Physics 1161: Lecture 17
Refraction & Lenses
• Textbook sections 26-3 – 26-5, 26-8
Indices of Refraction
Physics 1161: Lecture 17, Slide 2
Checkpoint
Refraction
When light travels from one
medium to another the speed
changes v=c/n, but the frequency
is constant. So the light bends:
n1
q1
q2
n2
1) n1 > n2
2) n1 = n2
Compare n1 to n2.
3) n1 < n2
Checkpoint
Refraction
Which of the following is correct?
n1
q1
q2
n1 sin(q1)= n2 sin(q2)
n2
Compare n1 to n2.
1) n1 > n2
q1 < q2
2) n1 = n2
sinq1 < sinq2
3) n1 < n2
n1 > n2
FST & SFA
• A ray of light crossing the boundary from
a fast medium to a slow medium bends
toward the normal. (FST)
• A ray of light crossing the boundary from
a slow medium to a fast medium bends
away from the normal. (SFA)
Snell’s Law Practice
Usually, there is both reflection and refraction!
A ray of light traveling through the air (n=1) is incident on water
(n=1.33). Part of the beam is reflected at an angle qr = 60. The
other part of the beam is refracted. What is q2?
q1
qr
n1
n2
normal
n 1 sin q 1  n 2 sin q 2
q
2
Snell’s Law Practice
Usually, there is both reflection and refraction!
A ray of light traveling through the air (n=1) is incident on water
(n=1.33). Part of the beam is reflected at an angle qr = 60. The
other part of the beam is refracted. What is q2?
q1 =qr =60
q1
qr
sin(60) = 1.33 sin(q2)
q2 = 40.6 degrees
n1
n2
normal
n 1 sin q 1  n 2 sin q 2
q
2
Refraction Applets
• Applet by Molecular Expressions -- Florida
State University
• Applet by Fu-Kwung Hwang, National Taiwan
Normal University
Parallel light rays cross interfaces from air
into two different media, 1 and 2, as shown
in the figures below. In which of the media
is the light traveling faster?
1. Medium 1
2. Medium 2
air
air
3. Both the same
1
2
0%
1
0%
2
0%
3
Parallel light rays cross interfaces from air
into two different media, 1 and 2, as shown
in the figures below. In which of the media
is the light traveling faster?
1. Medium 1
2. Medium 2
air
air
3. Both the same
1
2
The greater the difference in the speed
of light between the two media, the
greater the bending of the light rays.
0%
1
0%
2
0%
3
Parallel light rays cross interfaces from medium 1
into medium 2 and then into medium 3. What can
we say about the relative sizes of the indices of
refraction of these media?
1. n1 > n2 > n3
2. n3 > n2 > n1
3. n2 > n3 > n1
4. n1 > n3 > n2
5. none of the
above
1
2
3
0%
1
0%
0%
2
3
0%
0%
4
5
Parallel light rays cross interfaces from medium 1
into medium 2 and then into medium 3. What can
we say about the relative sizes of the indices of
refraction of these media?
1
1. n1 > n2 > n3
2. n3 > n2 > n1
3. n2 > n3 > n1
4. n1 > n3 > n2
2
3
5. none of the
above
Rays are bent toward the normal when crossing into #2, so n
0%
1
0%
0%
2
3
0%
0%
4
5
> n1 .
But rays are bent away from the normal when going into #3, so n3 < n2.
2
How to find the relationship between #1 and #3? Ignore medium #2!
So the rays are bent away from the normal if they would pass from
#1 directly into #3. Thus, we have: n2 > n1 > n3 .
Apparent Depth
• Light exits into
medium (air)
of lower index
of refraction,
and turns left.
Spear-Fishing
• Spear-fishing is made
more difficult by the
bending of light.
• To spear the fish in the
figure, one must aim at
a spot in front of the
apparent location of the
fish.
Apparent Depth
Apparent depth:
n2
d  d
n1
n2
n1
d
apparent fish
d
actual fish
50
To spear a fish, should you aim directly at
the image, slightly above, or slightly
below?
1. aim directly at the image
2. aim slightly above
3. aim slightly below
0%
1
0%
2
0%
3
To spear a fish, should you aim directly at
the image, slightly above, or slightly
below?
1. aim directly at the image
2. aim slightly above
3. aim slightly below
0%
1
Due to refraction, the image will
appear higher than the actual
fish, so you have to aim lower to
compensate.
0%
2
0%
3
To shoot a fish with a laser gun, should
you aim directly at the image, slightly
above, or slightly below?
1. aim directly at the image
2. aim slightly above
3. aim slightly below
0%
1
The light from the laser beam
will also bend when it hits the
air-water interface, so aim
directly at the fish.
0%
2
0%
3
Delayed Sunset
• The sun actually
falls below below
the horizon
• It "sets", a few
seconds before
we see it set.
Broken Pencil
Water on the Road Mirage
Palm Tree Mirage
Mirage Near Dana – Home of Ernie Pyle
Texas Mirage
Looming
Antarctic Looming
Looming
Looming
Types of Lenses
Lens Terms
Three Rays to Locate Image
• Ray parallel to axis bends through the focus.
• Ray through the focus bends parallel to axis.
• Ray through center of lens passes straight
through.
Characterizing the Image
• Images are characterized in the
following way
1. Virtual or Real
2. Upright or Inverted
3. Reduced, Enlarged, Same Size
Object Beyond 2f
• Image is
–Real
–Inverted
–Reduced
Object at 2f
• Image is
– Real
– Inverted
– Same size
Object Between 2f and f
• Image is
– Real
– Inverted
– Enlarged
Object at F
• No Image is
Formed!
Object Closer than F
• Image is
– Virtual
– Upright
– Enlarged
Converging Lens Images
Beacon
Checkpoint
A beacon in a lighthouse is to produce a parallel
beam of light. The beacon consists of a bulb
and a converging lens. Where should the bulb
be placed?
A. Outside the focal point
B. At the focal point
C. Inside the focal point
Lens in Water
Checkpoint
Focal point determined by geometry
and Snell’s Law:
n1 sin(q1) = n2 sin(q2)
P.A.
n1<n2
F
Fat in middle = Converging
Thin in middle = Diverging
Larger n2/n1 = more bending, shorter focal length.
n1 = n2 => No Bending, f = infinity
Lens in water has _________ focal length!
Lens in Water
Checkpoint
Focal point determined by geometry and
Snell’s Law: n1 sin(q1) = n2 sin(q2)
P.A.
n1<n2
F
Fat in middle = Converging
Thin in middle = Diverging
Larger n2/n1 = more bending, shorter focal length.
n1 = n2 => No Bending, f = infinity
Lens in water has larger focal length!
Half Lens
Checkpoint
A converging lens is used to project a real
image onto a screen. A piece of black tape
is then placed over the upper half of the
lens.
1. Only the lower half will show on screen
2. Only the upper half will show on screen
3. The whole object will still show on screen
How much of the image appears on the screen?
Half Lens
Checkpoint
A converging lens is used to project a real image onto a
screen. A piece of black tape is then placed over the upper
half of the lens.
Half Lens
Checkpoint
Still see entire image (but dimmer)!
Two very thin converging lenses each
with a focal length of 20 cm are are
placed in contact. What is the focal
length of this compound lens?
1. 10 cm
2. 20 cm
3. 40 cm
0%
1
0%
2
0%
3
Two very thin converging lenses each
with a focal length of 20 cm are are
placed in contact. What is the focal
length of this compound lens?
1. 10 cm
2. 20 cm
3. 40 cm
0%
1
0%
2
0%
3
Concave (Diverging) Lens
• Ray parallel to axis
refracts as if it
comes from the first
focus.
• Ray which lines up
with second focus
refracts parallel to
axis.
• Ray through center of
lens doesn’t bend.
Image Formed by Concave Lens
• Image is always
– Virtual
– Upright
– Reduced
Concave Lens Image
Distance
• As object distance
decreases
– Image distance
decreases
– Image size increases
Image Characteristics
• CONVEX LENS – IMAGE DEPENDS ON
OBJECT POSITION
– Beyond F: Real; Inverted; Enlarged, Reduced, or
Same Size
– Closer than F: Virtual, Upright, Enlarged
– At F: NO IMAGE
• CONCAVE LENS – IMAGE ALWAYS SAME
– Virtual
– Upright
– Reduced
Lens Equations
do
di
1
1

f
M 

do
hi
ho

1
di
di
do
• convex: f > 0; concave: f < 0
• do > 0 if object on left of lens
• di > 0 if image on right of lens
otherwise di < 0
• ho & hi are positive if above
principal axis; negative below
Which way should you move object so
image is real and diminished?
1. Closer to the lens
2. Farther from the lens
3. A converging lens can’t
create a real, diminished
image.
F
Object
P.A.
F
0%
1
0%
2
0%
3
Which way should you move object so
image is real and diminished?
1. Closer to the lens
2. Farther from the lens
3. A converging lens can’t
create a real, diminished
image.
F
Object
P.A.
F
0%
1
0%
2
0%
3
3 Cases for Converging Lenses
Past 2F
Image
Object
Between
F & 2F
Image
Object
Inside F
Image
Object
Inverted
Reduced
Real
This could be used in a
camera. Big object on
small film
Inverted
Enlarged
Real
This could be used as a
projector. Small slide on
big screen
Upright
Enlarged
Virtual
This is a magnifying glass
Diverging Lens Principal Rays
F
Object
P.A.
F
1) Rays parallel to principal axis pass through focal point.
2) Rays through center of lens are not refracted.
3) Rays toward F emerge parallel to principal axis.
Image is (always true):
Real
or
Upright or
Reduced
Imaginary
Inverted
or
Enlarged
Diverging Lens Principal Rays
F
Object
F
Image
1) Rays parallel to principal axis pass through focal point.
2) Rays through center of lens are not refracted.
3) Rays toward F emerge parallel to principal axis.
Image is virtual, upright and reduced.
P.A.
Which way should you move the
object to cause the image to be real?
1. Closer to the lens
2. Farther from the lens
3. Diverging lenses can’t
form real images
F
Object
P.A.
F
0%
1
0%
2
0%
3
Which way should you move the
object to cause the image to be real?
1. Closer to the lens
2. Farther from the lens
3. Diverging lenses can’t
form real images
F
Object
P.A.
F
0%
1
0%
2
0%
3
Multiple Lenses
Image from lens 1 becomes object for lens 2
1
2
f1
Complete the Rays to locate the final image.
f2
Multiple Lenses
Image from lens 1 becomes object for lens 2
1
2
f1
f2
Multiple Lenses: Magnification
1
do = 15 cm
2
L = 42 cm
f1
f2
f1 = 10 cm
di = 30 cm
Net magnification:
m1  
m2  
di
do

8.6 cm
12 cm
30 cm
15 cm
  .7 1 7
di = 8.6 cm
f2 = 5 cm
do=12 cm
mnet = m1 m2
 2
m net  m1 m 2  1.4