Download Optics Image Formation from Mirrors and Lenses

Optics
Image Formation from Mirrors and Lenses
Lana Sheridan
De Anza College
June 8, 2016
Last time
• refraction
• Huygens’ Principle
• dispersion
Overview
• mirrors
• image terminology
• images formed by spherical mirrors
• images formed by refraction
• images formed by lenses
• images formed by lens combinations
Ray Optics and Image Formation
Simple geometric ray optics can be used to understand how images
are formed by simple optical devices: mirrors and lenses.
1
Wikipedia user Heptagon.
Images Formed by Flat Mirrors
When we see an object “in the mirror”, we are not actually seeing
something that is behind the mirror.
We are seeing an image of and object that is placed in front of
the mirror.
The image seems to be the same distance behind the mirror as the
object is in front of it.
The image seems to be the same size as the object.
This is not true for all optical devices, but we can work out things
about how the image will form for many different optical devices.
Image Formation Terminology
object distance, p
The perpendicular (shortest) distance from the object to the
device.
image distance, q
The perpendicular (shortest) distance from where the image
appears to be to the device.
(lateral) magnification, M
The factor by which the image size exceeds the object’s size
M=
image height
object height
For mirrors and lenses
M=−
q
p
Image Formation Terminology
real image
An image that can be displayed on a screen formed when the light
rays pass through and diverge from the image point.
virtual image
An image that cannot be displayed on a screen, but can be seen
“in the device” because the light rays appear to diverge from the
image point.
The image in a flat mirror is virtual.
Image Formation Terminology
upright image
An image that appears to be right-side-up with respect to the
object.
inverted image
An image that appears to be upside-down with respect to the
object.
The image in a flat mirror is upright.
focal length, f
The distance from the optical device to where a set parallel rays
striking the device head on (perpendicularly) will be focused.
For a flat mirror, the f is infinite.
rays to deterand P'QR are congruent,
path
perpen| ! | q Mirror
| and h ! h".
Ray Diagram for a| pFlat
ay follows the
law of reflected rays back
p Q
q
P
P'
9 behind the
object would
h'
R
h
u
object behind
Image
Object u
, so |p | 5 |q |.
irror is as far
h equals the
Ray diagrams have
the optical
sketched
vertically in cross
Figure
36.2 device
A geometric
confollows:
section, the object on the left represented by an arrow pointing up.
struction that is used to locate the
image of an object placed in front
of a flat mirror.
The(36.1)
image is also represented by an arrow, but it may be to the
left or right, pointing up or down depending on how the image is
formed.
a spherical
d a concave
e R, and its
ction, and a
6.6a shows a
olid, curved
port for the
flecting suro a point as
Henry Leap and Jim Lehman
Images formed by Spherical Mirrors: Concave
Mirrors
Figure
36.7
focus
light. Reflection of paral6.6b,Concave
wheremirrorslelcanrays
from a concave mirror.
ght rays that
6
Images formed by Spherical Mirrors: Concave
If the rays diverge from
Mirrors
O at small angles, they
all reflect through the
same image point I.
Center of
curvature
Mirror
Mirror
R
C
C
V
O
I
Principal
axis
a
b
(a) A concave mirror of radius R. The center of curvature C is located on th
oint object placed at O in front of a concave spherical mirror of radius R, wher
principal axis farther than R from the mirror surface, forms a real image at I.
Assumption: paraxial rays
In studying curved mirrors and thin lenses we assume that all rays
are paraxial rays.
Paraxial rays are rays close to the principle optical axis of our
optical device. (Rays that strike close to the middle.)
For a spherical lens, rays that strike further from the axis are not
focused to the same point (spherical aberation).
Mirror Equation and Thin Lens Equation
The same equation can help us to find the location and
magnification of the image that will be formed by curved mirrors
and thin lenses.
1
1 1
= +
f
p q
Spherical Concave Mirrors Focal Length
For a spherical concave mirror of radius R
f =
R
2
Spherical Concave Mirrors
The focal point is where the detectors are place on satellite dishes
and radio telescopes.
1
Parkes telescope in Australia.
Concave Mirrors and the Mirror Equation
Simple geometry shows why the mirror equation is true for a
concave mirror.
ge Formation
The real image lies at the
location at which the
reflected rays cross.
h
a
O
C
V
I
u
a h' u
Principal
axis
q
R
p
Figure 36.9 The image formed by a spherical concave mirror when the object O lies outside the center of curvature
This this
geometric
construction is used to derive Equation 36.4.
We willC.use
ray diagram.
Ray Diagrams for Spherical Mirrors
For a ray diagram: draw at least two rays that you know the path
of accurately.
For Spherical mirrors:
1
Draw a ray from the top of the object parallel to the principle
axis reflected through the focal point F .
2
Draw a ray from the top of the object through the focal point
and reflected parallel to the principal axis.
3
Draw a ray from the top of the object through the center of
curvature C and reflected back on itself.
Where the lines meet, an image is formed.
ge Formation
Concave Mirrors and the Mirror Equation
The real image lies at the
location at which the
reflected rays cross.
h
a
O
C
V
I
u
a h' u
Principal
axis
q
R
p
Figure 36.9 The image formed by a spherical concave mirror when the object O lies outside the center of curvature
C. Thisof
geometric
construction
is used to derive
Equation
The angles
incidence
and reflection
are the
same36.4.
magnitude, θ.
h0
q
So,sign
− p , and because the image is inverted, so h9 is taken to be neganegative
is introduced
h =
q
tive. Therefore, from Equation 36.1 and
M =these
− results, we find that the magnificap
tion of the image is
Concave Mirrors and the Mirror Equation
Looking at the green triangle and the red triangle with angle α:
−h 0
h
=
R −q
p−R
which rearranges to
p−R
−h 0
=−
h
p−R
Using our magnification expression:
p−R
q
=−
p
p−R
Concave Mirrors and the Mirror Equation
p−R
q
=−
p
p−R
Cross-multiplying and rearranging gives
2
1 1
= +
R
p q
However, we already concluded that f = R/2, so
1
1 1
= +
f
p q
We have confirmed the mirror equation for spherical concave
mirrors, and it follows from simple geometry.
o not derive any equations for convex spherical mirrors because Eq
6.4,Convex
and 36.6Mirrors
can be used for either concave or convex mirrors if we
ct sign convention. We will refer to the region in which light rays or
ve toward the mirror as the front side of the mirror and the other sid
We can draw the ray diagrams for convex mirrors in the same way.
The image formed by the
object is virtual, upright,
and behind the mirror.
Front
Back
O
p
q
I F
C
Examples of Ray Diagrams
36.2 Images Formed by Sphe
F
When the object is located so that the center of
curvature lies between the object and a concave
mirror surface, the image is real, inverted, and
reduced in size.
T
S
J
2
3
F
C
O
Principal axis
I
Front
Back
a
When the object is located between the
focal point and a concave mirror surface,
the image is virtual, upright, and enlarged.
© Cengage Learning/Charles D. Winters
1
Front
Examples of Ray Diagrams
Back
a
When the object is located between the
focal point and a concave mirror surface,
the image is virtual, upright, and enlarged.
2
3
C
1
F
O
Front
I
Back
b
When the object is in front of a convex
mirror, the image is virtual, upright,
© Cengage Lea
Principal axis
Front
Back
Examples of Ray Diagrams
b
When the object is in front of a convex
mirror, the image is virtual, upright,
and reduced in size.
1
3
2
O
I F
Front
C
Back
c
r 3BZJTESBXOGSPNUIFUPQPGUIFPCKFDUUPXBSEUIFGPDBMQPJOUPOUIFCBDL
Mirror Question
Quick Quiz 36.31 Consider the image in the mirror in shown.
Based on the appearance of this image, which of the following
1098
Chapter 36 Image
should you conclude?
in
vi
ro
ve
(A) the mirror is concave and
the image is real
(B) the mirror is concave and
the image is virtual
Q
(C) the mirror is convex and the
image is real
(D) the mirror is convex and the
image is virtual
NASA
Q
1
Serway & Jewett, page 1098.
Figure 36.14 (Quick Quiz 36.3)
Mirror Question
Quick Quiz 36.31 Consider the image in the mirror in shown.
Based on the appearance of this image, which of the following
1098
Chapter 36 Image
should you conclude?
in
vi
ro
ve
(A) the mirror is concave and
the image is real
(B) the mirror is concave and
the image is virtual ←
Q
(C) the mirror is convex and the
image is real
(D) the mirror is convex and the
image is virtual
NASA
Q
1
Serway & Jewett, page 1098.
Figure 36.14 (Quick Quiz 36.3)
Sign Conventions for Mirrors!
1 1
1
= +
f
p q
Variable
is Positive
is Negative
p
object in front of mirror
—
q
image in front of mirror
(real)
image behind mirror
(virtual)
h 0 and M
image upright
image inverted
f and R
concave mirror
convex mirror
Example 36.4: Convex Mirror Image
An automobile rearview mirror shows an image of a truck located
10.0 m from the mirror. The focal length of the mirror is −0.60 m.
ple 36.4) An
en in a convex
e of an autoimage of the
e frame of the
monstrates
the same locaace.
. Bo Zaunders/Corbis
Find the position of the image of the truck and the magnification
of the image.
Example 36.4: Convex Mirror Image
An automobile rearview mirror shows an image of a truck located
10.0 m from the mirror. The focal length of the mirror is −0.60 m.
Find the position of the image of the truck. (where does it appear
to be?)
Example 36.4: Convex Mirror Image
An automobile rearview mirror shows an image of a truck located
10.0 m from the mirror. The focal length of the mirror is −0.60 m.
Find the position of the image of the truck. (where does it appear
to be?)
q = −0.57 m
Find the magnification of the image.
Example 36.4: Convex Mirror Image
An automobile rearview mirror shows an image of a truck located
10.0 m from the mirror. The focal length of the mirror is −0.60 m.
Find the position of the image of the truck. (where does it appear
to be?)
q = −0.57 m
Find the magnification of the image.
M=−
q
= +0.057
p
Images Formed by Refraction
When light rays change media they are bent.
This also can form images.
sider two transparent m
boundary between the
We assume the object a
principal
axis
diverge
from
a
point
We can find the location and size of the image formed
by
Let’s consider the parax
object at O and are refracted
considering paraxial rays.
through the image point I.
at the spherical surface
Figure 36.17 shows a
n1 ! n2
law of refraction applie
Images Formed byRaysRefraction
making small angles with the
n1
n2
R
O
I
p
q
Because u1 and u2 are a
tion sin u < u (with ang
We know that an exteri
interior angles, so apply
Figure 36.16 An image formed
by refraction
at a spherical surface.
For paraxial rays, Snell’s
law becomes:
n1 θ1 = n2 θ2
(1)
Images Formed by Refraction
n1
u1
36.3 Images Formed by Refraction
P
Figure 36.17 G
derive Equation 3
n 1 , n 2.
n2
d
u2
b
a
g
C
O
I
R
p
q
Looking at the diagram:
Combining all three expressions and eliminating u1 and u2 gives
a 1−n 2β
g 5 (n⇒
2 2 n 1)bθ1 = α + β
180◦ − θ1 = 180◦ −n 1α
(36.7)
(2)
Figure 36.17 shows three right triangles that have a common vertical leg of length
d. For paraxial rays (unlike the relatively large-angle ray shown in Fig. 36.17), the
likewise:
horizontal legs of these triangles are approximately p for the triangle containing
= γ + angle
θ2 b, and(3)
angle a, R for the triangleβ
containing
q for the triangle containing
angle g. In the small-angle approximation, tan u < u, so we can write the approximate(2)
relationships
follows:them and subtract (1):
Multiply
by n1 from
andthese
(3)triangles
by n2as
, add
tan a < a <
d
d
d
b<b<
tan g < g <
n2 β p= ntan
q
R + n2 γ
1 α + n1 β
Substituting these expressions into Equation 36.7 and dividing through by d gives
Rearranging:
n
n
n 2n
2
2
1
5 (n2 − n1 )β
n1 α + pn1 1
2 γq =
R
(36.8)
Relation betw
image distanc
Images Formed by Refraction
For small angles (paraxial approx):
α ≈ tan α =
d
p
Similarly,
β≈
d
d
and β ≈
R
q
So,
n1 n2
n2 − n1
+
=
p
q
R
Flat Refracting Surfaces
(Like a rectangular
1102
Chapterfish
36 tank.)
Image Formation
The image is virtual and on
the same side of the surface
as the object.
n1 ! n2
n1 n2
Flat Refracting Surfaces
If a refracting surface is flat, then R is infinite an
In this case R → ∞.n 1 5 2 n 2
q
p
n1 n2
n2
+
=0
q52 p
p
q
n
1
O
q
p
Figure 36.18 The image formed
And so we see that the sign of q
From this expression,
according to Table 36.2, the image
formed by a
n2
− object
p as illustrate
same side of the surfaceqas=the
n1
in which the object is in the medium of index n 1
case, a virtual image is formed between the objec
n 2, the rays on the back side diverge from one an
in Figure 36.18. As a result, the virtual image is fo
Q uick Quiz 36.4 In Figure 36.16, what happens
point O is moved to the right from very far awa
Sign Conventions for Refracting Surfaces!
n2 − n1
n1 n2
+
=
p
q
R
Variable
is Positive
is Negative
p
object in front of surface
[virtual object]2
q
image behind surface
(real)
image in front of surface
(virtual)
h 0 (and M)
image upright
image inverted
R
object faces convex surf.
(C behind surface)
object faces concave surf.
(C in front of surface)
C is the center of curvature.
0
M = hh = − nn12 qp (proof for homework).
2
Will be useful in derivations.
36
Refracting Surface Question
Quick Quiz 36.4 In the figure, what happens to the image point I
Rays making small angles with the
as the object point O is moved to the rightprincipal
from very
far away
to
axis diverge
from a point
object at O and are refracted
very close to the refracting surface?
through the image point I.
(A) It is always to the right of the
surface.
(B) It is always to the left of the
surface.
n1 ! n2
n1
n2
R
O
I
(C) It starts off to the left, and at
some position of O, I moves to
the right of the surface.
(D) It starts off to the right, and at
some position of O, I moves to
the left of the surface.
2
Serway & Jewett, page 1102.
p
In t
und
side
bou
We
Let’
at th
F
law
q
Figure 36.16 An image formed
by refraction at a spherical surface.
Bec
tion
We
inte
36
Refracting Surface Question
Quick Quiz 36.4 In the figure, what happens to the image point I
Rays making small angles with the
as the object point O is moved to the rightprincipal
from very
far away
to
axis diverge
from a point
object at O and are refracted
very close to the refracting surface?
through the image point I.
(A) It is always to the right of the
surface.
(B) It is always to the left of the
surface.
n1 ! n2
n1
n2
R
O
I
(C) It starts off to the left, and at
some position of O, I moves to
the right of the surface.
(D) It starts off to the right, and at
some position of O, I moves to
the left of the surface. ←
2
Serway & Jewett, page 1102.
p
In t
und
side
bou
We
Let’
at th
F
law
q
Figure 36.16 An image formed
by refraction at a spherical surface.
Bec
tion
We
inte
Refracting Surface Question
1102
Chapter 36
Image For
The image
is virtual
and on
Quick Quiz 36.5 In the figure, what happens
to the
image
point I
the same side of the surface
as the object point O moves toward the right-hand
as the object. surface of the
material of index of refraction n1 ?
(C) It approaches the surface and
then moves to the right of the
surface.
O
q
p
Figure 36.18 The image formed
by a flat refracting surface. All rays
are assumed to be paraxial.
2
Serway & Jewett, page 1102.
If a r
n1 ! n2
n1 n2
(A) It always remains between O and
the surface, arriving at the
surface just as O does.
(B) It moves toward the surface more
slowly than O so that eventually
O passes I.
Flat
From
acco
same
in wh
case,
n 2, t
in Fi
Q ui
po
su
th
rig
Refracting Surface Question
1102
Chapter 36
Image For
The image
is virtual
and on
Quick Quiz 36.5 In the figure, what happens
to the
image
point I
the same side of the surface
as the object point O moves toward the right-hand
as the object. surface of the
material of index of refraction n1 ?
(C) It approaches the surface and
then moves to the right of the
surface.
O
q
p
Figure 36.18 The image formed
by a flat refracting surface. All rays
are assumed to be paraxial.
2
Serway & Jewett, page 1102.
If a r
n1 ! n2
n1 n2
(A) It always remains between O and
the surface, arriving at the
surface just as O does. ←
(B) It moves toward the surface more
slowly than O so that eventually
O passes I.
Flat
From
acco
same
in wh
case,
n 2, t
in Fi
Q ui
po
su
th
rig
Images Formed by Thin Lenses
We will derive the thin lens equation
1
1 1
= +
f
p q
(Notice it is the same as the mirror equation!)
And the lens maker’s equation
1
= (n − 1)
f
1
1
−
R1 R2
We do this by considering each side of the lens as a refracting
surface.
Images Formed
by Thin
Lenses
The image
due to surface
1 is virtual,
A thick lens: so I1 is to the left of the surface.
n1 ! 1
R1
I1
Surface 1
R2
Surface 2
n
O
p1
t
C1
q1
refraction a
the notion t
object for th
let the thick
Consider
faces with r
R 1 is the rad
reaches first
lens.) An ob
Let’s beg
and assumin
the image I1
p2
a
We can imagine that there first surface forms an image I1 , and
where q 1 is
that image becomes The
the object
for the
second1 surface.
image due
to surface
is real,
so I is to the right of the surface.
The pink arrow shows I11
formed by su
if the image
First image formed
The image due to surface 1 is virtual,
Let n1 = 1, n2so=I1n.is to the left of the surface.
n1 ! 1
R1
I1
Surface 1
R2
Surface 2
n
O
t
p1
C1
q1
the image fo
refraction at
the notion th
object for the
let the thickn
Consider a
faces with rad
R 1 is the radi
reaches first a
lens.) An obje
Let’s begin
and assuming
the image I1 f
p2
a
1
n
n−1
+ due=to surface 1 is real,
The image
p
q
R
so I1 is1 to the1 right of1 the surface.
where q 1 is th
formed by sur
if the image i
Second image formed
n
1
1−n
+
=
p2 q2
R2
We must relate p2 to q1 . The new object is the image from the
previous surface so:
p2 = −q1 + t
where t is the lens thickness.
p2 could be positive (real object)...
The image
due to
surface
... if q1 is negative
(virtual
image
1). 1 is virtual,
so I1 is to the left of the surface.
n1 ! 1
R1
I1
Surface 1
R2
Surface 2
n
O
p1
q1
p2
a
t
C1
just learn
the image
refraction
the notion
object for
let the thi
Consid
faces with
R 1 is the r
reaches fi
lens.) An
Let’s be
and assum
the image
a
p2 could be negative (virtual object)...
The image
image 1).
due to surface 1 is real,
... if q1 is positive (real
so I1 is to the right of the surface.
n1 ! 1
R1
Surface 1
R2
Surface 2
n
where q 1
formed b
if the ima
Now le
(We mak
2 are in t
the objec
I1
O
p1
t
C1
p2
q1
b
Figure 36.21
To locate the image formed
We now i
face acts
is virtual
related to
q 1 is nega
a
p2 could be negative (virtual object)...
The image
image 1).
due to surface 1 is real,
... if q1 is positive (real
so I1 is to the right of the surface.
n1 ! 1
R1
Surface 1
R2
Surface 2
n
where q 1
formed b
if the ima
Now le
(We mak
2 are in t
the objec
I1
O
p1
t
C1
p2
We now i
face acts
q1
is virtual
related to
b
Either way, the sign is taken care of in the expression p2 = −q1 + t.
q 1 is nega
Figure 36.21 To locate the image formed
Relate the original object to the final image
surface 1
1
n
n−1
+
=
p1 q1
R1
n
surface 2
(−q1 + t)
+
(2)
1
1−n
=
q2
R2
For a thin lens, t is negligible, so take it to be zero.
−
n
1
1−n
+
=
q1 q2
R2
Add equation (2) to equation (3) to get
1
1
1
1
+
= (n − 1)
−
p1 q2
R1 R2
(3)
Thin Lens Equation
Lastly, measure the object distance to be p = p1 + t/2 and
q = q2 + t/2.
Since t is effectively zero, we have:
1 1
+ = (n − 1)
p q
1
1
−
R1 R2
The RHS is the inverse of the focal length (lens maker’s equation)
1
= (n − 1)
f
1
1
−
R1 R2
This gives the thin lens equation
1 1
1
+ =
p q
f
Magnification with a Lens
By definition,
M=
h0
h
And it follows from simple trigonometry that
M=−
Same as for a mirror!
q
p
Sign Conventions for Lenses!
1 1
1
+ =
p q
f
Variable
is Positive
is Negative
p
object in front of surface
[virtual object]3
q
image behind lens
(real)
image in front of lens
(virtual)
h 0 (and M)
image upright
image inverted
R1 and R2
object faces convex surf.
(C behind surface)
object faces concave surf.
(C in front of surface)
f
lens is converging
lens is diverging
3
Useful in derivations.
Understanding the Sign of f
1
= (n − 1)
f
1
1
−
R1 R2
Always n > 1, so (n − 1) is positive (if zero, there’s no lens!).
1
R1
1
R2
−
could be positive or negative, depending on the signs
and magnitudes of R1 and R2 .
applies to a refracting surface.)
Sign Examples: Converging Lenses
Biconvex
Convexconcave
Planoconvex
Figure 36
sign conven
refracting s
Various l
thicker at t
center than
Magnific
Ra1 R2
R1 +ve
Biconcave
R2 −ve
f +ve
R1
R2
R1 +ve
RConvex2 +ve
concave
R1 < R2
f +ve
R1 R2
R1 +ve
PlanoR2 ∞
concave
f +ve
Consider a
(Eq. 36.2),
image is
a
Sign Examples: Diverging Lenses
Biconcave
Convexconcave
Planoconcave
Consider a
(Eq. 36.2)
image is
From this e
the same s
and on the
b
R −ve
R +ve
R ∞
1
Figure
36.25 1Various lens1
R2 +ve
R2 +ve lenses
R2 −ve
shapes.
(a) Converging
have a positiveRfocal
1 > R2length and
are fthickest
at the
middle. f −ve
−ve
f −ve
(b) Diverging lenses have a
Ray Diag
Ray diagra
tems of len
diagrams f
Focal Length Question
Quick Quiz 36.6 What is the focal length of a pane of window
glass?
(A) zero
(B) infinity
(C) the thickness of the glass
(D) impossible to determine
Focal Length Question
Quick Quiz 36.6 What is the focal length of a pane of window
glass?
(A) zero
(B) infinity
←
(C) the thickness of the glass
(D) impossible to determine
Ray Diagrams for Converging Lenses
Again, draw rays whose behavior we know.
Rays parallel to the principle axis are refracted through the focal
point.
Rays that travel through the center of the lens are (effectively) not
refracted.
For Converging Lenses:
1
Draw a ray from the top of the object parallel to the principle
axis refracted through the focal point F .
2
Draw a ray from the top of the object through the focal point
(or back to the focal point) and refracted parallel to the
principal axis.
3
Draw a ray from the top of the object through the center of
the lens, and continuing in a straight line.
Wh
foca
the
tha
side
Ray Diagrams
Converging
Whenfor
the object
is in frontLenses
of and outside
the focal point of a converging lens, the
image is real, inverted, and on the back side
of the lens.
1
2
O
3
F2
I
I F
1
F1
Front
Back
Fron
b
a
Figure 36.26
Ray diag
Ray Diagrams for Converging Lenses
Wh
foca
the
tha
side
When the object is in front of and outside
the cases
focal of
point
of a converging lens, the
There are two
interest:
image is real, inverted, and on the back side
of the lens.
Object is beyond focal point
1
2
O
3
F2
I
I F
1
F1
Front
Back
b
a
Image is real and inverted.
Fron
Figure 36.26
Ray diag
Object Beyond Focal Point
The object is the Sun.
1
Photo from
http://www.mahalo.com/how-to-start-a-fire-with-a-magnifying-glass
6
than the
and on
the front
Ray Diagrams
forobject,
Converging
Lenses
o
l
side of the lens.
Object is closer than the focal point
2
1
I F
1
O
O
3
Front
b
F2
Image is virtual, upright, and magnified.
Back
c
Ray diagrams for locating the image formed by a
Object Beyond Focal Point
The object is the stamp.
Converging Lens Example
A converging lens has a focal length of 10.0 cm.
(a) An object is placed 30.0 cm from the lens. Construct a ray
diagram, find the image distance, and describe the image.
(b) An object is placed 5.00 cm from the lens. Find the image
distance and describe the image.
Converging Lens Example
A converging lens has a focal length of 10.0 cm.
mation
(a) An object is placed 30.0 cm from the lens. Construct a ray
diagram, find the image distance, and describe the image.
(b) An object is placed 5.00 cm from the lens. Find the image
distance and describe the image.
d by a Converging Lens
10.0 cm.
he lens. Constance, and
verging, the focal
e expect the poses.
F2 I
O
30.0 cm
a
I, F1 O
F1
10.0 cm
Figure 36.28
(Example 36.8) An
image is formed by a
converging lens.
The object is closer to
the lens than the focal
point.
The object is farther from the
lens than the focal point.
F2
10.0 cm
5.00 cm
15.0 cm
10.0 cm
b
Converging Lens Example
(a) An object is placed 30.0 cm from the lens.
q = +15.0cm
M = −0.500
(b) An object is placed 5.00 cm from the lens.
q = −10.0cm
M = +2.00
(c) An object is placed 10.0 cm from the lens. Find the image
distance and describe the image.
Converging Lens Example
(a) An object is placed 30.0 cm from the lens.
q = +15.0cm
M = −0.500
(b) An object is placed 5.00 cm from the lens.
q = −10.0cm
M = +2.00
(c) An object is placed 10.0 cm from the lens. Find the image
distance and describe the image.
q=∞
No image is formed!
Ray Diagrams for Diverging Lenses
For Diverging Lenses:
1
Draw a ray from the top of the object parallel to the principle
axis refracted outward directly away focal point F on the front
side of the lens.
2
Draw a ray from the top of the object toward the focal point
on the back side of the lens and refracted parallel to the
principal axis.
3
Draw a ray from the top of the object through the center of
the lens, and continuing in a straight line.
object, for
andDiverging
on the front
side of the
Ray Diagrams
Lenses
lens.
1
2
O
F1
Front
I
F2
3
Back
c
The image formed is upright, virtual, and smaller in size.
med by a thin lens.
Diverging Lenses
The image formed is upright, virtual, and smaller in size.
3
Photo from the Capilano University Physics Dept. webpage.
Combinations of Lenses
Two or more lenses can be used in series to produce and image.
This is used in
• eyeglasses (your eyeball lens is the second lens)
• refracting telescopes
• microscopes
• camera zoom lenses
Lenses can also be used together with curved mirrors, for example
in reflecting telescopes.
Two thin lenses placed right up against each other (touching) will
act like one lens with a focal length
1
1
1
= +
f
f1 f2
ual focal lengths by the expression
Combinations of Lenses Example 36.10
(36.19)
Focal length
for a cm and
Two thin converging lenses
of focal lengths
f1 = 10.0
combination of two thin
f2 = 20.0 cm are separated by 20.0 cm as illustrated. An object is
lenses in contact
her areplaced
equivalent
to to
a single
30.0 cm
the leftthin
of lens 1.
9.
Find the position and the magnification of the final image.
e?
.0 cm
rated
e left
of the
h the
f ) in
these
Lens 1
Lens 2
I2 I1
O1
6.67 cm
10.0 cm
30.0 cm
Figure 36.30
15.0 cm
20.0 cm
(Example 36.10) A combination of two converging lenses. The ray diagram shows the location of the final image
Summary
• image terminology
• images formed by spherical mirrors
• images formed by refraction
• images formed by lenses
• images formed by lens combinations
Collected Homework!
due Monday, June 13.
Homework Serway & Jewett:
• Carefully read all of Chapter 36.
• Ch 36, onward from page 1123. OQs: 1, 3, 5, 11, 13; CQs: 5,
9, 11; Probs: 1, 7, 9, 19, 21, 29, 37, 39, 43, 53, 55, 71, 73,
87, 89, (93)