Physics 1230: Light and Color
Ch. 3: Mirrors and Lenses Ivan Smalyukh
•
Building on concepts from the last
chapter
– Now that we know how rays and
images work we can understand how
curved mirrors and lenses produce
images
– Practical applications include
magnifying mirrors, rear-view mirrors,
glasses and contact lenses, and
cameras.
•
Spherical mirrors
– Significance of focal point
– Tracing reflected rays to find virtual
images in convex mirrors
– Tracing reflected rays to find virtual
and real images in concave mirros
• Lenses
– Significance of focal points,
focal lengths, focal planes and
power of a lens
– Tracing rays to find images
from thin convex and concave
lenses.
– Fresnel lenses
– Compound lenses and using
intermediate images to find
final images
1
What is a spherical convex mirror?
(Think of the rear view mirror on your car)
•
A convex mirror bulges outwards
towards you
Reflecting surface of
convex spherical mirror
Axis of mirror
Center of sphere
Mirrors curved like a cylinder can
make you look fat or skinny
Looking at yourself in a concave mirror, the image is
vertically expanded like a bathroom magnifying mirror
but in one dimension only. Your image looks skinny
Virtual image
on other side
of mirror is
compressed
vertically
Looking at yourself in a convex mirror,
the image is compressed vertically like
rear view mirror but in one dimension only.
Your image looks fat
Virtual image
on other side
of mirror is
stretched
vertically
A concave mirror bulges away from you
• The center of the sphere (circle) is in
front of the mirror
• The focal length is also in front of
the mirror, halfway between the
center and the mirror surface
Mirror
surface
Axis
Center
Focal
point
Why are the rays from a distant light source such
as the sun or a star essentially parallel?
Here the rays from a distant light source
such as the sun
By the time they arrive here
(into a camera or mirror) only
the nearby almost parallel rays enter
Convex mirror
Whenever we speak of
incoming parallel rays
you can always visualize
rays from the sun or a star.
Convex mirror
Rays which arrive at the mirror
from a close source, like Alex’s
nose are not almost parallel
http://micro.magnet.fsu.edu/primer/java/mirrors/concavemirrors/index.html
http://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors/index.html
http://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors3d/index.html
http://micro.magnet.fsu.edu/primer/java/mirrors/concave.html
http://micro.magnet.fsu.edu/primer/java/mirrors/convex.html
Here is how we use those rays to find the
image of an object in a concave mirror
•
•
Where is Alex's image when he is between the
center and it's focal point?
Let's find the image of his nose
–
–
Here is a ray of type 1 from his nose reflecting
off the mirror
Here is a ray of type 3 from his nose reflecting
off the mirror
The image of the nose is at the intersection of
reflected rays of type 1 and type 3. Why?
Can we use a ray of type 2?
–
–
Is it (a) real or (b) virtual?
Magnified or reduced?
–
–
•
Mirror
surface
Axis
Here is Alex's image
Center
Focal
point
11. From which points can your eye see his nose?
a)
b)
c)
d)
e)
From all points (A, B, C)?
From A and C, but not B?
From B and C, but not A?
From A only?
From C only
A
C
Clicker Question
B
Special rays 1, 2 and 3 are not necessary to find the image
of an object but are easier to work with than others
• Here are rays 1 and 3 used to
find the image Alex's nose
• Here are some other (less
convenient) incident and
reflected rays
– They all go through the same
image point as rays 1 and 3 after
satisfying the law of specular
reflection!
– Any two rays from an object
point will intersect at the image
point but rays 1, 2 and 3 are the
easiest to use to find the image
point
What happens to his image if Alex is
closer to the mirror than the focal point?
•
•
•
Draw rays of type 1 and type 3 from
his nose to the mirror
The reflected rays will never intersect
behind Alex since they diverge behind
him
However we can extend the reflected
rays behind the mirror where they do
intersect
–
The image of Alex's nose is at the
intersection of the backward extended
reflected rays
Mirror
surface
Center
Focal
point
12. Is it real (a) or virtual (b)?
13. Reduced (a) or magnified (b)?
14. Right side up (a) or inverted (b)?
Clicker question
• Here is the whole image
–This is a magnifying mirror
(for shaving or cosmetics)
Informatiuon
• HW # 4 due on
Thursday, Sept. 24
• Exam #1 on Tuesday,
September 29 –
Chapters 1-3 of the
textbook;
• Exam preparation:
http://www.colorado.edu/physic
s/phys1230/phys1230_fa09/Exa
ms.htm
• On Thursday:
questions/answers &
exam #1 material
overview + new
material
HW#2
Concept question
• Is your makeup/shaving mirror
A.Flat
B.Concave
C.Convex
Lenses
Focal point of a converging lens is where
image of a distant object (sun) lies
• The focal point of a mirror was at the
image point of a distant star (or the sun)
seen in that mirror
• This is true for both convex mirrors
(virtual image of star) and for concave
mirrors (real image of star).
• The reason for involving the sun or a
distant star is to guarantee that the
incoming rays are essentially parallel.
• The same is true for lenses
– The converging lens of a magnifying glass
brings the sun’s (parallel) rays to a focus at
the “hot spot” which can produce a fire.
The location of the hot spot is at the focal
point of the lens.
(Parallel) rays from the sun
Converging
lens
Focal length
Focal point
Unlike a mirror there are two focal
points of a lens, one on either side
• Here is a converging lens
with its axis and BOTH
focal points shown
Incoming ray is bent
towards nornal
Outging ray is bent
away from nornal
– Both sides of a converging
lens bulge outwards
• Parallel incoming rays are
refracted so that they ALL
pass through the focal
point F'
F
F'
F
F'
– This fact defines F' and can
be used to find its location!
• Rays diverging from F and
passing through the lens
ALL come out parallel
Can we be more precise with
distances and magnifications?
• Do we have to draw a to-scale
ray-tracing for each different
object location to find the
image location?
• No, you don't have to do any
ray-tracing at all (Thank
goodness!!)
• But you have to use some
algebra
• The distance between the center
of the lens and either F or F' is
the same.
– It is called the focal length, f
F
F'
These distances are the
SAME. Either one is
called the focal length, f
How to use rays of type 1, 2 and 3 to find the
image produced by a thin converging lens
•
•
•
•
•
The image produced by a thin
converging lens whose axis and focal
points are known can be found by
replacing the lens by a plane through its
center
A ray of type 1 goes from Alex's nose
parallel to the axis to the plane and then
refracts through the focal point F'
A ray of type 2 goes from his nose
through the center of the lens (plane)
The image of the nose is at the
intersection of the two rays
You can check your results with a ray
of type 3 which goes from his nose
through focal point F and then parallel
to the axis
F
F'
Is the image upside down or right side up?
Use the axis to help determine
the image by locating its bottom!
Where is the image when Alex is close to the
converging lens (closer than F)?
• We draw incident and transmitted rays
of type 1 and 2.
– Note that the transmitted (bent) rays
diverge on the other side of the lens so
the image cannot be there.
• Try extending the transmitted rays
backwards.
– The image is where they intersect
– Note rays of type 3 are in agreement
with the image point determined by rays
1 and 2 even though they don't pass
through the lens!
– Is it inverted (a) or right-side-up (b)?
– Is it magnified (a), reduced (b) or
unchanged (c) in size?
F
F'
This is a magnifying glass!!
Compare the results of ray tracing to find an
image of Alex for each of his two positions.
F
F'
Alex is further from the lens
than the focal point F
Image is real
F
F'
Alex is closer to the lens
than the focal point F
Image is virtual
What happens if we put a translucent screen
at the location of the real image?
• Each image point is now
scattered from the screen in all
directions
• Hence, you eye can see the
image not only by looking into
rays 1, 2 and 3, but also by
looking at the screen into the
new (scattered) rays
• We can even see the image
from the other side of the
screen since rays are scattered
there too!
F
F'
New rays scattered off translucent
screen particles at each point
How does a diverging lens differ from a
converging lens?
• A diverging lens has concave
surfaces (one or both)
• The focal points on a diverging
lens are reversed.
Incoming ray is bent
towards nornal
Outging ray is bent
away from nornal
– Parallel incoming (type 1) rays
come out the other side
diverging as though they came
from the first focal point which
is called F'.
– Incoming (type 3) rays aimed
at the focal point behind the
lens (F) come out parallel
• Diverging lenses deflect rays so
that they point further away
from the axis
F'
F'
F
F
Computer simulation of light focusing/defocusing by lenses:
http://micro.magnet.fsu.edu/primer/java/lenses/converginglenses/index.html
http://micro.magnet.fsu.edu/primer/lightandcolor/lenseshome.html
http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html
http://micro.magnet.fsu.edu/primer/java/lenses/diverginglenses/index.html
http://micro.magnet.fsu.edu/primer/java/components/perfectlens/index.html
As a pencil moves closer to a converging lens
what happens to its image?
See the following online interactive
experiment to learn the answer
http://www.colorado.edu/physics/phet/simulations/lens/lens.swf
Sept 24
The lens equation gives the same results
as ray-tracing but without any rays!
•
1/x0 + 1/xi = 1/f
•
f is focal length of lens
•
xo = positive distance from object
to center of lens (when object is left
of the lens)
•
•
•
f
xi = distance from image to lens
center
•
xi is a positive number for (real)
image on other side of lens from
object.
•
xi is a negative number for
(virtual) image on same side as
object.
Given two, find the third.
Can use the lens equation to find
image position if know object
position or vice versa, without any
rays
f
xo
xi
(will be positive
number for this case)
Here is one example of how to use the
lens eqn with a converging lens
• Given:
• f = 10 cm
• Object is 15 cm in front of
lens: x0 = 15
• Find:
– Where is image and is it
real or virtual?
• Solve equation for xi:
– Substitute numbers for
letters
– Subtract 1/15 from both
sides
– Arithmetic on calculator
– Multiply by xi/0.033
1
1
1
—=—+—
10
f
xo
15
xi
1
1
1
—-—=—
xi
10 15
1
0.033 = —
xi
1
xi = .033
— = 30 cm
Image is 30 cm to right of center of lens
and is real because xi is positive
Here is a sketch to show the previous
result
• We can verify our result
by ray-tracing
f = 10
xo = 15
xi = 30
Here is an example of how to use the lens eqn
with a diverging lens (see Fig. 3.28)
• Given:
• f = —5 cm
• NOTE, THE FOCAL
LENGTH OF A DIVERGING
LENS IS NEGATIVE
• Object is 12 cm in front of
lens: x0 = 12
• Find:
– Where is image and is it real or
virtual?
1
1
1
—=—+—
-5
f
xo
12
xi
1
1
1
—-—=—
xi
-5
12
1
-0.283 = —
xi
• Solve equation for xi:
–
–
–
–
Substitute numbers for letters
Subtract 1/12 from both sides
Arithmetic on calculator
Multiply by xi/(-0.283)
1
xi = -0.283
— = -3.53 cm
Image is 3.53 cm to left of center of lens
and is virtual because xi is negative
Summary of the meaning of negative number
in the lens and magnification equations
• Negative focal length, f,
means the lens is diverging.
– Otherwise it is converging.
• Negative object distance, xo,
means the object seen by the lens
is on the wrong side of the lens
(to right of lens)
• Negative image distance, xi,
from image to lens means the
image is to left of the lens, on • Negative magnification, M,
same side as object and
means the image is upside down
virtual (rays coming from it
(inverted) relative to the object.
never really meet)
– Otherwise image is real
1/f = 1/xi + 1/xo
M = -xi/xo
We can find the magnification of the image
relative to the object by using another formula
– The magnification of the image is
defined by the following eqn.
f = 10
• M = Si/So
• So = size of object
• Si = size of image
• M is negative for image inverted
relative to object.
• M is positive for image not inverted
relative to obect.
• The magnification can found in
terms of the distances of the image
and object from the lens
• M = -xi/xo
– For the situation at the right,
M = -30/15 = -2
– Hence, the image is twice as big as
the object and is upside down
xo = 15
xi = 30
Now find the magnification using
the magnification equation
• M = -xi/xo = -(-10)/5 = 2
– Hence, the image is twice
as big as the object and is
rightside up
• Compare with our earlier
ray-tracing
– The image is the same by
both methods
F
F'
The "power," P, of a lens is equivalent to the focal
length, f. One can be found from the other.
• Definition of P in terms of
f is
1
P (in diopters) 
f (meters)
• Meaning of P
• P is a measure of the ray
bending power of the lens
• Large P means the lens
bends rays more than if P
were small
• Your eyeglass or contact
lens prescription is usually
given in diopters (P)
• The power of a converging lens is
always positive because f is a
positive number for a converging
lens
– The converging lens always bends
rays towards the axis behind the lens
• The power of a diverging (concave)
lens is always negative because f is
negative for a diverging lens
– The diverging lens always bends rays
away from the axis behind the lens
Demonstration
Demonstration
Question:
The infrared camera in
demonstration allows one to see
people in the classroom because:
• (A) This is a magic camera;
• (B) It detects infrared rays
emitted by people;
• (C) Camera is based on total
internal reflection;
• (D) Because of dispersion;
• (E) This is the camera obscura
Question for class (doesn't count)
•
What is your opinion of
the class textbook, Seeing
the Light?
a) It is clear and useful
b) I sometimes have trouble
understanding it
c) I can't understand anything
in it
d) I don't read (or have) it
e) I “sort of” read it
sometimes but don’t spend
that much time trying to
understand it.
How do you feel about using the clickers
(doesn't count)?
a)
b)
c)
d)
e)
Scares me to death
Ruins the class
Not fair
Keeps me on my toes
Good method to remind
me what I need to know
for exams but I wish
there was some other way
Aside from whether or not you like them, do you
find the clicker questions
a) Helpful in learning
b) Especially helpful
because I go to the
library to review them
afterwards
c) Neither helpful nor
unhelpful
d) Not too helpful in
learning the concepts
e) They confuse me more
than they help me
Which phrase best describes your
opinion of the class lectures?
a)
b)
c)
d)
e)
Can't understand them
Not worthwhile
Somewhat helpful
Very helpful
Boring, since I know this
stuff!
Exam: September 29, regular class time
• No homework assignment
this week;
• Exam: book chapters 1-3;
To practice, use homeworks;
Use web-based tutorials:
• Will need a calculator;
• The exam will be worth 100
points. There are 10 multiple
choice questions worth 4 points
each and 3 problems worth a total
of 60 pts.
• Example of Fall 2007 assignment
& solutions for exam #1:
Especially to learn properties of mirrors,
eclipses, and lenses. Examples:
http://www.colorado.edu/physics/phys1230/phys1230_fa08/X1WhiteSoln.htm
http://www.colorado.edu/physics/phys1230/phys1230_fa08/WebTutorials.htm
http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html
http://antwrp.gsfc.nasa.gov/apod/image/0311/112003lunareclipse_koehn.gif
http://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors/index.html
http://micro.magnet.fsu.edu/primer/java/mirrors/concave.html
http://micro.magnet.fsu.edu/primer/java/components/perfectlens/index.html
http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html
Can verify your calculations and ray diagrams.
Exam information
• Test of understanding;
• One page of info allowed;
• No need to memorize
equations (all needed equations will be
provided, see an example on the left).
Use web-based tutorials:
http://www.colorado.edu/physics/phys1230/phys1230_fa08/WebTutorials.htm
Especially to learn properties of mirrors,
eclipses, and lenses. Examples:
http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html
http://antwrp.gsfc.nasa.gov/apod/image/0311/112003lunareclipse_koehn.gif
http://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors/index.html
http://micro.magnet.fsu.edu/primer/java/mirrors/concave.html
http://micro.magnet.fsu.edu/primer/java/components/perfectlens/index.html
http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html
Can verify your calculations and ray diagrams.