Find the image formed by two
lenses in combination
Combinations of lenses
Image formed by lens 1
q1 =
• When two lenses are used in combination,
the image of the first lens is the object for
the second lens.
• The total magnification is the product of
the magnifications of the first and second
lens.
p1f1
(30)(10)
=
= 15cm
p1 − f1 30 − 10
Image 1
Image 2
from lens 1
Image formed by lens 2
p2 =20-q1 = 20-15= 5 cm
q2 =
p2 f2
(5)(20)
=
= −6.7cm
p2 − f2
5 − 20
from lens 2
Magnification
Inverted
Reduced
⎛ q ⎞⎛ q ⎞ ⎛ 15 ⎞ ⎛ −6.7 ⎞
M = M1M2 = ⎜ − 1 ⎟⎜ − 2 ⎟ = ⎜ − ⎟ ⎜ −
⎟ = −0.67
5 ⎠
⎝ p1 ⎠⎝ p2 ⎠ ⎝ 30 ⎠ ⎝
For two lenses in contact the total power is
the sum of powers of the individual lenses
p
For two lenses in contact the total power is
the sum of powers of the individual lenses
1 1 1
+
=
p1 q1 f1
q
f1
f2
1 1 1
= +
f f1 f2
P = P1 + P2
−
1
1 1
+
=
p2 q2 f2
1 1 1
+
=
q1 q2 f2
substitute
p2 = −q1
Eliminate q1
1 1 1 1
+
= +
p1 q2 f1 f2
p1
=
1
f
q2
f1
f2
q1
The image of the first
lens is the object for the
second lens. Virtual object
has a negative sign.
Magnifiers
How do we image small objects?
5.2 Optical Instruments
• Simple magnifier
• Compound microscope
• Telescope
• We can image a small object by bringing it close to
our eye.
• But we cannot bring it closer than the near point.
(we can’t focus on it).
• A magnifier can produce a larger image of the
object at the near point (or farther away) that can be
focused on by the eye.
• Key concept - Angular magnification
1
Angular size
Simple Magnifier
Not to scale
25 cm
unfocused
A converging lens in combination with the lens of
the eye forms an image on the retina from an object
closer than the near point of the eye.
Ignore the distance
between the lens and
the eye
25 cm
θ
h
P
The angle θ increases as p decreases
The image size increases
image on
retina
Objects closer than the near point are not in
focus.
How does it work?
Produces an enlarged virtual image at a distance from
the eye (from 25 cm to infinity) that the eye can focus on.
h’
25 cm
object
virtual
image
θ
Angular Magnification
The angular magnification is the ratio of θ for the magnified
image compared to value of θo for the object at the near
point of the eye. (25 cm)
θ
m=
θo
25 cm
h’
Magnified
object
virtual
image
h
θ
h
•
q
θ Increases
f
•
f
p
25 cm
Simple Magnifier
Unmagnified
θo
h
Angular magnification
Angular magnification
Simple Magnifier
25 cm
Magnified
object
θ
The angular magnification for the simple
magnifier can have a range of values
because the focal length of the eye can
vary due to accommodation.
h
•
f
q = -infinity
p
25 cm
Unmagnified
The simplest case is the magnification for
the relaxed eye. (focused at infinity)
for small angles
tan θ ~ θ
Angular magnification
h
θo
θ≈
h
f
m=
θo ≈
h
25cm
θ 25cm
=
θ0
f
f is the focal length
of the lens in cm.
2
Simple magnifier
The maximum magnification of a simple magnifier.
The virtual image is at q=-25 cm
Simple magnifier.
A simple magnifier with a focal length of 5.0
cm is used to view an insect. What is the
angular magnification for a relaxed eye?
25 cm
object
θ
m=
h
•
f
q=-25 cm
m = 1+
p
25cm 25cm
=
= 5.0
f
5.0cm
What is the magnification for the accommodated eye?
25cm
f
m = 1+
25cm
25cm
= 1+
=6
f
5cm
Simple magnifiers.
Compound Microscopes.
The angular magnification for a single lens is
limited by aberration to about 4.
Combination lenses can have
magnification to about 20.
Magnification by 2 lenses.
Objective lens – Produces an enlarged
real image of the object.
Eyepiece – Used like a simple magnifier to
view the image.
The net angular magnification of the product
of the two magnifications.
Compound microscope
Compound microscope
Eyepiece
Objective
Not to scale
The objective lens produces a magnified real image I1
The image is viewed through the eyepiece.
M
a
3
Two stages of magnification by
1) objective and 2) eyepiece.
Total Magnification is the product
Not to scale
The objective lens produces a magnified real image I1
The image is viewed through the eyepiece.
M
Moa = −
q1
L
≈−
p1
fo
me =
25cm For relaxed eye
fe
Magnification
L (25cm)
fo
fe
Magnification increases when fe and fe get smaller.
Refracting Telescope
A compound microscope has an objective
lens and eyepiece with a focal lengths of
1.5 cm and 2.0 cm respectively. The
microscope is 20 cm long. Find the
angular magnification
m=−
m = M1me = −
Two lenses
Objective lens – produces a reduced image
of a distant object near the focal point.
Eyepiece – used to magnify the image.
L (25cm)
20 (25cm)
=−
= −167
fo
fe
1.5 2.0
Angular magnification - ratio of the focal length of
the objective and the eyepiece
Angular magnification
m=
θ fo
=
θo fe
θ=
h'
fe
θo =
h'
fo
Telescope
The Hubble space telescope has an
objective mirror with a focal length of 57.8
m viewed with optics equivalent to an
eyepiece with a focal length of 7.2x10-3m
What is the angular magnification?
m=
fo
57.8
=
= 8.0x103
fe 7.2x10 −3
focus at infinity
4
Hubble Telescope Image of M100 Spiral Galaxy (NASA)
Limits to magnification
Why can’t we use light microscopes to see atoms?
• For refracting optics there are problems of
chromatic and spherical aberration.
• Problems in precision in constructing the
refracting and reflecting surfaces.
• Diffraction – A basic problems having to
do with the wave nature of light (discussed
next week)
5