HOMEWORK V with Solutions
1.
(A) From the given location of C1 and C2 and the values of R1, R2, n, and d of the thick lens shown in
Fig.1, determine its focal length, the location of its focal points, and principal planes. (Use the concepts
and relations in Handout I, taking care of proper signs.)
(B) Draw your results into the diagram and obtain graphically the location and size of the image. What is
so and si and the magnification M in numerical value and sign?
(A) n = 2.0 , d = 3.0 cm, R1 = – 10.0 cm (Left of V1), R2 = – 4.0 cm (Left of V2)
1
f = (n – 1)
1
1
(n – 1) d
–
+
R1
R2
n R1 R2
[
]
 f = 5.3 cm
h1 = –
f (n – 1) d
R2 n
= + 2.0 cm (Rightside from V1)
h2 = –
f (n – 1) d
R1 n
= + 0.8 cm (Rightside from V2)
so = + 3.5 cm 
M= –
so
si
=
si = – 10.2 cm
+ 2.9
(Virtual image at 10.2 cm from H2)
(Up-right and about 3 times magnified image)
H1
Ff
Image
H2
Object
Fb
so
si
From the ray-tracing,
si = 10.2 cm on the left side from H2 (Virtual image)
yo = 1.1 cm and yi = 3.3 cm  M = 3.0
···········
Good agreement with calculation.
2.
The eye of a particular person is described (strongly over-simplified) by a positive lens with a focal length
which can be varied between 1.4 and 1.5 cm, and a distance of this lens to the retina of 1.6 cm. Besides
giving the answers to questions (A) and (B), try to make some schematic sketches for illustration.
(A) What is the range (far point and near point) of good sight for this eye?
Lens equation for a positive lens:
1
feye
=
1
so
+
1
si
For this eye, si = 1.6 cm and feye can vary from 1.4 to 1.5 cm. Therefore,
so can vary from 11.2 cm to 24.0 cm to make clear image on the retina.
Near point = 11.2 cm, Far point = 24.0 cm
when f eye= 1.5 cm
Far Point
when f eye= 1.4 cm
Near Point
(B) What spectacles (focal length? refractive power?) will correct the eyesight as well as possible to the
"normal" one – – and how does this eye-glass change far point and near point vision of the person?
* Normal eye usually means that its far point is at infinity.
Since the far point of this eye is 24 cm (not infinity), the eye is nearsighted. Therefore, a
negative lens (f = fcor) is needed to correct this eye. When s o = ∞, the lens combination of
feye = 1.5 cm and the corrective lens fcor should make si = 1.6 cm, that is, clear image on
the retina. (We ignore the distance between the eye lens and the corrective lens.)
1
feye
+
1
=
fcor
1
so
+
1
si

fcor = – 24 cm
Refractive power = fcor-1 = – 4.2 Diopters
Then, the near
point (feye = 1.4 cm) changes to 21.0 cm.
fcor= – 24 cm
Far Point
at infinity
when f eye= 1.5 cm
when f
= 1.4 cm
eye
Near Point
at 21 cm
(C) What would be the visual magnification if the person uses a magnifying glass of f = 2 cm with or
without spectacles?
Magnifying Glass: M =
do
f
where f = 2 cm
Without spectacles: do = Near point = 11.2 cm

M = 5.6
With spectacles:
do = 21 cm

M = 10.5
(D) What would be the visual magnification of very distant objects if the person uses an objective of f=30cm
together with the magnifiying glass under C to form a telescope:
M 
fo
 15
fe
(E) What minimum diameter should the objective have to collect the maximum possible light?
M
f o Dentrance Dobject


 15
fe
Dexcit
Dpupil
Dobjective  15  0.8cm  12cm

3. The objective of a microscope lens has a focal length of 12 mm. The image formed by this objective
is 210 mm from its second focal point. The eyepiece has a focal length of 35 mm.
(A) What is the magnification of the microscope?
For the object lens,
si = 12 + 210 = 222 mm, fo = + 12 mm  so = 12.7 mm
Microscope magnification: ( do = 250 mm for normal eyes)
si
M = Mobjective · Meyepiece = – s
o
·
do
fe
=
– 124.3
*** When calculating the Mobjective, we can use the following relation:
si
L 210
Mobjective = – s
= –

17.5
o
f o 12
where L is the distance from the second focal point of the objective to the image of the
objective lens. In this problem, L = 210 mm as given in question.
(B) The unaided eye can distinguish two points as separate if they are about 0.1 mm apart. What is the
minimum separation that can be perceived with the aid of that microscope?
Unaided eye can distinguish the separation by 0.1 mm.
With this microscope ( | M | = 175) , the eye can distinguish the separation by
∆ =

0.1mm
=
124.3
0.0008 mm = 0.8 m
Problem 4
In this solution the notation with nr’ was used. This gives for the a
thin lens the ABCD matrix:
 1
 1

 f
0  1 0  1 0

 

1 
D
1
P
1
  i 
 
no difference to our notation
