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2.71/2.710 Optics Spring 2014
Solution for HW
1. Modified from Pedrotti 18-
a) The schematic of the system is given below.
b) Using matrix method from point A to B,
1
2 1
2
1 2 1 2 2
lens
A B
M
C D
d d
d f
d d
f f
f f f f f
^
From the above matrix, we see that A and D are unitless, B is in the unit of length, and C has the unit of 1/length.
The effective focal length and location of the principle planes can be calculated from Table 18-2 of Pedrotti:
1 2
eq
EFL f cm
C
D A
FFP p cm BFP q cm
C C
D A
PP r cm PP s cm
C C
Here, PP1 represents the location of the starting plane of the system with respect to the original frontal plane A.
PP2 represents the location of the modified ending plane of the system with respect to the original ending plane B.
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2.71/2.710 Optics Spring 2014
Solution for HW
c) An example is sketched with the object placed at 20cm to the left of the first lens. The object is located 10cm
left to PP1.
From the lens law, we get:
i
i
s cm
s
We find that the image is located 20cm left to the 2
nd
principal plane, which is 25cm left to L1.
2
1
2
2 1
1 2 1 2 2
lens
d d
d f
M
d d f f
f f f f f
^
2.71/2.710 Optics Spring 2014
Solution for HW
3. A Telephoto Lens
a) The focal length of each individual lens can be determined as the following.
The ABCD matrix for two thin lens is:
.
From the matrix we can set the following parameters to be:
eq
EFL f m
C
A
BFL q mm
C
Plugging in the two values and using d=120mm, we get
b) the principal planes can be obtained from:
2
lens
A B
M
C D
^
0
1
2
f
n
D
n
PP mm
C
A
PP mm
C
4
1
2
mm
f mm
f
tan( ) 0. approx
x EFL EFL mm
in in
t in in
out
ou
x BFL A B x B D BF x
C D C D
L
A
B BFL B D B B C E
C C
D FL
C
AD BC 1
2.71/2.710 Optics Spring 2014
Solution for HW
PP1 is 240mm left of L1 and PP2 is 200mm left of L2.
c) The entrance pupil is identical to the aperture stop. The diameter can be obtained from the equation for F-
number:
f f mm N D mm
D N
d) The most efficient way to predict the answer is to graphically think about how much light propagates with the
given angle variation. We can convert the input angle into output vertical distance by recognizing that the
effective distance of propagation until the focus is EFL:
.
Is it really true? Let’s check mathematically with the rigorous matrix method. We can construct a new system
matrix that includes the propagation to the BFP:
,
since BFL=-A/C. This can be further simplified by:
.
Indeed! Here, comes from the fact that the determinant of the system matrix is equal to the ratio
between the initial and final refractive indices,. Therefore the distance between the two focuses is 0.3mm. 0
f
n n
5
( s 30)(s 6) 0
( )sin 5. o
x d s mm
o o i
x x x
0.873 mm
2.71/2.710 Optics Spring 2014
Solution for HW
5. Camera lens and Image Stabilizer system
a)
1 2
n
f R R mm
The transfer matrix from A to A’ can be found as:
s s s d s
f f s d s A B
C D
f
f
d s
f
^
^
The distance is d=36nm. The imaging condition is B=0, which can be solved for s to yield.
Which is the correct answer? To have a demagnified image (s<d-s), the valid answer is s=6cm.
b) After the rotation of the camera,
Then we move the lens up by distance : ,
.
i
o
x s
x d s
Therefore the lens must be moved by upward to compensate for the shaking.
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