MASSACHUSETTS INSTITUTE OF TECHNOLOGY
2.710 Optics
Solutions to Problem Set #3
Spring ’09
Due Wednesday, March 4, 2009
Problem 1: Wanda’s world
a) The geometry for this problem is shown in Figure 1. For part (a), the object
(Wanda) is located inside the bowl and we are interested to …nd where the image is
formed. We start by using the matrix formulation to analyze the given system,
i
xi
=
=
"
1 0
s 1
R
n
1
n
+
(1 n)
R
1
0
s
n
1
1
(1 n)
R
+ s(1R n)
#
1 0
1
n o
xo
R
n
n o
xo
(1)
:
To …nd the imaging condition, we note that all the rays, regardless of their departure
angle o , arrive at the image point xi (i.e. @xi =@ o = 0),
R s
= 0)
+
n n
s =
R:
(2)
We see that the image is formed at the center of the bowl and is virtual. Using this
result in equation 1,
i
=
xi
1
n
(1 n)
R
0
n
n o
xo
(3)
:
The lateral magni…cation is,
xi
= n;
xo
ML =
(4)
and therefore, the image is erect.
b) For this part, the object (Olive) is located outside the bowl and we are interested to
…nd where the image is formed. Again, we solve this part using the matrix formulation,
n i
xi
=
=
"
1 0
1
s0
n
s0
n
1
+s
1
0
(n 1)
R
s(n 1)
R
ss0 (n 1)
nR
1
1
1
1 0
s 1
(n 1)
R
s0 (n 1)
nR
o
#
(5)
xo
o
xo
:
From the imaging condition (@xi =@
s0
+s
n
o
= 0); we solve for s0 ,
ss0 (n 1)
= 0)
nR
s0 =
(6)
snR
s (n 1)
The lateral magni…cation is (for an on-axis ray,
o
R
:
= 0),
s0 (n 1)
nR
snR (n 1)
=
(s (n 1) R) nR
xi
=1
xo
ML =
= 1
(7)
R
s (n 1)
R
:
From equations 6 and 7, the following cases arise:
1. If R > s(n 1) ! s0 < 0; the image is virtual, erect and is located at a distance js0 j
outside the bowl.
2. If R > s(n 1) ! s0 > 0; the image is real, inverted and is located at a distance
js0 j inside the bowl.
c) If we were to consider the glass container of thickness t as well as the inner, R1 ; and
outer, R2 , radii, the matrix formulation becomes,
n i
xi
=
1 0
s0
1
n
=
1 0
s0
1
n
=
"
"
1
0
(n ng )
R1
1
t(n ng )
R1 ng
t
ng
1
t
ng
1
P
1
t(ng 1)
R2 ng
+
t(n ng )
R1 ng
0
s snP +
1 + Ps
0
s
n
1
t(n ng )
R1 ng
+
t
ng
0
1
#
(ng 1)
R2
1
0
1 0
s 1
1
1 0
s 1
o
n o
xo
P
1
(8)
xo
t(ng 1)
R2 ng
1+
s0 P
n
t(ng 1)
R2 ng
#
o
xo
where,
P =
(n
ng )
R1
+
(ng 1)
R2
t
(n
n g R1 R2
ng ) (ng
(9)
1) :
For the case of uniform glass: R1 = R2 = R. The imaging condition is,
0
s
+s
n
ss0 (n 1)
nR
0
s t (n ng )
t
+
Rng n
ng
st (ng 1) ss0
+
Rng
nR
2
t (n
ng ) (ng
Rng
1)
=0
;
Figure 1: Wanda’s world problem
0
s
+s
n
ss0 (n 1)
+
nR
g
(10)
= 0;
where,
0
g
=
t
s t (n ng )
+
Rng n
ng
st (ng 1) ss0
+
Rng
nR
t (n
ng ) (ng
R ng
1)
(11)
:
Comparing equations 10 and 6, we see that in order to neglect the aquarium walls we
require that g << 1;
0
t
s (n ng )
1
+
Rng n
ng
t <<
R2 n + ss0 (n
s (ng 1) ss0
+
Rng
nR
ng ) (ng
(n
R 2 ng n
1) s0 R (n
ng ) (ng
Rng
ng )
1)
sRn (ng
<< 1 )
1)
:
(12)
Problem 2: Ball lens magni…er
a) In class we derived the composite matrix for a thick lens surrounded by air and
is given by,
h
i #
"
d(n 1)2
d
n 1
1
1
(n 1) R1 R2 + nR1 R2
1 + R2 n
Mthick =
:
(13)
d
1 Rd1 nn 1
n
For a ball lens magni…er, the lens thickness and radii are: d = 2R; and R1 =
Using these values on equation 13 we get,
3
R2 = R:
1
Mball =
2
2R
n
2 (n 1)
R n
2 nn 1
n 1
n
1
(14)
;
so the EFL is,
Rn
1
=
:
P
2 (n 1)
EF L =
(15)
b) To calculate the BFL we consider an on-axis object source at in…nity (i.e. so = 1;
and o = 0) which focuses by the ball lens at the optical axis (xi = 0). Using the
matrix formulation,
i
0
=
1
0
BF L 1
=
1
BF L 1
1
2
2R
n
2
2
BF L
+1
EF L
n 1
n
n 1
n
2
n 1
n
1
+
2R
n
n
1
n
2 (n 1)
R n
2 nn 1
BF L
EF L
0
xo
1
EF L
+1
2
n 1
n
= 0)
BF L =
(16)
0
xo
)
(17)
R(n 2)
:
2(1 n)
By symmetry, the F F L = BF L. Since 1 < n < 4=3, BF L > 0. The location of the
2nd principal plane respect to the back of the lens is,
x2P P = BF L
EF L =
R:
(18)
Again, by symmetry the location of the 1st principal plane is also at the center of the
sphere as shown in Figure 2.
c) As shown in Figure 3, an object is located at a distance d to the left of the back
surface of the ball lens, where
d=R
4 3n
:
4(n 1)
(19)
The distance of the object to the 1st principal plane is,
so = d + (EF L
F F L) =
4
1 Rn
1
= EF L:
2 2(n 1)
2
(20)
Figure 2: Location of principal planes: ball lens magni…er.
Figure 3: Imaging with a ball lens magni…er.
Since the object is located between the principal plane and the focal point, as indicated
by equation 20, the image formed is virtual. To …nd the image position we use the
imaging condition,
1
1
1
+
=
)
so si
EF L
si =
EF L;
(21)
so the image is formed to the left of the 2nd principal plane (i.e. to the left of the
center of the ball lens).
d) The lateral magni…cation is,
5
si
= 2;
so
ML =
(22)
and therefore, the image is virtual and erect.
e) The aperture stop (AS) is given by the …nite size of the ball lens and thus is located
at the center of the lens. The numerical aperture, N A; is,
N A = nair sin
=
= tan ;
where we have used the paraxial approximation and
point object and the edge of the AS,
NA =
R
=4 1
d+R
is the angle formed by an on-axis
1
n
:
(23)
f) By observing the object through the lens, our eye’s lens is e¤ectively imaging the
image generated by the sphere. So, if our eye is located at a distance L behind the
lens, the new object distance is so = L + R + EF L. The crystalline lens of the eye
accomodates by changing the eye’s focal length f in order to still satisfy a positive
image distance of si =5
~ cm. From the imaging condition, we …nd that the resulting
focal length after accomodation is,
2(n 1)
1
1
=
+ :
f
2(L + R)(n 1) + Rn si
(24)
To be able to see an object at a relaxed state (i.e. without accomodation of the eye),
the image generated by the spherical lens should be at in…nity, so the object should be
placed at a distance equal to EF L to the left of the 1st principal plane.
Problem 3: Telephoto lens design
a) For this problem, we are given two requirements of a telephoto lens system: i)
for an object at in…nity, = 10 2 radians and h = 5 10 2 f (cm); ii) The location of
the real image produced by the system. The telephoto lens system is shown in Figure
4. By inspection, we see that the …rst lens, L1 , focuses the object at in…nity to a point
of height xi = f , at a distance of f to its right. This image point now acts as the
object source for the second lens, L2 . The object and image distances are,
so = d f; and
si = 3f d:
(25)
We use the equation for the lateral magni…cation to …nd the separation distance d,
6
Figure 4: Telephoto lens system.
h
5 10 2 f
=
=5
xi
10 2 f
3f d
si
=
)
=
so
d f
ML =
3f
f
(26)
d
= 5)
d
f
:
d =
2
We use the imaging condition to …nd the focal length of L2;
1
f
2
1
1
1
+
=
)
so si
fo
1
1
+ 5
=
)
f
f
o
2
5
fo =
f
8
b) Referring to Figure 5, we see that,
7
(27)
Figure 5: Locating 2nd principal plane.
Figure 6: Locating 1st principal plane.
BC =
1
1
EF = AO )
2
2
OD = 2CD = 2 3f
OF = 5f
(28)
f
2
= 5f )
3f = 2f:
So the 2nd principal plane is located to the left of L1 at a distance of 2f:
To …nd the 1st principal plane, we reverse the propagation direction of the rays and
consider a point at in…nity coming from the right as shown in Figure 6. The parallel
rays will have a virtual image at point G produced by L2. This point image then acts
as a point object for L1 and we use the imaging condition to …nd the location of the
image (point D),
8
1
1
1
+
=
)
so si
f
1
1
1
+
=
)
9
si
f
f
8
si = 9f:
(29)
From Figure 6 we see that,
BC
EF
=
5
f
8
9
f
8
=
5
AO
=
)
9
EF
(30)
DO
5
=
) DO = 5f )
9
DF
OF = 4f:
So the 1st principal plane is located to the left of L1 at a distance of 4f:
c) From part (b) we know that,
(31)
EF L = DO = 5f:
Or even from the problem we can know the EF L without any calculations: because
the parallel ray bundle with angle passing through the system will focus at the focal
plane with height h, so the e¤ective focal length is EF L = h= = 5f:
d) This part can be answered by using only the principal planes and applying the
imaging condition,
so = 24f 4f = 20f )
1
1
1
+
=
)
so si
5f
20
si =
f)
3
1
ML =
:
3
The image plane is located to the right of L1 at a distance of 20=3 f
(32)
2f = 14=3 f:
An alternative way to solve this problem is by using the matrix formulation,
9
out
xout
1
=
2
= 4
3f
1
d + (3f
xout = d + (3f
1
0
0
d 1
d) 1
1
fo
1
0
h
1 0
d 1
1
d
fo
d) 1
d
fo
d
fo
in
d
f
1
+
in
1
1
f
+
d
f fo
d)
1
f
(3f
d)
(33)
xin
i
1
fo
(3f
d
f
1
1
f
+
1
fo
3
5
d
f fo
1
1
+
f
fo
d
f fo
in
xin
;
xin :
(34)
Because all the parallel rays with angle will focus at the point at the image plane
with h = 5 f , xout is independent of xin ;
d + (3f
1
d
f
d) 1
d
fo
d)
1
1
+
fo
f
(3f
(35)
= 5f;
d
f fo
(36)
= 0:
From equations 35 and 36 we have,
1
=
fo
1
d
f
(3f
d)
1
1
+ 1
f
5f
3f
5f
3f
d
d
1
)
d
d
d
1
d
(37)
1
d
f
= 0)
d =
f
;
2
which is consistent with the result of equation 26.
Problem 4: Microscope design
a) The geometry of this problem is shown in Figure 7. To let the human’s unaccommodated eye to focus the image on the observer’s retina, L3 should form an image
at in…nity (i.e. for a given point in the object the corresponding output is an o¤-axis
parallel ray bundle). The …rst lens, L1; forms an intermediate image at the S2 plane,
1
1
1
+
=
)
so 170
10
so = 10:625mm.
10
(38)
Figure 7: Microscope design problem.
b) If we place a small object at so in front of L1; the instrument acts as a microscope.
The magnifying power, M P , is the ratio of the image size formed on the human’s retina
when using the instrument and the image size when viewed with the naked eye,
(39)
M P = ML1 MA3
170
=
10:625
= 203X,
254
20
where, ML1 is the lateral magni…cation of the objective (L1) and MA3 is the angular
magni…cation of the eyepiece (L3).
c) According to part (a), S2 is at an intermediate image plane, so it is not the aperture
stop (we will see later that S2 is the …eld stop). Potential aperture stops are then the
rims of either L1 or L3. The aperture stop should be the smallest of the two. As shown
in Figure 8, the input angle 1 admitted by L1 is smaller than the angle admitted by
L3, 2;
1
5
170
<
<
(40)
2
1
20
;
so the aperture stop is the rim of L1. Since there are no optical components presiding
L1, the aperture stop is also the entrance pupil. To …nd the exit pupil the aperture
stop through L3 and …nd the location of the image using the imaging condition,
11
Figure 8: Finding the aperture stop.
1
1
1
)
+
=
20
190 si
si = 22:35mm,
si
ML3 =
so
=
0:117:
(41)
So the exit pupil is located at a distance of 22.35mm to the right of L3, and its radius
is aep = a1 ML3 = 0:588mm.
S2 is the …eld stop which limits the intermediate image size. The maximum lateral
size of an object that can be imaged by this instrument is restricted by the size of the
entrance window. The radius of the entrance window is,
aew = a2 ML1
10:625
= 16
170
(42)
= 1mm,
so the maximum lateral size that can be viewed is 2mm (diameter).
d) In traditional microscopes, the aperture stop is located at the objective’s rim; therefore, the subsequent optics create an image of the stop (i.e. the exit pupil) that is
located to the right of the eyepiece. The observer’s eye can be comfortably located
such that the eye’s pupil coincides with the exit pupil and the image can be observed
without vignetting. If the radius of L1 becomes a1 = 10mm, the aperture stop would
be the rim of L3 and would be collocated with the exit pupil. To avoid vignetting, the
12
eye pupil would have to be adjacent to the eyepiece, which is of course infeasible because (a) the eye pupil is located behind the cornea, and (b) even if the small distnace
between the cornea and pupil could be neglected, it would be really uncomfortable for
the viewer to place his or her eye in contact with the eyepiece. One remedy to this
problem is to stop down the objective (i.e. reduce its radius so that it becomes the
aperture stop instead of the eyepiece); that is not a good solution because, as we will
see when we do wave optics, this solution reduces the overall numerical aperture of the
system and, hence the resolution of the microscope. A better remedy is to replace the
eyepiece with one that has larger radius (assuming we can a¤ord one). Then again,
the objective becomes the aperture stop as desired.
13
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