Physics 212
Lecture 26: Lenses
Physics 212 Lecture 26, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Ramblin’ Jack Elliott
Arlo Guthrie
Pete Seeger
Phil Ochs
U. Utah Phillips
Why?
Last time we talked about the Brewster
angle diagram with “circles & arrows”
Physics 212 Lecture 26, Slide 2
Your Comments
“would like to go over the diagrams much more slowly, I’ve
never been very good at understanding them.”
“i really dont understand what a virtual image is can you
explain that in detail please”
We’ll show the key points with
examples
“It was good and going along at a nice pace, but the lensmaker’s formula must have been on a slide with a smaller
index of refraction than the others because it went
super fast.”
“This prelecture could not have better timing (I’m actually
doing a project about lenses and mirrors in another class)! I
am wondering how a flat lens, such as a fresnel lens, works
though.”
“This prelecture reminded me of my childhood frying ants in the sunlight. Ahh, memories.”
“The demonstraion will be very valuable for this lecture. Choose wisely, or a kitten will die ...”
Hour Exam 3: 1 week from tomorrow (Wed. Apr. 25)
• covers L19-26 inclusive (LC circuits to lenses)
•
Sign up for conflicts, etc. no later than Mon. Apr. 23 at 10:00 p.m.
Physics 212 Lecture 26, Slide 3
Refraction
Snell’s Law
n1sin(1) = n2sin(2)
1
n1
n2
2
That’s all of the physics –
everything else is just geometry!
Physics 212 Lecture 26, Slide 4
air
water
i
i
glass
2
1.3
2
1.5
glass
1.5
Case I
Case II
In Case I light in air heads toward a piece of glass with incident angle i
In Case II, light in water heads toward a piece of glass at the same angle.
In which case is the light bent most as it enters the glass?
I or II or Same
(A)
(B)
(C)
The angle of refraction in BIGGER for the water – glass interface:
n1sin(1) = n2sin(2)
sin(2)/sin(1) = n1/n2
Therefore the BEND ANGLE (1 – 2) is BIGGER for air – glass interface
Physics 212 Lecture 26, Slide 5
Checkpoint 2
What happens to the focal length of a converging lens when it is placed under water?
A. increases
B. decreases
C. stays the same
“Under water, the index of refraction is greater and closer to the index of refraction
of the lens; this makes the angle of refraction smaller by Snell's law so the focal
length will increase.”
“The index of refraction will increase so the focal length will decrease.”
“The lens makers formula shows focal length only depends on N and R”
Physics 212 Lecture 26, Slide 6
Checkpoint 2
What happens to the focal length of a converging lens when it is placed under water?
A. increases
B. decreases
C. stays the same
The rays are bent more from air to glass than from water to glass
Therefore, the focal length in air is less than the focal length in water
We can see this also from Lensmaker’s Formula
Physics 212 Lecture 26, Slide 7
Object Location
• Light rays from sun bounce off object and go in all
directions
– Some hits your eyes
We know object’s location
by where rays come from.
We will discuss eyes in lecture 28…
Physics 212 Lecture 26, Slide 8
Waves from object are focused by lens
Physics 212 Lecture 26, Slide 9
Two Different Types of Lenses
Physics 212 Lecture 26, Slide 10
Converging Lens: Consider the case where the shape of the lens is
such that light rays parallel to the axis of the mirror are all
“focused” to a common spot a distance f behind the lens:
f
f
Physics 212 Lecture 26, Slide 11
Recipe for finding image:
1) Draw ray parallel to axis refracted ray goes through focus
2) Draw ray through center refracted ray is symmetric
object
f
image
You now know the position of the same point on the image
Physics 212 Lecture 26, Slide 12
S > 2f
image is:
real
inverted
smaller
Example
1 1 1
 
S S f
S
M 
S
object
f
image
f>0
S>0
S’ > 0
Physics 212 Lecture 26, Slide 13
S=f
Example
1 1 1
 
S S f
image is:
at infinity…
S
M 
S
object
f
f>0
S>0
Physics 212 Lecture 26, Slide 14
Example
0<S<f
1 1 1
 
S S f
image is:
virtual
upright
bigger
S
M 
S
image
f
object
S>0
S’ < 0
f>0
Physics 212 Lecture 26, Slide 15
Checkpoint 1a
A
B
C
D
“The object is real, therefore its image is real. Real things make real things.”
“The screen is to the downstream of the lens, so s prime must be positive and the image must
be real. Since both s and s prime are positive, the magnification will be inverted.”
“The image is virtual because it is behind the lens. It is upright because it is always upright. ”
“converging lens always give the virtual image and if it can be projected on the screen, it is
inverted”
Physics 212 Lecture 26, Slide 16
Checkpoint 1a
A
B
C
D
Image on screen
s  0
M 
s
0
s
MUST BE REAL
MUST BE INVERTED
Physics 212 Lecture 26, Slide 17
Diverging Lens: Consider the case where the shape of the lens is
such that light rays parallel to the axis of the lens all diverge but
appear to come from a common spot a distance f in front of the lens:
f
Physics 212 Lecture 26, Slide 18
Example
image is:
virtual
upright
smaller
1 1 1
 
S S f
S
M 
S
object
f
image
f<0
S>0
S’<0
Physics 212 Lecture 26, Slide 19
Executive Summary - Lenses:
S > 2f
real
inverted
smaller
2f > S > f
real
inverted
bigger
f >S>0
virtual
upright
bigger
S >0
virtual
upright
smaller
converging
1 1 1
 
S S f
diverging
f
S
M 
S
f
Physics 212 Lecture 26, Slide 20
It’s always the same:
1 1 1
 
S S f
S
M 
S
You just have to keep the signs straight:
The sign conventions
S: positive if object is “upstream” of lens
S’ : positive if image is “downstream” of lens
f: positive if converging lens
Physics 212 Lecture 26, Slide 21
Checkpoint 1b
A converging lens is used to project the image of an arrow onto a screen as shown above.
A piece of black tape is now placed over the upper half of the lens. Which of the following is true?
A. Only the lower half of the object (i.e. the arrow tail) will show on the screen
B. Only the upper half of the object (i. e. the arrow head) will show on the screen
C. The whole object will show on the screen
“The light from the bottom half only makes it through the lens.“
“When the top half is covered, the image would cover the lower
end of the screen. The image is inverted but real size.”
“The entire image will still be projected onto the screen but there
will not be as many rays so the intensity will be lower.”
Physics 212 Lecture 26, Slide 22
Cover top half of lens
Light from top of object
Cover top half of lens
Light from bottom of object
object
image
object
image
What’s the Point?
The rays from the bottom half still focus
The image is there, but it will be dimmer !!
A converging lens is used to project the image of an arrow onto a screen as shown above.
A piece of black tape is now placed over the upper half of the lens. Which of the following is true?
A. Only the lower half of the object (i.e. the arrow tail) will show on the screen
B. Only the upper half of the object (i. e. the arrow head) will show on the screen
Physics 212 Lecture 26, Slide 23
C. The whole object will show on the screen
Calculation
A magnifying glass is used to read the fine print on a
document. The focal length of the lens is 10mm.
At what distance from the lens must the document
be placed in order to obtain an image magnified by a
factor of 5 that is NOT inverted?
• Conceptual Analysis
• Lens Equation: 1/s + 1/s’ = 1/f
• Magnification: M = -s’/s
• Strategic Analysis
• Consider nature of image (real or virtual?) to determine relation between
object position and focal point
• Use magnification to determine object position
Physics 212 Lecture 26, Slide 24
A magnifying glass is used to read the fine print on a
document. The focal length of the lens is 10mm.
At what distance from the lens must the document
be placed in order to obtain an image magnified by a
factor of 5 that is NOT inverted?
Is the image real or virtual?
(A) REAL
(B) VIRTUAL
A virtual image will be upright
A real image would be inverted
h
h’
f
h’
h
f
Physics 212 Lecture 26, Slide 25
A magnifying glass is used to read the fine print on a
document. The focal length of the lens is 10mm.
At what distance from the lens must the document
be placed in order to obtain an image magnified by a
factor of 5 that is NOT inverted?
How does the object distance compare
to the focal length?
(A)
s f
Lens
equation
(B)
s  f
(C)
1 1 1
 
s f s
s 
Virtual Image fl s’ < 0
Real object fl s > 0
Converging lens fl f > 0
s  f
fs
s f
h’
s’
h
s
f
s f 0
Physics 212 Lecture 26, Slide 26
A magnifying glass is used to read the fine print on a
document. The focal length of the lens is 10mm.
At what distance from the lens must the document
be placed in order to obtain an image magnified by a
factor of 5 that is NOT inverted?
s 
fs
s f
What is the magnification M in terms of s and f?
(A) M 
Lens
equation:
1 1 1
 

s
f s
fs
s 
s f
s f
f
(B) M 
f s
f
(C) M 
f
s f
(D) M 
f
s f
Magnification
equation:
s
M 
s
h’
M
f
s f
s’
h
s
f
Physics 212 Lecture 26, Slide 27
A magnifying glass is used to read the fine print on a
document. The focal length of the lens is 10mm.
At what distance from the lens must the document
be placed in order to obtain an image magnified by a
factor of 5 that is NOT inverted?
(A) 1.7mm
(B) 6mm
(C) 8mm
f
M
s f
(D) 40 mm
(E) 60 mm
M  5
f  10 mm
M
f
s f
s
M  1

s f
M
4
f  8 mm
5
h’
h
s   sM  40 mm
f
Physics 212 Lecture 26, Slide 28
Follow Up
Suppose we replace the converging lens with a diverging
lens with focal length of 10mm.
If we still want to get an image magnified by a factor
of 5 that is NOT inverted, how does the object sdiv
compare to the original object distance sconv?
(A) sdiv < sconv
(B) sdiv = sconv
(C) sdiv > sconv
EQUATIONS
M
f
s f
M  5
f  10 mm
(D) sdiv doesn’t exist
PICTURES
s f
M 1
M
4
s  f  8 mm
5
s negative fl not real object
h
h’
s
f
s’
Draw the rays: s’ will always be smaller than s
Magnification will always be less than 1
Physics 212 Lecture 26, Slide 29
Follow Up
Suppose we replace the converging lens with a diverging
lens with focal length of 10mm.
M
What is the magnification if we place the object at
s = 8mm?
(A) M 
1
2
(B) M  5
EQUATIONS
M
f
s f
s  8 mm
f  10 mm
(C)
M
3
8
(D)
M
5
9
(E)
f
s f
M
4
5
PICTURES
10
10 5
M 
 
8  (10) 18 9
h
h’
f
Physics 212 Lecture 26, Slide 30