ kV/m exists for z > 0. Find

(a) E 2 for z < 0
(b) The angles E1 and E 2 make
 with the interface 
3
(c) The energy densities in J m in both dielectrics
(d) The energy within a cube of side 2 m centered at (3, 4, -5).


Solution:
(a) E1  E1t  E1n
, where E1t  5iˆ  2 ˆj and E1n  3kˆ
Boundary conditions: (2 points)



E 2t  E1t  5iˆ  2 ˆj
D2n  D1n  K E2 E 2n  K E1 E1n


K E1
ˆ
E1n  4 k (2 points)
As a result, E 2n 
K E2

Thus , E 2  E 2t  E 2n  5iˆ  2 ˆj  4 kˆ kV/m.
1
1
(b) 1  tan E1n E1t  29.1 (1 point);  2  tan E 2n E 2t  36.6 (1 point)





2
1
1
1 E1   4  8.85 10 12  38 10 6  6.72 10 4 J m 3 (1 point)
2
2
2
1
1
  2 E 2   3 8.8510 12  45 10 6  5.97 10 4 J m 3 (1 point)
2
2
(c) uE1 
uE2
(d) The cube centered at (3, 4, -5) is in region 2. Hence
u  5.97 10 4  8  4.776 10 3 J (2 points)
Problem 3.
The 50-year-old man uses +2.5-D lenses to read a newspaper 25 cm away. Ten years
later, he must hold the paper 32 cm away to see clearly with the same lenses. What power
lenses does he need now in order to hold the paper 25 cm away? (Distances are measured
from the lenses.)
Solution:
We will use here equation for converging lenses
P
1 1
1


(4 points)
f d 0 d1
First we calculate the near point with P1  2.5 -D lenses and d01 = 0.32 m:

1
1
P 1 
 1.6m (3 points)
d1
d 01
Now we can calculate the power that is needed to hold the paper at d02 = 0.25 m:
P2 





1
1
1 
1


 P1 
 3.4D (3 points)
d 02 d1 d 02
d 01
Problem 4
Light strikes an equilateral right angle prism in a direction parallel to the prism’s base.
The point of incidence is high enough that the refracted ray hits the opposite sloping side.
(a) Through which side of the prism does the beam emerge?
(b) Through what angle has it been deflected?
Solution:
(a) The angle of refraction can be found from Snell’s law
2  sin 1 (sin 1 n)  sin 1 (sin 45 1.52)  27.7 (2 points)
The angle of incidence on the opposite side
 2  90  2  62.3 (1 points)
This angle is larger than the critical angle for total reflection from glass-air surface
c  sin 1 (1 1.52)  41.1 . (3 points)
Therefore, the incident light is totally reflected in the glass and hits the base at an
incidence angle  2   2  45  17.3 (1 point)
So the light emerges through the base. (no extra point for this answer without
considering total reflection)
1
(b) Its angle of refraction to the air is 1  sin (1.52sin  2 )  26.8 . (3 points)
Anet deflection is 90   2  63.2
Problem 5
 find the intensity at the first of secondary maxima in
In a single-slit diffraction picture
terms of thecentral peak intensity I 0 . (Hint: Assume that an angle
Solution:
The first minimum angle position is given by
Dsin 1  D1    1   D
The second minimum 
angle position is given by
Dsin 2  D2  2  2  2  D
 is small)

The position of the first secondary maxima is a midway between them:




  (3 2)  D (5 points)
Intensity of this maximum is
sin (Dsin   ) 2
sin (D  ) 2
sin (3 2) 2
I  I 0 
  I 0 
  I 0 
  0.045I 0

Dsin



D


3

2






(5 points, the form with a square will get at least 1 point)
Problem 6
Earth and Sun are 8.3 light-minutes apart, as measured in their rest frame.
(a) What is the speed of a spacecraft that makes the trip in 5 min according to its onboard clock?
(b) What is the trip time as measured by clocks in the Earth-Sun frame?
Solution:
v  x t  x (t ' ) (2 points)
v
8.3
v  x t '   
(2 points)
c
5
v
 0.857 (2 points for solving the result)
c
 1.94 (1 point)
t  t' 9.69 min


(1 point for the formula, 2 points for the result)
Problem 7
A particle of mass m whose total energy is twice its rest energy collides with an identical
particle at rest. If they stick together, what is the mass of the resulting composite particle?
What is its velocity?
Solution:

Momentum and energy of the system are conserved.
Initial momentum:
E12  p12 c 2  m 2 c 4  (2mc2 ) 2 (1 point)  p1  3mc  p  p1  p2  3mc (2 points
for using momentum conservation)
2
Initial energy: E  E1  E 2  3mc (2 points for using energy conservation)
Mass of the resulting composite particle:
M 2 c 4  E 2  p 2 c 2  9m 2 c 4  3m 2 c 4  M  6m (2 point)
Velocity of the composite particle:

p

Mv
1  v2 c2
 3mc  v  c
3 (3 points)