HW Set 1 - Solutions
PHYS 320, Spring 2013
Due: January 23, 2013
SHORT ANSWER: (3 points each)
S1)
Pedrotti3 1-3
hc
  hv 

  380 nm  
6.6 1034  3 108 
380 10
9
1.6 10 
19
 3.26 eV
  770 nm   1.6 eV
S2)
Pedrotti3 1-7

hc

 Å  1010 meters
 Å 
 Å 
S3)
6.6 1034  3 1018  Å/sec
 1.6 10-19 
 12375 Å
12,375 12, 400

  eV    eV 
Pedrotti3 1-9
 p  2 109  eV, Proton mass = 1.67 10-27  kg
a)
mc 2  Ek  0.94  2  2.94 GeV  4.7 10 10   E
P
E 2  mc 2

c
P  1.49 1018 
4.7 2  1.52 1010 
kg  m
sec
3 108 
HW Set 1 - Solutions
PHYS 320, Spring 2013
S3)
Pedrotti 1-9 continued…
b)
c)
hc
=


E 2  m2 c 4
6.6 1034 
1.49 1018 
h
P
 4.45 1016  m
h
; m  0;
P
6.6 1034  3 108 
6.6 1034  3 108 
hc
or


Ek 2 109  1.6  1019 
4.7 1010 

  6.18 1016  m or   4.22 1016  m
Depending on interpretation of energy
S4)
Due: January 23, 2013
Pedrotti3 1-11
1 
hc
; 2 
hc
 energy per photon
1
2
1   2  energy of beans
N  number of photons
1 11

hc
1


N 2# (photons)= 2  2 2
hc
2
N1 11 Q2 2
N1# (photons) 
N2

hc
hc
N1 1

QED
N 2 2
; since 1   2
PHYS 320, Spring 2013
S5)
HW Set 1 - Solutions
Due: January 23, 2013
Pedrotti3 1-14

 length
4
v  90 106  Hz
8
c 3 10 
 
 3.33 m
v 90 106 

4
S6)
 0.833 m  83.3 cm
What is a diopter?
A Diopter is the unit of refracting power of an optical system when expressed in
meter-1. The refracting power of the lens is calculated by taking the inverse of
the focal length in meters of the lens. The refracting power is the change of the
curvature of the wave front from object space to image space due to the lens.
S7)
Describe the difference between aperture stop and field stop.


S8)
Aperture determines the amount of optical energy an optical system collects
Field stop determines the scene coverage or viewed region.
Describe the solid angle, linear angle, and the difference between the two.
The solid angle typically represented by Ω is the angle in three-dimensional
space that an object subtends at a point given in steradians or degrees squared.
Larger area at a further distance can subtend the same solid angle as a smaller
area nearer. It is usually thought of the area seen from the center of unit radius
sphere which would be 4 π sr, since the surface area is 4πr².
Linear angle is in a 2 dimensional plane, within a circle and has a maximum
value of 360 degrees or 2 π radians.
The difference is 2-dimensional space for linear angle vs 3 dimension geometry
for solid angle.
HW Set 1 - Solutions
PHYS 320, Spring 2013
S9)
Due: January 23, 2013
A one-meter (diameter) parabolic telescope (F#/1) mirror has a 1˚ full field of
view. What is the size of the detector array dimensions in the x and y directions
required.
1M
F/1
1˚
Using the Aproduct which is a constant value thru out the system, We know what A is given
by the entrance pupil given and the field of view.
A
 (1)2
4

)2
  2 (1  cos  )   2   (1/ 2)(
180
A 
3
8(180)
2
 1.88(104 ) m-sr
Since the object is at infinity,
At the focal plane angle is from F/1 beam, so:
tan  
0.5
at image plane
1
  26.6 degrees
Solid angle for cone of light is:   2 (1  cos 26.6)  0.67 sr
A  1.88  10 4  0.665*Aarray
Aarray  2.83  104 m 2
x  y  2.83  104  0.017 m  1.7 cm
Also, an easier way is to take Field of view times focal length:
2  100  tan 0.5   1.7 cm
HW Set 1 - Solutions
PHYS 320, Spring 2013
Due: January 23, 2013
S10) The Abbe number of a glass provides what general information about the F, d
and C wavelength of light?
The Abbe number is the inverse of the dispersive power of the material. The
Abbe number is calculated by:  
nD  1
. It is a quantitative measure of the
nF  nC
average slope of the dispersion (λ vs. n) curve. Smaller values of abbe # have
more spread of color that larger numbers.
PROBLEMS: (7 points each)
P1)
Pedrotti3 2-8
sin(θ1  θ 2 ) 
s

t

Rearranging and using trig. identity
cos θ 2 
 sin  a  b     sin a  cos b    cos a  sin b  :
sin 1 cos 2  cos 1 sin 2  sin  1  2  
s s t s
   cos 2
  t t
 cos 1  s
sin 1  sin 2 


t
cos

2 
Snell's Law
 cos 1  s
n
n
sin 2  sin 1 ;
sin 1  sin 1 

n'

n'
t
cos
2 

Solve for s :

n cos 1 
s  t sin 1  1 

 n 'cos 2 
2  30.71
n 'sin 2  n sin  50 
cos 50 

s  3sin 50  1 

 1.5cos 30.71 
s  1.153 cm
PHYS 320, Spring 2013
P2)
HW Set 1 - Solutions
Due: January 23, 2013
Pedrotti3 2-9
A meter stick lies along the optical axis of a convex mirror of focal length 40 cm, with its
nearer end 60 cm from the mirror surface. How long is image of the meter stick?
Focus:
Given: f = -40 cm, s1 = 60 cm, s2 = 160 cm
Find: L’ = s’
1
1
1


Plan: Use eqn (2-31),
, to find image location of each of the meter stick and
s
s'
f
then the image length.
Execute:
1
1
1


60 s'1  40
1
1
1


160 s' 2  40
s '1  24 cm
s '1   32 cm
L '  24  (32)  8 cm
P3)
Pedrotti3 2-10
 Image 1 in glass
5


 3.33 cm  reduced distance
n 1.5
 Image 2 in reflection from mirror
through flat surface
n ' n n ' n
 
s' s
R
1.5 1.5 1.5  1.5


s'
2.5
7.5
1.5 1.5
1
0.5
1
=



s'
2.5 2.5
2.5
5
s '  7.5 in glass
flat surface effects going to air
1.5
n' n
 0 
15
s' s
1 1.5

s ' 15
s '  10 cm
HW Set 1 - Solutions
PHYS 320, Spring 2013
P4)
Pedrotti3 2-16
n2  n1 1

R
f
R
f 
n2  n1

R   n2  n1  f  1.52  1 25 
R  13 cm
P5)
Pedrotti3 2-19
a)
t0

1
1
3
1



20 5
20 f eff
f eff 
b)
20
3
t  10
1 1 1 1
   10
20 5 5 20
1
4
2
1




20 20 20
20

f eff  20 cm
Due: January 23, 2013
HW Set 1 - Solutions
PHYS 320, Spring 2013
P6)
Due: January 23, 2013
Pedrotti3 2-31
Show that the minimum distance between an object and its image, formed by a thin lens, is
4f. When does this occur?
D  s  s'  s 
fs

s f
 s  f  f  fs
dD
 1
0
2
ds
s  f 
s  0, 2 f
This occurs when s  s '
P7)
Pedrotti3 3-4
a)
b)
1 1 1
  ;
'
s1 s1 f1
1
1
1
2



s1' 40 40 3 40
s1'  20 cm
1
1 1
1
1
1





s2' s2 f 2 10 20 3 20
s2'  20
PHYS 320, Spring 2013
c)
HW Set 1 - Solutions
Due: January 23, 2013
Image of A in object space
1
1 1


'
s A s A f1
1
1
1
2
3
1






20  40 3 40 40
40
s A'
40
  0.5   1 cm @source
20
 Entrance pupil & aperture are Lens 1, L1
s A'  40 cm; size =
d)
Exit pupil is image of entrance pupil/stop in image space
1
1 1


'
s x sE f 2
2 9
1
1
1
7





sx' 30 20 3 60 60 60
sx'  8.57 cm 
Dx 
60
cm
7
4
cm
7
e)
Field stop is at the Aperture A.
f)
 0.25 
FOV   tan 1 
  0.716
 20 
FOV  0.716
PHYS 320, Spring 2013
P8)
HW Set 1 - Solutions
Pedrotti3 2-32
a)
b)
Each layer : displacement is
d1  t1 tan 
d 2  t2 tan 2
d3  t3 tan 3
d 4  t4 tan 4
Summation :
 d   ti tan i
i
Due: January 23, 2013
PHYS 320, Spring 2013
P9)
HW Set 1 - Solutions
Due: January 23, 2013
Pedrotti3 3-10
D
nF  nc
nd  1
 Find index @ wavelengths
  A
 A
n sin    sin  n

2
 2 
   A
sin
2
n
sin  A 2 
A  60
 98.33  
38 20 ' 60 
sin 

sin
2

  1.5132
2

nc 
sin 30
0.5
 38 33' 60 
 98.55 
sin 
sin 


2
2



  1.5157
nd 
0.5
sin 30
 99.2 
3912 ' 60 
sin 

sin
2

  1.523
2

nF 
0.5
sin 30
1.5230  1.5132
D
1.5157  1
D  0.019
PHYS 320, Spring 2013
P10) Pedrotti3 3-24
HW Set 1 - Solutions
Due: January 23, 2013
PHYS 320, Spring 2013
a)
HW Set 1 - Solutions
Due: January 23, 2013
Objective lens :
1
1
1

  0.1666
s ' 1.2 1
h' 6
s '  6 cm;

 5  MT
h 1.2
Eyepiece :
1 1 1
 
s ' s fe
1
1 1
 
25 s 3
1
1 1
 
25 3 s
28
3 25 1

 
75 75 s
75
h ''
25
s  2.68,

 9.33
h ' 2.68
Total Magnification  5  9.33  46.66
b)   6  2.68  868 cm
Separation of Lens is 8.68 cm