Physics 5B
Lecture 17, February 22, 2012
Chapter 33, Thin lenses
Second Midterm Exam
The second midterm exam is scheduled
for next Monday, February 27.
 Be
e su
suree to bbringg bot
both a ca
calculator
cu ato and
a a
straightedge (small ruler).
 Last year
year’ss exam
exam, with solutions
solutions, is posted
on eCommons.
 The exam will cover Sections 14.8
14 8
through 33.4 (up through this week’s
homework assignment).
assignment)

Ray Diagrams (principal rays)
http://phet.colorado.edu/en/simulation/geometric-optics
Lens Sign Conventions
f  0 converging lens
f  0 diverging
di
i lens
l
d o  0 object
j on side of enteringg light
g
do  0
object on side of exiting light (virtual object)
di  0 image
i
on side
id off exiting
i i light
li h
di  0
image on side of entering light (virtual image)
m  0 non - inverting
m  0 inverting
Suppose that a convex lens is used to form a real image of an
object, with the image projected on a screen as illustrated below.
Th a barrier
b i is
i place
l
i front
f
h upper half
h lf off the
h lens,
l
Then
in
off the
blocking all light from passing through that half. What happens
to the image?
A It
A.
I gets di
dimmer on the
h screen bbut iis otherwise
h
i unchanged.
h
d
B. It vanishes.
C. The upper
pp half of the image
g vanishes.
D. The lower half of the image vanishes.
barrier
screen
Where are the images of the letter A
in the two situations depicted in the
photograph of a page being viewed
through a magnifying glass? “Above”
means closer
l
to you and
d “below”
“b l ”
means on the side with the paper.
A. Above the lens in (a) and below
the
h lens
l
in (b)
B. Below the lens in (a) and above
the lens in (b)
C. Above the lens in (a) and in (b)
D. Below the lens in (a) and in (b)
Problem 33
33--11
How far from a converging
g g lens with a focal
length of 25 cm should an object be placed
to p
produce an image
g that is the same height
g
as the object?
Multiple Lenses
In the case that two lenses are present, first
treat the lens that the light encounters first
and find its image.
That intermediate image becomes the
object
b
ffor the
h second
d lens.
l
When figuring the object distance for the
second lens, keep in mind that it may be
negative
ti (virtual
( i t l object).
bj t)
33--21
Problem 33
A diverging lens with f=35 cm is placed 14.0 cm behind
a converging lens with f=20
f=20.0
0 cm
cm.
a) Where will an object at infinity be focused?
b) Wh
Where will
ill an object
bj
at 40 cm bbe ffocused?
d? Wh
What iis
the magnification in that case? Draw a ray diagram.
c) Where will an object at 15 cm be focused? Is the
image real or virtual? What is the magnification?
p.s. I added
dd d parts
t (b) and
d ((c)) tto the
th problem.
bl
Problem 3333-21, ppart b
Here is an accurate ray diagram for part b of the problem on the previous slide, with
an object distance of 40 cm. The principal rays for the first lens are in blue; those for
the second lens are in magenta. The image from the first lens never actually forms,
because the second lens gets in the way.
way Therefore, the intermediate image become a
virtual object for the second lens. To draw principal rays for the virtual object, I draw
rays on the opposite side (the side of the incoming light) heading toward the virtual
object, and then draw how they are refracted by the diverging lens, as follows:
(1) A ray through the center of the lens is not refracted.
(2) A ray parallel to the axis refracts away from the axis, as if it were coming from the
focus F2 on the left.
(3) A ray heading toward F2 on the right gets refracted to become parallel to the axis.