Announcements 11/2/12
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Prayer
Slinkies! (Seth, Ryan, William B, Clement)
Progress Report due a week from Saturday
Pearls
Before
Swine
From warmup
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Extra time on?
a.(nothing in particular)
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Other comments?
a.Do astronomers have to worry about the
fact that air doesn't have an n of exactly
1?
Clicker question:
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A converging lens should be:
a. convex
b. concave
Convex/converging Lens
at each interface,
Snell’s law is
satisfied
f
parallel rays
converge to the
focal point
if not parallel, they
converge (image formed)
at a different point
Where will image be formed?
focus
focus
Ray diagram just like one for concave mirror, except
rays continue to right instead of reflecting back!
The equation:
1 1 1
 
p q f
positive q = right
hand side of lens
Magnification same:
q
M 
p
From warmup

The mirror eqn and the thin lens eqn look
identical. How exactly are they different?
a.The sign conventions are different, a
positive focus means a focus that is
"behind" the lens. The same is true of q.
b.(from my answer) Also--how f is calculated
is different. For mirrors, it's simply related to
the radius of curvature, but for lenses it
relates to the curvature as well as the index
of refraction of the lens glass.
Clicker quiz
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Just out of curiosity… have I ever used your
warmup answer?
a.yes
b.no
Concave/diverging Lens
f
parallel rays seem to be
coming from the focal point
Diverging lenses
virtual image!
The equation:
1 1 1
 
p q f
(negative f)
Negative q: means image on same side
of lens as object (virtual)
Demos
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Demo: “Real image or not?” Class poll: What
kind of image is this?
a. Real
b. Virtual
Sign conventions: summary
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Assuming that the light rays are coming from the left:
quantity
positive
+
negative

p
if on the left of the
lens/mirror
if on the right of the lens/mirror
(can only happen in a
compound problem)
q
if on the right of the lens
(left side of mirror),
means a real image
if on the left of the lens
(right side of mirror),
means a virtual image
f
if converging: convex
lens/concave mirror
if diverging: concave
lens/convex mirror
M
if image is right-side up
if image is upside-down
“Lensmaker’s Eqn”
Lens-makers’ eqn:
 1
1
1 
  n  1  

f
 R1 R2 
From warmup

When you look straight
down at an object under
water, will it appear to be
closer to you or farther
away than actuality? Why?
a. Because you are looking
at a flat refracting
surface the image of the
object forms closer to the
surface than the actual
object is (figure 36.18).
This means the object
appears closer to you
than it actually is.
Clicker Quiz
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Is that a real or a virtual
image?
a. real
b. virtual
Thick lenses and surfaces
n1 n2 n2  n1 Careful with signs!!!


p q
R
(see table 36.2)
Clicker question:
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Can a converging lens ever produce a virtual
image?
a. Yes
b. No
Clicker question:
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What happens if I cover up the upper half of lens?
a. Nothing
b. Upper half of image disappears
c. Lower half of image disappears
d. Intensity of image goes down by 50%
Clicker question:
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Can a diverging lens ever produce a real image?
a. Yes
b. No
Multiple element problems
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Image of one becomes object of next
Worked problem: Find qfinal, Mtot
f = 20 cm
f = -60 cm
60 cm
15 cm
Answers: image is 20 cm to right of -60cm lens; M = -0.67