Physics 102
Homework set 9
Solutions
L. Weinstein

Exercises:
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28.4: To shoot his assailant, he should aim his gun directly at the
assailant's image in the mirrored metal plate.  The bullet will bounce
off the plate at equal angles (angle of  incidence = angle of
relection) and so does the light.

Note that there is only one bullet and it hits the plate in only one
location.  There are lots of light rays, but the  only ray that
bounces from the plate and hits the cowboy's eyes is the one that
reflects at an equal angle.


28.6: The lettering on the front of some vehicles is backwards so that
it can be read by drivers looking at it through their  rear-view
mirrors.  


28.10: The person in daylight is well illuminated by sunlight.  The
person in the dark room has  almost no light illuminating her.  Light
reflecting from the person outside will reach the  person in the dark
room and there will be little other light.  The very little light
reflecting from the person in the dark room will be so much less than the
sunlight reflecting from the window  that the person outside will not
be able to see the person inside.


28.14: Half your height.  We did this in class.  Let's assume that
your eyes are 6 feet from the floor.  A light ray travels from your
feet to your eyes.  It travels 3' vertically as it goes from your feet
to the mirror and then the remaining 3' vertically as it travels from
the mirror to your eyes.  Thus, you do not need the bottom 3' of the
mirror.  (If you are 5' tall, then you do not need the bottom 2.5' of
the mirror.)


28.32: When you shine a light from the air to the water, there is no
angle where there is total internal reflection.  No matter what angle
you aim the light at, some light will always be transmitted
(refracted) into the water.

When you shine the light from water to air, the refracted ray is at a
larger  angle than the incident ray.  This means that there is some
angle (the critical angle) where the  refracted  ray is  horizontal
(ie: parallel to the water surface).  At larger angles, the refracted
ray has nowhere to go and so the light is totally internally
reflected.  See fig 28.36.

When you shine light from air to water, the refracted ray is at a
smaller angle than the incident ray.  Therefore, there is never an
angle at which the refracted ray is horizontal.

Another way to look at this is to use the 'I see you, you see me'
theorem.  This says that if light can go from point A to point B along
a certain path, then light can also go the other direction.  Thus, if
you look at figure 28.36 (and reverse the direction of the light), no
matter what angle light travels in the air, there is always a
refracted beam in the water.


28.40: No.  The swimming pool cover is the same size as the swimming
pool.  Therefore the same amount of energy from sunlight reaches the
pool in either case.  The tiny lenses just redistribute the light
hitting the pool so that there are light spots and dark spots as in
figure  28.45.  You would do just as well to use a simple sheet of
plastic.


28.46: Covering the top half of a camera lens would just reduce the
light available for the picture by one-half.  You would still have the
same picture.  We did this in class where I covered half of a lens and
the projected image did not change (except for getting a little
dimmer).


28.50: When you hold a magnifying glass in air, the light refracts as
it goes from air to glass and back to the air.  This allows the
magnifying glass to focus the light.  If you put the same magnifying
glass in water, then the light will refract as it goes from water to
glass and back to the water.  However, because the change in the speed
of light from water to glass is less than the change in the speed of
light from air to glass, the light will refract less and the
magnifying glass will have less effect.  It will magnify less.

If you put the magnifying glass in a liquid with exactly the same
index of refraction as the glass (like we saw in the movie in class),
then light will not refract at the boundary and the glass will not
magnify at all.


28.55: The image is one focal length behind the lens for distant
objects.  Light from a distant object is almost parallel when it
enters the lens  and so it focuses at the focal point.


29.2: Diffraction is more evident for sound wave because the
wavelength of sound waves ranges from about 10 cm to 10 m.  The
wavelength of light is about 5*10^(-7) m (or 1/100th the diameter of
human hair).  Diffraction occurs when a wave passes through a hole
about the same size as a wavelength.


29.12: Blue light will produce narrower-spaced fringes because blue
light has a smaller wavelength than yellow light.


Problems:
---------------------

28.2: The image of the butterfly is 20 cm behind the mirror.  Thus,
the image is 50 cm + 20 cm = 70 cm from your eyes.  Another way to
look at the problem is that light  must travel 20 cm from the
butterfly, reflect from the mirror, and travel 50 more cm to your eyes
for a total distance of 70 cm.


28.4: You walk toward the  mirror at 2 m/s.  You image appears to walk
toward the mirror at 2 m/s.  Therefore, you approach your image  at 4
m/s.


Estimation:
--------------------

a) The time it takes cars and trucks to kill 50 Americans is

t = 50 Americans * (1 year / 4*10^4 Americans) = 10^(-3) year 
  = 10^(-3) year * (365 day/year) * (24 hour/day) = 9 hours

(Yes, that is sloppy arithmetic.  Your answer should be between 8 and
12 hours.)

b) Since tobacco kills ten times more Americans per year than cars do,
it takes  ten times less time to do so, so the answer will be 1 hour.
We can also recalculate the answer:

t = 50 Americans * (1 year / 4*10^5 Americans) = 10^(-4) year 
  = 10^(-4) year * (365 day/year) * (24 hour/day) = 1 hour