Multiple Lenses • We determine the effect of a system of lenses by considering
the image of one lens to be the object for the next lens.
For the first lens: s1 = +1.5m, f1 = +1m
For the second lens: s2 = +1m, f2 = -4m
∴
∴
f = +1m f = -4m
-1 +3+10 +2 +6+5+4
Draw Rays !
m3'1 =s 2
1
'1
1 −=−=ssm
m31
m5.11
m11111
11'1
=−=−=sfs
m8.0'2 −=s 5
4
2
'2
2 +=−=ssm
m45
m11
m41111
22'2
−=−−
=−=sfs
58
21 −== mmm
Multiple Lenses • Objects of the second lens can be virtual. Let’s move the
second lens closer to the first lens (in fact, to its focus):
∴
∴
Note the negative object distance for the 2nd lens.
f = +1m f = -4m
-1 +3+10 +2 +6+5+4
421 −== mmm2
2
'2
2 +=−=ssmm4'
2 +=s
m41
m21
m41111
22'2
=−
−−
=−=sfs
For the first lens: s1 = +1.5m, f1 = +1mm31
m5.11
m11111
11'1
=−=−=sfs
m3'1 =s 2
1
'1
1 −=−=ssm
For the second lens: s2 = -2m, f2 = -4m
The eye
The Eye• The “Normal Eye”
– Far Point ≡ distance that relaxed eye can focus onto retina = ∞– Near Point ≡ closest distance that can be focused on to the retina
~ 25 cm
2.5cm
25cm
This is called “accommodation”Diopter: 1/f Eye = 40 diopters, accommodates by about 10%, or 4 diopters
cm 5.2=f
cm 3.2=f
cm 5.210111
' +=+=ssf
5.21
251111
' +=+=ssf
Therefore the normal eye acts as a lens with a focal lengthwhich can vary from 2.5 cm (the eye diameter) to 2.3 cm which
allows objects from 25 cm → ∞ to be focused on the retina!
Eye corrections (glasses, contacts)Near-sighted eye is elongated, image of distant object forms in front of retina
Add diverging lens, image forms on retina
Far-sighted eye is short, image of close object forms behind retina
Add converging lens, image forms on retinapower = 1/f; f in meters
Near vision Near point changes with age:
7 cm → 200 cm7 years 60 years
Most distinct vision is at near point. Image is largest.
Example: Reading glasses. The near point of a person’seye is 75 cm. What power reading glasses should be usedto bring the near point to 25 cm?
dioptersmf
cmcmfss
cmscms
67.2375.0
11751
251111
7525
==
−+==
′+
−=′=
ss’
oi F F
Assumes eye very close. Results are slightly different when distancebetween them is taken ino consideration.
virtual
A person with normal vision (near point 28cm) is standing in front of a plane mirror. What is the closest distance to the mirror the person can stand and still see himself in focus?
a) 14 cmb) 28 cmc) 56 cm
Object for eye = Image of self• Distance from eye to object = s+s’• Set s+s’ = near point
s s’
Old Preflight
Two people who wear glasses are camping. One of them is nearsighted and the other is farsighted. Which person's glasses will be useful in starting a fire with the sun's rays?
Old Preflight
a) the farsighted person's glasses
b) the nearsighted person's glasses
What do you need to start a fire?• REAL IMAGE ! (light is focused to a point)
• Converging lens gives REAL IMAGE• Far-sighted people need converging lenses!
Special Lens CombinationsIf two thin lenses are close together, they act effectively as a single lens. The focal length of the “doublet” is given by
1 2
1 1 1doubletf f f
= +f1 f2
fdoublet
Note the power (=1/f) of the combination is just Pdoublet = P1 + P2
Lecture 27, ACT 2• Hildegard’s retina is 2.5 cm behind
the lens, which has a minimum focal length of 2.6 cm.
1. What does the focal length fcl of her contact lens need to be?
(a) 65 cm (c) -0.1 cm(b) -65 cm
2. What is the power Pcl of the contact lens?
(a) 1.5 D (c) 1000 D(b) -1.5 D
2.5 cm
feye = 2.6 cm
Lecture 27, ACT 2• Hildegard’s retina is 2.5 cm behind
the lens, which has a minimum focal length of 2.6 cm.
1. What does the focal length fcl of her contact lens need to be?
(a) 65 cm (c) -0.1 cm(b) -65 cm
2.5 cm
feye = 2.6 cm
1 1 1eye clf f f
= +1 1 1
2.5 cmcl eyef f= −
(2.5 cm ) 2.6 2.5 65 cm2.5 cm 2.6 2.5
eyecl
eye
fff
×= = =
− −
Lecture 27, ACT 2• Hildegard’s retina is 2.5 cm behind
the lens, which has a minimum focal length of 2.6 cm.2. What is the power Pcl of the contact lens?
(c) 1000 D(b) -1.5 D
2.5 cm
feye = 2.6 cm
1 1 1.5 D0.65 mcl
clP
f≡ = =
40 D 38.5 Dcl need eyeP P P≡ − = −
Note: We could have solved for the power directly:
(a) 1.5 D
Angular Magnification• Our sense of the size of an object (in the absence of other
clues) is determined by the size of image on the retina.• This is proportional to the angle subtended by the object:
α1h
Bigger image
out
in
M θθ
≡
• The magnification of an optical system can then be equivalently defined as the ratio of the output angular spread to the input angular spread:
h’1 ~ α1
α2hh’2 ~ α2
Smaller image
Magnifying Glass• Our sense of the size of an object is determined by the size
of image on the retina. – If the object were closer to our eye, it would subtend a larger
angle.– However, we can only focus on an object if it is no closer than
the near-point distance Lnp (~25 cm).– We can use a simple magnifier to create an enlarged virtual
image outside Lnp.
α
Lnp
h
Object at Near Point - can’t get nearer
~fβh
Object just inside Focal Pointof simple magnifier
Positive “f” lens
npLh
≈α
fh
≈β
Define Angular Magnification: fL
M np≈≡αβ
⇒ Choose f < Lnp
Ma gnifierAt normal distances a small object (small letter print) subtends a small angle θ. with a lens or magnifier we can magnify the subtended angle to θ′.
The ratio is call angularmagnification
Which is NOT the same aslateral magnification.
θθ ′
=M
Ma
Telescopes• The purpose of a telescope is to gather light from distant
objects and produce a magnified image.– Refracting telescopes use lenses so that the objects can be viewed
directly.
– Reflecting telescopes use mirrors to create the image» Most astronomical telescopes are reflectors, since the most
important feature for these telescopes is the light gathering ability, and it is easier to make a large mirror than it is to make large lenses.
detector
θ1
θ1
θ2
h’
f1 f2
2 2 1
1 1 2
'/'/
h f fMh f f
θθ
= = =
11
'hf
θ ≈ 22
'hf
θ ≈
Telescope
21
12
//M
ff
fyfy
−=′
−=′
=θθ
Angular magnification
Note, light gathering power depends on objective lensDiameter.
Hubble Space TelescopeThe HST was launched in 1990; it was discovered that a lens had been ground incorrectly, so all images were blurry!A replacement “contact lens”, COSTAR, was installed in 1993.
Aperture of primary mirror: 2.4 m (~8 ft.)Mass of primary mirror: 828 kg (~1800 lbs)
Before COSTAR After COSTAR
(Now, Hubble’s instruments have built-in corrective optics, so
COSTAR is no longer needed.)