Example: What kind of lens must be used, in order to give an
erect image 1/5 as large as an object placed 15 cm in front
of it?
M = -q/p  -q/p=1/5
So q = -p/5 = -15/5 = -3 cm
1/p + 1/q = 1/f  1/15 - 1/3 = 1/f
1/f = (1-5)/15
f = -15/4 = -3.75 cm
Diverging lens
Magnifier


Consider small object held in front of eye
• Height y
• Makes an angle  at given distance from the
eye
Goal is to make object “appear bigger”: ' > 
y

Magnifier

Single converging lens
• Simple analysis: put eye right behind lens
• Put object at focal point and image at infinity
• Angular size of object is , bigger!
Outgoing
rays
Rays seen coming
from here

f
y

f
Image at
Infinity
1 1 1
 
q f p
Angular Magnification
(Standard)




Without magnifier: 25 cm is closest distance to view
• Defined by average near point. Younger people do better
•   tan  = y / 25
With magnifier: put object at distance p = f
• '  tan ' = y / f
Define “angular magnification” m = ' / 
Note that magnifiers work better for older people because
near point is actually > 25cm
~y/25
’~y/f
M= ’/  = 25/f
Example

Find angular magnification of lens
with f = 5 cm
25
m 
5
5
25
m 
1  6
5
Standard
Maximum
Optical Instruments
Eye Glasses
Perfect Eye
Nearsighted
Nearsighted can be corrected with a diverging lens.
 A far object can be focused on retina.
Farsighted
A
Power of lens: diopter = 1/f (in m)
(+) diopter  converging lens
(-) diopter  diverging lens
Larger diopter
 Stronger lens (shorter f)
Material
n
Cornea
1.38
Aqueous
Humor
Lens
1.331.34
1.411.45
1.34
Vitreous
Humor
Air
Water
1.00
1.33
Combinations of Thin Lenses




The image produced by the first lens is
calculated as though the second lens
were not present
The light then approaches the second
lens as if it had come from the image of
the first lens
The image of the first lens is treated as
the object of the second lens
The image formed by the second lens is
the final image of the system
Combination of Thin Lenses, 2

If the image formed by the first lens
lies on the back side of the second
lens, then the image is treated at a
virtual object for the second lens
• p will be negative

The overall magnification is the
product of the magnification of the
separate lenses
Combinations of Thin Lenses




The image produced by the first lens is
calculated as though the second lens
were not present
The light then approaches the second
lens as if it had come from the image of
the first lens
The image of the first lens is treated as
the object of the second lens
The image formed by the second lens is
the final image of the system
Combination of Thin Lenses, 2

If the image formed by the first lens
lies on the back side of the second
lens, then the image is treated at a
virtual object for the second lens
• p will be negative

The overall magnification is the
product of the magnification of the
separate lenses
What is the combined focal length?
1/p1 +1/q1 =1/f1
1/p2 +1/q2 =1/f2
p2=-q1 !!!
1/p1 +1/q2 =1/f1 +1/f2
1/f = 1/f1 +1/f2
Combination of Thin Lenses,
example
Q. Two converging lenses are placed 20.0 cm apart. If the
first lens has focal length of 10.0 cm and the second has a
focal length of 20.0 cm, locate the. final image formed of a
object 30.0 cm in front of the first lens. Find the
magnification of the system
1/30 + 1/q = 1/10
M1 = -q/p = -0.5
For second lens:
P= 20 –15 = 5 cm
1/5 +1/q = 1/20
M2 = -q/p = 1.33
M=M1 M2 =-0.667
q=+15 cm
q=-6.67 cm
Virtual, inverted
Compound Microscope
(two converging lenses)
objective
po
qo
eyepiece
pe
qe
Magnified inverted virtual image
Real image formed by the objective lens  an object for the eyepiece lens
Each set follows lens
Equations!!!
fo
po
qo
Mo
fe
pe
qe
Me
M = MoMe = (-qo/po)(-qe/pe)
= (qo/po)(qe/pe)
M = MoMe = (-qo/po)(-qe/pe)
= (qo/po)(qe/pe)
Magnification becomes larger qo >> po:
1/po + 1/qo = 1/fo
1/po ≈ 1/fo
The object should be put near the focal point of
the object lens.
Q. In a compound microscope, the obj. lens has a focal length of 8 mm,
And the eyepiece has a focal length 40 mm. The distance between the
lenses is 200 mm. If the object is placed 8.4 mm from the obj. lens,
What would be the magnification?
M = (qo/po)(qe/pe) = (168/8.4)(-160/32) = -100
po = 8.4
qo = ?
fo = 8
D = 200
pe = ?
qe = ?
fe = 40
1/po + 1/qo = 1/fo
1/qo = 1/fo – 1/po = 1/8 – 1/8.4
Inverted (M<0)
but
Virtual image
(qe<0)
qo = 168
pe = D – qe = 200 – 168
= 32
1/qe = 1/fe – 1/pe
= 1/40 – 1/32
qe = -160
Telescope




View Distant Objects
(Angular) Magnification M=fobj/feye
Increased Light Collection
Large Telescope use Mirror
Spherical Aberration


Results from the
focal points of light
rays far from the
principle axis are
different from the
focal points of rays
passing near the
axis
For a mirror,
parabolic shapes
can be used to
correct for
spherical
Chromatic Aberration

Different wavelengths of
light refracted by by a
lens focus at different
points
• Violet rays are refracted
more than red rays
• The focal length for red
light is greater than the
focal length for violet light

Chromatic aberration
can be minimized by the
use of a combination of
converging and
diverging lenses
Multiple Lenses in Cameras

Multiple lenses correct for various aberrations
• Spherical aberration (poor focus at edge of lens)
• Chromatic aberration (index of refraction varies with )
• Gauss arrangement probably most common

Actual arrangements are compromises!
• No perfect corrections for all factors
• Balance of many factors, including cost