nL;

negative if the image is located in
front of the lens. Converging lenses
(thicker in the center than at the
edges) have positive l, while diverging
lenses (thinner at the center)
have negative /.
Another useful relationship is the
lens maker's formula. For the special
case when one of the surfaces is planar,
it tells us that
1= r_r
f R'
where n is the index of refraction of
the lens material andR is the radius
o{ the curved surface.
Now that we have completed this
very brief review, let's take a look at
our problem. A concave mirror of
radius R resting face up on a table
top has been filled with a small
amount of water (index of refraction
n = alSl as shown in the figure below.
A small object is located a distanced
= SRIZfromthe mirror along
the optic axis. Where is the image 1ocated?
In the spirit of the "thin lens
approximation" often used in such
problems, we will neglect the thickness
of the water.
A. Let's begin by using a technique
used by eye doctors. Often the
doctor will place a lens in front of
your glasses to show you how the
new lenses will work. This works
because the effective focal length f'
of two (or more) lenses (or mirrors)
in close proximity is given by
t _rr I
f,- L f,
is, the focal lengths add as re-
-that
ciprocals. Therefore, the mirrorwater
combination can be replaced
by a mirror with an effective focal
length and you can use the mirror
formula given above. Does the water
lens appear in the sum once or
twice? Use the other methods to
check yourself.
B. You can also obtain the effective
focal length by tracing aray parallel
to the optic axis as it enters the
water and bends according to Snell's
law, reflects from the mirror surface,
and exits the water again. Don't forget
to make suitable approximations.
C. Our third method makes use
of the observation that images
formedby one optical element act as
objects for subsequent optical elements.
Begin by finding the location
of the image formed by the air-water
interface. Use this image as the object
for the mirror (without the water)
and find the new image location.
Then find the image of this image
formed by the water-air interface
when the light exits the water. This
is the finai image produced by the
combination.
D. The trickiest method treats
the combination as a water lens, a
mirror, and a water lens in combination.
Find the location of the image
produced by each element and then
use it as the object for the next element.
This is trickybecause it's very
easy to make mistakes with the sign
conventions.
Please send your solutions to
Quantum, 1840 Wilson Boulevard,
Arlington VA2220l-3000 within a
month of receipt of this issue. The
best solutions will be noted in this
space and their authors will receive
special certificates from Quantum.
Shp on red, Uo on Ul'solt...
Quantum readers were asked to
determine when it's sa{e to go
through a yellow light and when it's
safe to apply the brakes at the yellow
light. We hope many of our
readers thought about the problem
as they waited at an intersection for
the light to change.
An excellent solution was submitted
by Ophir Yoktan of Israel.
Unfortunately, Yoktan provided no
biographical information and so we
don't know if Yoktan is a professor
or a student. Irrespective of that, the
solution presented here closely follows
Yoktan's submission.
A. (a) In the "go zofle" a person
wi1lbe able to continue at the traveling
speed and get through the intersection
within the time that the
yellow iight is illuminated. This
depends on the velocity of the clt vst
the yellow light time t.,,, the width of
the intersection ra4, ancl the length of
the carl. (The go zone is cluite different
for a stretch limo and a compact
car.) This gives us
drv^t Ur
--z ,o
-2,
(c) Whether we have a "dilemma
zolne" or an "overlap zofre" depends
on the difference between the go
zone and the stop zone.If the zone
is defined as the go zone minus the
stop zone, a negative value will indicate
a dilemma zone and a positive
value will indicate an overlap zone:
zone = voty -w - 1-votr+fi
1"t
=
*"'r* (r, - ,,)ro-(w+ /).
B. We can find the conditions for
which there will always be a dilemma
zoneby requiring that the
zone be negative:
o, j"'o* (r, - r,)ro - @ + 1).
See the equation in the box on the
next page. A dilemma zone willalways
exist if the terms within the
radical sign are negative. This occurs
34
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