Homework 3 Solutions
Problem 1
If we convert the wavefront to unnormalized coordinates (r,q), then the wavefront is given
by
W  , q = 0.001 Z40  , q = 0.001
r
3
5 6   - 6   + 1
r 4
3
r
3
r 2
3
Plotting the wavefront gives
PlotSqrt5  1000  6  r  3 ^ 4  6  r  3 ^ 2  1, r,  3, 3
0.0020
0.0015
0.0010
0.0005
-3
-2
1
-1
2
3
-0.0005
-0.0010
We expect for a smaller diameter pupil that the wavefront remains unchanged, just the edges get
clipped. Next, we want to normalize the coordinate system to the new pupil radius rmax = 1.5.
W  ApartSqrt5  1000  6  r  3 ^ 4  6  r  3 ^ 2  1 . r  3  2  
1
200

5
3 2
400
3 4

5
1600
5
As in problem 1, this is a rotationally symmetric function with a maximum power of r of 4. We can
therefore represent the wavefront as
W40 = c00 Z00 + c20 Z20 + c40 Z40 =
W40 = c00 + c20
1
3 2 r - 1 + c40
2
Equating like powers of r gives
200
-
3 r2
+
3 r4
5 6 r - 6 r + 1 =
5
400
4
5
1600
2
5
1
200
5
-
3 r2
400
5
+
3 r4
1600
5
2
HW3_13 Solutions.nb
Solvec00  c20  Sqrt3  c40  Sqrt5  1  200  Sqrt5,
2  Sqrt3  c20  6  Sqrt5  c40   3  400  Sqrt5,
6  Sqrt5  c40  3  1600  Sqrt5, c00, c20, c40
c00 
3
5
3
3
1600
3200
5
1
, c40 
, c20  
16 000

c00  0.000838525, c20   0.000726184, c40  0.0000625
N
W  Simplifyc00  c20
c00 
3
3
1600
3 2 2  1  c40
5 6 4  6 2  1 .
3
5
, c40 
, c20  
3200
5
1
16 000
,   2  3  r
27  18 r2  2 r4
5400
5
PlotW, r,  3  2, 3  2
0.0020
0.0015
0.0010
0.0005
-1.5
-1.0
-0.5
0.5
1.0
1.5
Problem 2
Part (a)
The power of the cornea is given by
1  1000  1.336  1  7.8
43.0769
For an object at infinity, the cornea (by itself) forms an image
HW3_13 Solutions.nb
Solve1.336  L1p  1, L1p
L1p  0.0310143
L1p  0.031014285714285706`
0.0310143
We can then use the imaging equation do determine where the IOL sits
Solve1.336  0.024  d  1.336  L1p  d  20.0, d
d  0.00557875, d  0.0494355
Choose the first solution since it is in the eye. So the IOL sits 5.58 mm behind the cornea.
Part (b)
For the object at 33cm, the cornea (by itself) forms an image at
ClearL1p
Solve1.336  L1p  1  .33  1, L1p
L1p  0.0333611
L1p  0.033361117578579735`
0.0333611
Calculate the IOL position d for the new object distance
Solve1.336  0.024  d  1.336  L1p  d  20.0, d
d  0.00323984, d  0.0541213
The IOL now sits 3.24 mm behind the cornea, so it had to move 2.34 mm
Part (c)
PMMA has an index of n = 1.49. The thickness t = 1 mm. Equiconvex means that the front and back
radii R1 = -R2. The power of each surface is given by
n  1.49
1  n  1.336  R1
2   1.336  n  R1
1.49
0.154
R1
0.154
R1
Next solve the thick lens power equation for R1
3
4
HW3_13 Solutions.nb
t  0.001
  20
Solve  1  2  t  1  2  n, R1
0.001
20
R1  0.0000518524, R1  0.0153481
So R1=15.35 mm. In air, the surface powers would be
1  n  1  0.015348147558319293`
2   1  n  0.015348147558319293`
31.9257
31.9257
and the total power (D) would be
  1  2  t  1  2  n
63.1673
In[1]:=
Problem 3
Part (a)
The Tscherning ellipse describes solutions for the front surface power f1 given the total power f of the
spectacle lens, the refractive index of the spectacle lens material (n = 1.4914 for PMMA), and the
distance between the spectracle and the center of rotation of the eye (27 mm in this case). Using these
values and solving for f1 gives
In[4]:=
n  1.4914;
  4;
q  0.027;
Solve1 ^ 2  n  2  1  2  q  n ^ 2  1    n  2  n    n  1  q ^ 2  0, 1
1  11.2332, 1  18.7412
So front surface powers of f1 = 11.23D and f1 = 18.74D are solutions for zero astigmatism. For question 4, it will also be useful to calculate the radii of these surfaces.
In[7]:=
n  1  11.2332
n  1  18.7412
Out[7]=
0.0437453
Out[8]=
0.0262203
The corresponding front surface radii which give zero astigmatism are 43.75 mm and 26.22 mm.
HW3_13 Solutions.nb
Part (b)
The Tscherning ellipse derivation assume a thin lens. Consequently, the total lens power is just the
sum of the front and back surface powers. From this, the back surface powers are
In[9]:=
Out[9]=
Out[10]=
  11.2332
  18.7412
 7.2332
 14.7412
So back surface powers of f2 = -7.23D and f2 = -14.74D are needed for zero astigmatism.
5
Question 4
This problem seeks to create a +4.00 D spectacle lens that minimizes astigmatism. The basic
setup is described in the problem. I added two features to the layout to facilitate the calculation.
First, I made radius of the posterior spectacle lens surface a variable and second I added the merit
function operand shown below, which drives the height of the on-axis marginal ray to zero at the
image plane.
I then wrote a little macro to export the astigmatism values:
for R1=10,100,1
radi(1)=R1
optimize
update all
code=ocod("ASTI")
astig=opev(code,0,1,0,1,0,0)
print R1,",",astig
next
The for loop sets the radius of the anterior spectacle lens surface to R1 and then optimizes the
system based on the merit function above. The optimization effectively sets the posterior radius
to a value that keeps the lens power constant. The ocod and opev lines in the macro extract the
Seidel astigmatism from Zemax and finally, the results are printed. A plot of the results is shown
below. The approximate solutions for R1 are 29 mm and 48.5 mm which are close to the values
in Q3.