# Class 7 - Telescope and microscope ```Class 7 - Telescope and microscope
Before we analyze the principle of a telescope, we have to understand how our eye creates images from
distant objects because the eye plays an integral role in the device. First, rays from a single point of a
distant object arriving essentially parallel to our eyes. Our eye, consisting of a lens, is mapping parallel rays
approaching at some angle to a single point on a retina proportional to the angle.
As it happens, the smaller the object the smaller the angles at which the rays incident upon our eyes.
Thus to magnify the image on the retina we have to enlarge the angles. This is what telescope does - maps
small angles to larger angles
1
1
Refracting Telescope
The optical setup is composed of two lenses placed at distance d from each other:
Paraxial rays are entering from left and exiting from right.
1. Write the equivalent matrix M of the whole system
rout
1
rout
M11
M21
M12
M22
rin
1
rin
(1.1)
where the prime denotes the angle from the optical axis.
2. What are the conditions for a telescope? What is d in this case?
3. Assume d f1 f2 . If one places an object at distance u from left to the system, find the distance v
from right to the system where the image is created.
4. What are the linear and angular magnifications?
5. Explain how this system works as telescope, namely enables one to see distant objects.
Solution:
1. The optical system is composed of three matrices: first is the lens of focal lens f1 , then it is propagation
in medium of constant index of refraction and the last is the lens of focal lens f2
rout
1
rout
1
f
1
2
0
1
1
0
d
1
rout
1
rout
1
0
1
f1
1
rin
1
rin
1
1 1
f
f
M
2
1 1
f
f
1
2
d
f1
1
d
f2
d
d
f1 f2
d
f2
d
d
f1 f2
1
1
2
d
f1
1
2
1
f
d
1
1
d
f2
rin
1
rin
1
f1
1
0
1
rin
1
rin
(1.2)
(1.3)
(1.4)
2. Telescopes work on angular magnification. In other words, it maps rays incoming at some angle to
another bigger angle. In addition, telescope’s action should be independent on rin
1
rout
f1
2
&ntilde; f1 f1
2
3. In case of d f1
d
f1 f2
1
f1
d
f1 f2
1
rin
&ntilde; d f1
0
f2 the equivalent matrix is
ff
M
0
f1
2
ff
1
1
d
f2
f2
Meq
1
0
v
1
ff
2
1
0
f1
ff
f2
1
2
(1.5)
f2
(1.6)
1
(1.7)
1
2
f
f
0f v f ff1
f
u
1
1
0
rin
Thus the total matrix including free propagation from u to v
2
1
1
2
f2
1
2
1
0
u
1
f
f
f
0f f u vf f
f
2
2
1
1
1
2
1
2
(1.8)
Now we demand imaging condition, namely the position of the ray on the imaging plane is independent
of the angle of the ray on the object plane
rout
&ntilde; ff2 u v ff1
1
2
M11 rin
f1
f2
1
M12 rin
0
v
,
&ntilde; M12 0
f2
pf1
f1
f2 q (1.9)
2
f2
f1
u
(1.10)
We got the imaging condition that is essentially different from that of single lens. v is linear in u instead
of the condition for single lens f1 v1 u1 . For instance if the object is very distant u &Ntilde; 8 , the image
will be formed at v &Ntilde; 8 rather than at f as it happens with single lens. Let us look at a simple case
with f1 f2 and u 2f1 . Then v 0 so the image is formed on the lens itself.
4. In the imaging condition, the equivalent matrix is
ff 0
Meq.img. 0
ff
2
1
1
(1.11)
2
That is object of height hin will be imaged with height
hout
ff2 hin mlinear hin
(1.12)
1
with mlinear ff21 the linear or transverse magnification. Angular magnification, however, is defined
as the ratio of the angles. For instance, rays coming from object at angular range ∆αin will be magnified
as
f1
∆αout ∆αin
(1.13)
f2
3
f1
f2
with mangular ff12 the angular magnification. The image formed at v is inverted and if the image is
smaller (mlinear 1) then there is an angular magnification mangular &iexcl; 1. Also the magnification is
independent of u and v. Linear magnification can be used only for real images.
5. Principle of operation of this telescope is as follows. First lens, called an objective, creates a real
inverted intermediate image for the next lens. The next lens, called eyepiece, is creating a virtual,
inverted magnified image which can be seen with an eye.
2
Microscope
The simplest microscope is composed of a single lens of
4
What is the condition to have linear magnification mlinear &iexcl; 1?
Solution: We first find the equivalent matrix of the system
M
1
0
v
1
1
f1
0
1
u
1
1
0
From imaging condition
u
v
1
0
v
1
1
u
1
f1
1
uv
f
0
&ntilde;
uf
uf
v1
&ntilde; v uuf
f
1
f
u
f
u1
1
f
1
v
v
f
u
v
1
u
f
uv
f
(2.1)
(2.2)
(2.3)
Linear magnification in this case is
mlinear
For |mlinear | &iexcl; 1 or in this case mlinear
M11 1 fv uv u f f
(2.4)
1 and u &iexcl; f
u f f 1
&ntilde;
2f
&iexcl;u
(2.5)
for 0 u f it also holds (check at home). In this case the image is real and can be seen on a screen.
In order to achieve higher magnification more optical elements are added. In particular we can add
another lens to the previous setup as in case of the telescope, but without demanding mapping of the angles
&ntilde; d f1 f2 . The first lens is referred to as the objective, it forms a real, inverted, magnified image
of the object. The second lens is an eyepiece and its role is to convert the diverging rays to parallel ones.
Finally, the human eye is converting the beams to an image on the retina. The magnifying power of the
system is the product of linear magnification of the objective and the angular magnification of the eyepiece:
mtot mlinear,objective mangular,eyepiece .
5
```