1. No category

ECE181 Homework Set 3 Solution
April 27, 2015
1
Problem 1
(a) For plano-convex glass, we have R2 = 1. Using lens maker equation, we
have:
1
= (ng
f
1)
1
R1
(1)
Where ng = 1.6 and f = 40mm. We can get that R1 = (ng
So the thickness of the lens is:
r
p
d
d=R
R2 ( )2 = 24
242 52 = 0.53mm
2
(b) We can use the equation f =
ing between the lenses.
D = f1 + f2
f1 f2
,
f1 +f2 D
f1 f2
= 7.27mm
f
1)f = 24mm.
(2)
where D is the physical spac-
(3)
1
2
Problem 2
First, we calculate the matrix for every element system. It is consist of two
free space with matrix:

1 d
Ms =
(4)
0 1
and two lenses with di↵erent focal length:

1 0
Mf =
1
1
f
(5)
We can calculate the matrix of this element system:
M = Ms Mf1 Ms Mf2
"
1 fd2
=
1
1
+ f1df2 1
f1
f2

A B
=
C D
d2
2d
f2
2
d
+ fd1 f2
f2
2d
f1
#
(6)
(7)
(8)
The determinant of the matrix det[M] == 1. So the ray trajectory is
bounded if
1
|A + D|  1
2
d
d2
+
 1 plus 1
(9)
f2 2f1 f2
d
d
d2
) 02
+
 2 divided by 2
f1 f2 2f1 f2
(10)
d
d
d2
) 01
+
 1 factor
(11)
2f1 2f2 4f1 f2
d
d
) 0  (1
)(1
)1
(12)
2f1
2f2
)
11
d
f1
2
3
Problem 3
(a) The system is showed in figure1, snell‘s law tells us that:
Figure 1: Problem 3a
sin ✓ = n sin ✓1
n sin ✓2 = sin ✓3
(13)
We also have ✓1 = ✓2 . From snell‘s law, we can get ✓ = ✓3 , this indicates the
ray is parallel to its initial direction.
The displacement of ray is
sin ✓
D = d tan ✓1 = d p
n2 sin2 ✓
(14)
3
(b) The system is showed in figure2. From (a), we know that ✓ = ✓0 = ✓00 , so
using snell‘s law, we have:
Figure 2: Problem 3b
n1 sin ✓1 = sin ✓
n2 sin ✓2 = sin ✓00 = sin ✓
(15)
So the transmitted ray is always parallel to the incident ray and nm sin ✓m =
sin ✓
4
4
Problem 4
Here we derive the general equation for ball lens‘ back focal length (the length
between the rear optical surface and the focus):
Figure 3: Ball Lens System
Incident angle ✓ is equal to arcsin( Ry ) The whole system is showed in fig3.
Assume ✓ is the incident angle, ✓0 is the refraction angle. According to snell‘s
law, they have the relationship:
nair sin ✓ = n sin ✓0
(16)
The back focal length of this ball equals to f = a + b
can be calculated by the equation:
= 2✓0
R where x = R cos .
✓
(17)
Now we calculate f , we know that y = l tan ✓00 where l = R sin and ✓00 =
90 +
✓. Using these equations we can calculate the focal length:
f = R cos(2✓0
= R cos(2✓0
✓) + R sin(2✓0
✓)
R sin(2✓0
sin(90 +
✓)
R
cos(90 +
✓)
cos(2✓0 2✓)
✓)
R
sin(2✓0 2✓)
✓)
5
(18)
(19)
(a) Apparently, we can not use paraxial approximation, because the largest
incident angle is:
✓max = sin 1 (
0.7
) = 44.43
1
(20)
We can see the di↵erence between paraxial approximation and general result,
we assume:
✓ = n✓0
(21)
The back focal length will be:
f =R
R
2✓0
2✓0
✓
2✓
R=
2 n
R
2(n 1)
(22)
Figure 4 shows the di↵erence. We can see that only when y < 0.2mm, we
can use paraxial approximation.
Figure 4: Paraxial Approximation Error for ball lens system
(b) based on the calculation above, we know the back focal length when
6
y = 0.7mm is f = 0.025mm
(c) Now we derive the general equation for e↵ective focal length (the distance
between the intercept with projected parallel ray and the focus) fEFL = a +
b r. We already know about a and b and r = tany✓000 where ✓000 = +90 ✓0 ,
so we have
f =R
R
2✓0
2✓0
✓
2✓
y
cos(90 + ✓0
sin(90 + ✓0
When y = 0.7, fEFL = 0.75mm
7
✓)
✓)
(23)