•Your eyes tell you where/how big an object is
•Mirrors and lenses can fool your eyes – this is
sometimes a good thing
Flat mirror images Ch
36
•Place a point light source P in front of a mirror
•If you look in the mirror, you will see the object as if it were at the
point P’, behind the mirror
•As far as you can tell, there is a “mirror image” behind the mirror
•For an extended object, you get an extended image
P’
•The distances of the object
P
from the mirror and the image
from the mirror are equal
•Flat mirrors are the only
perfect image system
Image
Object
(no distortion)
Mirror
p
q
p  q
Image Characteristics and Definitions
h
M 
h
h
h’
Object
p
q Image
Mirror
•The front of a mirror or lens is the side the light goes in
•The object distance p is how far the object is in front of the mirror
•The image distance q is how far the image is in front* of the mirror
•Real image if q > 0, virtual image if q < 0
•The magnification M is how large the image is compared to the object
•Upright if positive, inverted if negative
*back for lenses
If you place an object in front of a flat mirror, its image will be
A) Real and upright B) Virtual and upright
C) Real and inverted D) Virtual and inverted
CT - 1 In the morning you look at yourself in the mirror and you cannot see your feet. In
order to see them in the mirror you should
A. Move closer to the mirror and look down.
B. Move backward.
C. Give it up - you will never be able to see your feet in the mirror.
Spherical Mirrors
•Typical mirrors for imaging are spherical mirrors – sections of a sphere
•It will have a radius R and a center point C
•We will assume that all angles involved are small sin   tan   
•Optic axis: an imaginary line passing through the center of the mirror
•Vertex: The point where the Optic axis meets the mirror
The paths of some rays of light are easy to figure out
•A light ray through the center will come back exactly on itself
•A ray at the vertex comes back at the same angle it left
•Let’s do a light ray coming in parallel to the optic axis:
•The focal point F is the place this goes through
•The focal length f = FV is the distance to the mirror
X

•A ray through the focal point
comes back parallel
C
F
FC  FX  12 CX  12 R
V
f
f  FV  CV  FC
f  12 R
R
Spherical Mirrors: Ray Tracing
1. Any ray coming in parallel goes through the focus
2. Any ray through the focus comes out parallel
3. Any ray through the center comes straight back
f  12 R
•Let’s use these rules to find the image:
F
Do it again, but harder
•A ray through the center won’t
hit the mirror
•So pretend it comes from the center
•Similarly for ray through focus
•Trace back to see where they came from
C
F
P
C
Spherical Mirrors: Finding the Image
•The ray through the center comes straight back VP  p PX  h
•The ray at the vertex reflects at same angle it hits VQ  q QY  h
•Define some distances:
X
CV  R
•Some similar triangles:
VQY VPX
h
Q
V
CQY CPX
P
h’
C
Y
h
VQ q
CQ R  q




h
VP
CP p  R
p
f  12 R
•Cross multiply q  p  R   p  R  q 
Magnification
•Divide by pqR: 2 pq  pR  qR
2 1 1
 
R q p
1 1 1
 
p q
f
•Since image upside
down, treat h’ as negative
h q
 
h
p
q
M 
p
Convex Mirrors: Do they work too?
•Up until now, we’ve assumed the mirror is concave – hollow on
the side the light goes in
•Like a cave
f 
•A convex mirror sticks out on the side the light goes in
•The formulas still work, but just treat R as negative
•The focus this time will be on the other side of the mirror
•Ray tracing still works
Summary:
1
2
R
F
•A concave mirror has R > 0;
1
1
1
convex has R < 0, flat has R = 
 
p q
f
•Focal length is f = ½R
•Focal point is distance f in front of mirror
q
•p, q are distance in front of mirror of image, object
M 
p
•Negative if behind
C
Mirrors: Formulas and Conventions:
•A concave mirror has R > 0; convex has R < 0, flat has R = 
•Focal length is f = ½R
f  12 R
•Focal point is distance f in front of mirror
•p, q are distance in front of mirror of object/image
1 1 1
•Negative if behind
 
p q
f
•For all mirrors (and lenses as well):
•The radius R, focal length f, object distance p, and image distance
q can be infinity, where 1/ = 0, 1/0 = 
Light from the Andromeda Galaxy bounces off of a concave mirror with
radius R = 1.00 m. Where does the image form?
A) At infinity
B) At the mirror
C) 50 cm left of mirror
D) 50 cm right of mirror
•Concave, R > 0
f  12 R  50 cm
p  2 Mly  
1
1 1 1
0
  
q
f
p 50 cm
q  50 cm
Ex- (Serway 36-25) A spherical mirror is to be used to form, on a screen located 5 m
from the object, an image 5 times the size of the object. (a) Describe the type of mirror
required (concave or convex). (b) What s the required radius of curvature of the mirror?
(c) Where should the mirror be placed relative to the object?
Solve on board
Images of Images: Multiple Mirrors
•You can use more than one mirror to make images of images
•Just use the formulas logically
Light from a distant astronomical source reflects from an
R1 = 100 cm concave mirror, then a R2 = 11 cm convex
mirror that is 45 cm away. Where is the final image?
1
1
1


p1 q1
f1
1
1
1


p2 q 2
f2
1 1
1


 q1 50 cm
q1  50 cm
f1  50 cm
f 2  5.5 cm
5 cm
45 cm
10 cm
p2  5 cm
1
1
1



5 cm q2
5.5 cm
q2  55 cm
Refraction and Images
•Now let’s try a spherical surface between two regions with
different indices of refraction
n1 sin 1  n2 sin  2
•Region of radius R, center C, convex in front:
n1 tan 1  n2 tan 2
Two easy rays to compute:
h h
h
n1q
•Ray towards the center continues straight n1  n2

q
h
n2 p
•Ray towards at the vertex follows Snell’s Law p
•Small angles, sin tan
R
X
n
1
•A similar triangle:
q
h
CQY CPX
C
Q
1
P
h’
p
n1q
h CQ
qR


n2
2

Y
p  R n2 p
h
CP
•Magnification:
•Cross multiply: n2 p  q  R   n1q  p  R 
•Divide by pqR:
n1q
n1 n2 n2  n1
n2 pq  n1qp  n2 pR  n1qR p  q  R
M 
n2 p
Comments on Refraction
•R is positive if convex (unlike reflection)
n1 n2 n2  n1


•R > 0 (convex), R < 0 (concave), R =  (flat)
p q
R
•n1 is index you start from, n2 is index you go to
•Object distance p is positive if the object in front (like reflection)
•Image distance q is positive if image is in back (unlike reflection)
We get effects even for a flat boundary, R = 
•Distances are distorted:
R
X
n
1
n1 n2
q

0
h
p q
Q
P
n2
p
q
p
n1q
n2
2
n1
M 
Y
n p
2
•No magnification: M   n1   n2 p   1


n2 p 
n1 
Warmup 25
Flat Refraction
A fish is swimming 24 cm underwater (n = 4/3). You are
looking at the fish from the air (n = 1). You see the fish
A) 24 cm above the water
B) 24 cm below the water
C) 32 cm above the water
D) 32 cm below the water
E) 18 cm above the water
F) 18 cm below the water
1  24 cm 
q
43
 18 cm
24 cm
18 cm
•R is infinity, so formula above is valid
•Light comes from the fish, so the water-side is the front
p  24 cm
•Object is in front
•Light starts in water
n1  4 3
•For refraction, q tells you
distance behind the boundary n2  1
n2
q
p
n1
CT – 2 A parallel beam of light is sent through an aquarium. If a convex glass lens is held
in the water, it focuses the beam
A. closer to the lens than
B. at the same position as
C. farther from the lens than
outside the water.
Double Refraction and Thin Lenses
•Just like with mirrors, you can do double refraction
•Find image from first boundary
•Use image from first as object for second
n1
n2
We will do only one case, a thin lens:
•Final index will match the first, n1 = n3
•The two boundaries will be very close
p
Where is the final image?
n1 n2 n1
•First image given by:
n1 n2 n2  n1


p q1
R1
•This image is the object for the second boundary:
n2 n1 n1  n2
•Final Image location:


p2 q
R2
•Add these:
 1
n1 n1
1 
   n2  n1   

p q
R
R
 1
2 
n3
q1   p2
 1
1 1  n2
1 
    1 

p q  n1
R
R
 1
2 
Thin Lenses (2)
 1
1 1  n2
1 
    1 

p q  n1
R
R
 1
2 
 1
1  n2
1 
   1 

f  n1
R
R
 1
2 
•Define the focal length:
•This is called lens maker’s equation
•Formula relating image/object distances
•Same as for mirrors
n1q1
M1  
Magnification: two steps
n2 p
•Total magnification is product
•Same as for mirrors
M  M 1M 2 
qq1
pp2
q1   p2
q
M 
p
1 1 1
 
p q
f
n2 q
M2  
n1 p2
Using the Lens Maker’s Equation
 1
1  n2
1 
   1 

f  n1
R
R
 1
2 
•If you are working in air, n1 = 1, and we
normally call n2 = n.
•By the book’s conventions, R1, R2 are
positive if they are convex on the front
•You can do concave on the front as well, if
you use negative R
•Or flat if you set R = 
If the lenses at
right are made of
A B
glass and are used
in air, which one
definitely has f < 0?
D
C
•If f > 0, called a
converging lens
•Thicker in middle
•If f < 0, called a
diverging lens
•Thicker at edge
•If you turn a lens
around, its focal length
stays the same
Light entering on the left:
•We want R1 < 0: first
surface concave on left
•We want R2 > 0: second
surface convex on left
Ray Tracing With Converging Lenses
•Unlike mirrors, lenses have two foci, one on each side of the lens
•Three rays are easy to trace:
1. Any ray coming in parallel goes through the far focus
2. Any ray through the near focus comes out parallel
3. Any ray through the vertex goes straight through
F
F
f
f
•Like with mirrors, you sometimes have to imagine a ray coming
from a focus instead of going through it
•Like with mirrors, you sometimes have to trace outgoing rays
backwards to find the image
Ray Tracing With Diverging Lenses
•With a diverging lens, two foci as before, but they are on the wrong side
•Still can do three rays
1. Any ray coming in parallel comes from the near focus
2. Any ray going towards the far focus comes out parallel
3. Any ray through the vertex goes straight through
F
F
f
f
•Trace purple ray back to see where it came from
Lenses and Mirrors Summarized
•The front of a lens or mirror is the side the light goes in
R>0
p>0
q>0
Concave Object Image
mirrors
front in front in front
lenses
f
f  12 R
 1
1 
Convex Object Image 1  n2
   1 

front in front in back f  n1
 R1 R2 
1 1 1
 
p q
f
Variable definitions:
•f is the focal length
•p is the object distance from lens
•q is the image distance from lens
•h is the height of the object
•h’ is the height of the image
•M is the magnification
h
q
M  
h
p
Other definitions:
•q > 0 real image
•q < 0 virtual image
•M > 0 upright
•M < 0 inverted
Warmup 25
Ex- A transparent sphere of unknown composition is observed to form an image of the
Sun on the surface opposite to the Sun. What is the refractive index of the sphere?
Ex - A transparent photographic slide is placed in front of a converging lens that has a
focal length of 2.44 cm. The lens forms an image of the slide 12.9 cm from the slide.
How far is the lens from the slide if the image is (a) real and (b) virtual.
Solve on Board
Imperfect Imaging
•With the exception of flat mirrors, all imaging systems are imperfect
•Spherical aberration is primarily concerned with the fact that the small
angle approximation is not always valid
F
•Chromatic Aberration refers to the fact that different colors refract
differently
F
•Both effects can be lessened by using combinations of lenses
•There are other, smaller effects as well
I  A d2
I  d f
f -number  f d
I   f -number 
Film
Aperture
Lens
•Real cameras use a lens or combination of lenses for focusing
•The aperture controls how much light gets in
•The shutter only lets light in for the right amount of time
•The film (or CCD array) detects the light
Focusing: Film must be at distance q:
1
•Adjust position of lens for focus 1 1
 
•Typically, p  , q  f
p q
f
Exposure:
•The more the object is magnified, the dimmer
it is I  1 h2 1 M 2  1 q 2 1 f 2
q
•The larger the area of the aperture, the 2more light
Shutter
Cameras

•The ratio of the diameter to the focal length is called the f-number
2
t   f -number 
•The exposure time will be inversely proportional to Intensity
2
Eyes
•Eyes use a dual imaging system
•The Cornea contains water-like fluid that does most of the refracting
•The Lens adds a bit more
•The iris is the aperture
•The eye focuses the light on the retina
•Neither the cornea nor the lens moves
•The shape (focal length) of the lens
is adjusted by muscles
•Over time, the lens becomes stiff
and/or the muscles get weak
•A healthy eye can normally focus on
objects from 25 cm to 
•If it can’t reach , we say someone
is nearsighted
•If it can’t reach 25 cm, we say
someone is farsighted
Adjusting Eye Problems
•To make the eye work, just put a lens that turns the object (p) you want
to see into an image at a distance (q) where you can see it
A farsighted person can’t see objects closer than 1.00 m away. What
focal length lens would adjust his eyesight so he can read 0.50 m away?
A) +1 m
B) -1 m
C) +3 m
D) -3 m
•The object will be 0.50 m in front of the lens
•p = +0.50 m
•The image will be 1.00 m in front of the lens
•q = -1.00 cm
Opticians give the inverse
focal length, f -1, which is
given in diopters (= m-1)
1 1 1
 
p q
f
1
1
1


0.5 m 1 m
f
1
1m 2 m
f  1 m
1
1

f
Angular Size & Angular Magnification
•To see detail of an object clearly, we must:
•Be able to focus on it (25 cm to  for healthy eyes, usually  best)
•Have it look big enough to see the detail we want
•How much detail we see depends on the angular size of the object
0
h
0  h d
d
Two reasons you can’t see objects in detail:
1. For some objects, you’d have to get closer than your near point
• Magnifying glass or microscope
2. For others, they are so far away, you can’t get closer to them
Angular Magnification:
• Telescope
how much bigger the
Goal: Create an image of an object that has
angular size of the image is
• Larger angular size
• At near point or beyond (preferably )
m  
0
The Simple Magnifier
•The best you can do with the naked eye is:
•d is near point, say d = 25 cm
•Let’s do the best we can with one converging lens
•To see it clearly, must have |q| d
1 1 1
h’
p

q

f
1 1 
 1 1
h
h

  h     h   
q 
q p
 f q
 f
0  h d
  h  q 
p
h
-q
F

d d
m
 
0
f
q
•Maximum magnification when |q| = d
d
•Most comfortable when |q| = 
m

1

max
•To make small f, need a small R:
f
•And size of lens smaller than R
•To avoid spherical aberration, much smaller
•Hard to get m much bigger than about 5
d
m
f
The Microscope
A simple microscope has two lenses:
•The objective lens has a short focal length and produces a large,
inverted, real image
•The eyepiece then magnifies that image a bit more
Fe
Fo
•Since the objective lens can be small, the magnification can be large
•Spherical and other aberrations can be huge
•Real systems have many more lenses to compensate for problems
•Ultimate limitation has to do with physical, not geometric optics
•Can’t image things smaller than the wavelength of light used
•Visible light 400-700 nm, can’t see smaller than about 1m
The Telescope
A simple telescope has two lenses sharing a common focus
•The objective lens has a long focal length and produces an inverted,
real image at the focus (because p = )
•The eyepiece has a short focal length, and puts the image back at 
(because p = f)
fe
fo
F
Angular Magnification:
0




h
f
•Incident angle:
m   0
0
o
m  fo fe
•Final angle:
  h f e
•The objective lens is made as large as possible
•To gather as much light as possible
•In modern telescopes, a mirror replaces the objective lens
•Ultimately, diffraction limits the magnification (more later)
•Another reason to make the objective mirror as big as possible