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\noindent {\bf PHYSICS 1020}
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{\Large \bf Homework set 8\\}
{\large 10 February 1997\\}
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\item[{\bf [8.5]}] If you doubled the wavelength of a wave without changing its
wave speed, what would happen to the frequency? What if you halved the
wavelength?
  \\
Answer:\\
From the relation between speed, wavelength and frequency:
$$ v = f \cdot \lambda ~~~\Rightarrow ~~~ f~ = ~ {v\over \lambda} ,$$
we conclude that, with the wavespeed kept fixed, 
 doubling the wavelength will halve the frequency, while halving
the wavelength will double the frequency.

\item[{\bf [8.8]}] What happens to the energy of the two waves in Fig. 8.9
when they
interfere destructively, as shown in the second of the three sketches? Did the
energy vanish? \\
(Hint: what are the parts of the rope doing as the wave moves?
Which type of energy is the wave carrying?
  \\
Answer:\\
The wave is carrying kinetic energy corresponding to the up-and-down 
oscillatory motion of the elements of the rope. At the moment when the two
opposite wave crests meet, the net displacement is zero, but the speed is not.
The elements of the rope move through the zero displacement position at the
instant when the displacements due to the two waves extinguish each other, but
they still move. The energy does not disappear.

\item[{\bf [8.12]}] You shine two flashlights on the wall. Why don't you see an
interference pattern?
  \\
Answer:\\
The light emitted by  flashlights is not a continuous stream of
electromagnetic waves, but a sequence of rather short ``wavetrains'', each of
which is independent of the others. Therefore they have no well-defined phase
relation -- they are not synchronized -- they  are ``incoherent''. Even though 
there is interference between these wavetrains, there is no point in space where
they all interfere constructively or destructively. The resultant light
intensity is the result of a random superposition of all these wave trains
(or ``wave packets''), causing a rather uniform intensity distribution.

\item[{\bf [8.18]}] How does the operation of the electroscope (previous
exercise) demonstrate electrical conduction?
  \\
Answer:\\
In order for the charge deposited onto the electroscope to reach the leaves 
of the electroscope, charge must have moved from the charged rod (or whatever
was used to deposit charges) through parts of the electroscope to the leaves.
This motion (flow)  of charges is called electrical conduction.
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\item[{\bf [8.26]}] Making estimates: About how many atoms thick is a sheet of
paper? 
  \\
Answer:\\ 
A 500-page stack (one package) of typing (or Xerox) paper is about 5 cm thick;
so the thickness of one sheet is about 5 cm/500 = 0.1 mm = $10^{-4}{\rm m}$.
The diameter of an atom is about $10^{-10}{\rm m}$, so the number of atoms
across the thickness of one sheet
$$=~~ {10^{-4}{\rm m}\over 10^{-10}{\rm m}} = 10^6. $$
So, one sheet of paper is about 1 million atoms thick.

\item[{\bf [8.27]}] Making estimates: Which is bigger, an atom or a wavelength
of light? Roughly how much bigger? 
  \\
Answer:\\
The diameter of an atom is about $10^{-10}{\rm m}$, while the wavelength of
visible light (e.g. blue-green light) is about 
500 nm = $5\times 10^{-7}{\rm m}$.
The ratio between these two numbers 
$$=~~{5\times 10^{-7}{\rm m}\over 10^{-10}{\rm m}}~~=~~ 5000.$$
So a typical light wave is about 5000 times bigger than an atom.

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