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\begin{center}
{\Large {Solution Derivations for Capa \#13\\[20pt]} }
\end{center}
1) A super nova releases $1.3 \times 10^{45}\ J$ of energy. It is $1540\ ly$
from earth. If you were facing the star in question, and your face was a circle
$7\ cm$ in radius, how much energy would reach your face?
\begin{quote}
$U$ = Given\\
$r_{1}$ = Given\\
$r_{2}$ = Given\\
This problem can be done in the same way as on the previous CAPA involving
the Poynting vector. You are given the total energy. Calculate the
intensity (energy per unit area) and find the proportion of that occupied by
the detector. In other words, the amount of energy you feel on your face
is the percentage of the total surface area which the energy is spread about.
\[
S=\frac{energy}{area}=\frac{U}{4\pi r_{1}^{2}}
\]
The area calculated is the surface area of a sphere since the energy was
released in all directions. Also note that the radius was given in light
years and you need it to be in meters. To convert, use
$1\ m=9.46 \times 10^{15}\ ly$.
\[
U_{detected}=S\ast A_{detector}
\]
Since the assumption is being made that your face is a circle of radius
$r_{2}$, the area is then $A_{detector}=\pi r_{2}^{2}$. Thus, the amount of
energy that reaches your face is
\[
U_{detected}=\frac{U}{4\pi r_{1}^{2}}\pi r_{2}^{2}=\allowbreak \frac{%
U}{4r_{1}^{2}}r_{2}^{2}
\]
\end{quote}
2) A point source of light is located $4.52\ m$ below the surface of a large
lake of clear toxic fluid (Lake Ontario, where $n = 1.50$). Find the area
of the largest circle on the pool's surface through which light coming
directly from the source can emerge.
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.1704in]{capa13-02.png}
\end{center}
\end{figure}
\begin{quote}
$d$ = Given\\
$n_{1}$ = Given\\
$n_{2}$ = Given (air)\\
First, we must find the critical angle as this will be the point at which no
additional light will emerge. From Snell's Law,
\begin{eqnarray*}
n_{1}\sin \theta _{c} & = & n_{2}\sin \theta _{2}=n_{2}\sin 90^{\circ} = n_{2}\\
\theta _{c} & = & \sin ^{-1}\left( \frac{n_{2}}{n_{1}}\right) =\sin ^{-1}\left(
\frac{1}{n_{1}}\right)
\end{eqnarray*}
Where the last step follows since the ray is emerging into air. Now, we
can find the length of the triangle which will be the radius of the largest
circle of which light can still emerge. Thus,
\begin{eqnarray*}
\phi & = & 90^{\circ}-\theta _{c}\\
\tan \phi & = & \frac{d}{r}\\
r& = & \frac{d}{\tan \phi } = \frac{d}{\tan \left( 90^{\circ}-\theta_{c}\right) }
\end{eqnarray*}
The area of a circle is
\[
A=\pi r^{2}=\pi \left( \frac{d}{\tan \left( 90^{\circ}-\theta
_{c}\right) }\right) ^{2}
\]
\end{quote}
3) A laser beam enters a $14.0\ cm$ thick glass window at an angle of $46^{\circ}$
(from the normal). The index of refraction of the glass is $1.50$. At what angle
from the normal does the beam travel through the glass? Use units of ``deg.''
\begin{quote}
$w$ = Given (width)\\
$\theta _{1}$ = Given\\
$n_{2}$ = Given\\
$n_{1}$ = Given (air)\\
Thus, using Snell's Law,
\begin{eqnarray*}
n_{1}\sin \theta _{1} & = & n_{2}\sin \theta _{2}\\
\theta _{2} & = & \sin ^{-1}\left( \frac{n_{1}}{n_{2}}\sin \theta _{1}\right)\\
& = & \sin ^{-1}\left( \frac{1}{n_{2}}\sin \theta _{1}\right)
\end{eqnarray*}
\end{quote}
4) How long does it take the beam to pass through the plate?
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.1704in]{capa13-04.png}
\end{center}
\end{figure}
\begin{quote}
This one is a bit trickier since the ray does not travel straight
through the glass. The index of refraction for a material comes from the
ratio of speed that light travels through it. That is,
\[
n=\frac{c}{v}
\]
Thus, the speed is
\[
v=\frac{c}{n}
\]
Since velocity is distance over time,
\[
t=\frac{d}{v}=\frac{d}{\frac{c}{n}}=\frac{dn}{c}
\]
Since $\theta _{2}$ was calculated in the previous problem, the distance is
calculated by
\begin{eqnarray*}
\cos \theta _{2} & = & \frac{w}{d}\\
d & = & \frac{w}{\cos \theta _{2}}
\end{eqnarray*}
Thus, the time required for the light to travel through the glass is
\[
t=\frac{dn}{c}=\frac{w\ast n}{c\ast \cos \theta _{2}}
\]
\end{quote}
5) Light in air enters a stack of three parallel plates with indices of
refraction $1.38$, $1.52$, and $1.98$, respectively. The incident beam makes
a $74.0^{\circ}$ angle with the normal to the plate surface. At what angle
with respect to the stack normal does the beam emerge into the air after
passing through the stack? Use units of ``deg.'' (If you do a lot of
calculations for this one, you will kick yourself when you realize the
answer.)
\begin{quote}
$n_{1}$ = Given\\
$n_{2}$ = Given\\
$n_{3}$ = Given\\
$\theta _{1}$ = Given\\
When light travels through different media but emerges back into the
original medium, it will be traveling in the same direction; thus the angle
at which it emerges will be the same as that when it entered. Thus,
\[
\theta _{f}=\theta _{1}
\]
\end{quote}
6) A woman stands between a vertical mirror $0.40\ m$ tall and a distant tree
whose height is $H$. She is $2.30\ m$ from the mirror, and the tree is
$17.0\ m$ from the mirror. If she sees the tree just fill the mirror, how
tall is the tree?
\begin{figure}[h]
\begin{center}
\includegraphics[width=6.3114in]{capa13-06.png}
\end{center}
\end{figure}
\begin{quote}
$d$ = Given\\
$h^{\prime }$ = Given (image height)\\
$l^{\prime }$ = Given (image distance)\\
From the magnification equation, we can find the height of the tree, $h$.
\begin{eqnarray*}
\frac{h^{\prime }}{h} & = & \frac{l^{\prime }}{l}\\
h & = & \frac{l\ast h^{\prime }}{l^{\prime }}
\end{eqnarray*}
Note that since the image fills the mirror, the image height is the size of
the mirror. Also, the distance to the object is the distance from the
object to the mirror plus the distance back to the observers eye. Thus,
\[
l=d+l^{\prime }
\]
So,
\[
h=\frac{\left( d+l^{\prime }\right) \ast h^{\prime }}{l^{\prime }}
\]
\end{quote}
7) A student uses a semicircular flat plastic box to observe the optical
properties of liquids. As shown in the figure, she sets two pins to observe
through the water. When the box is filled with water (index of refraction =
1.33) she observes the images of the pins defining the paths $X$ and $Y$ and
measures the angles $\alpha$ and $\beta$. They are $\alpha = 49.7^{\circ}$,
and $\beta = 35.0^{\circ}$. Replacing the water with another liquid (index of
refraction = 1.48), she remeasures the values for angles $\alpha$ and $\beta$.
What is the value of $\alpha$ in the presence of the new liquid? Use units of
``deg.''
\begin{quote}
$n$ = Given (original liquid)\\
$n_{1}$ = Given (new liquid)\\
$\vspace{1pt}\theta _{1}$ = Given\\
$\alpha $ = Given\\
$\beta $ = Given\\
All you need to know for this problem are $n_{1}$ and $\theta _{1}$. Snell's
Law tells us
\[
n_{1}\sin \theta _{1}=n_{2}\sin \theta _{2}
\]
Since the light ray is travelling from the liquid into the air, $n_{1}$ is
the refractive index of the liquid and $n_{2}$ is that of air; namely 1.
The outgoing angle is labeled $\alpha $ in the diagram. Substituting this
back in,
\[
n_{1}\sin \theta _{1}=n_{2}\sin \alpha =\sin \alpha
\]
Thus,
\[
\alpha =\sin ^{-1}\left( n_{1}\sin \theta _{1}\right)
\]
\end{quote}
8) A light ray strikes the surface between two transparent materials as shown.
Each angle is labeled with a letter. Give the letters for the angle of
incidence, angle of refraction and angle of reflection, in that order (e.g.
dbb.)
\begin{quote}
Note the direction of the rays. The ray approaching the boundary is the
incident ray. The angle of incidence is measured between this ray and the
normal line to the boundary. The reflected ray travels away from the
boundary, but on the same side as the incident ray. The angle of reflection
is the angle between this ray and the normal line. The refracted ray also
travels away from the boundary, but it is inside the other medium at the
boundary.
\end{quote}
9) A beam of light, colored as indicated, strikes one face of a glass prism at
various angles, as shown in each choice above. Which choices correctly
represent the passage of the light through the prism? Note carefully whether
the beams within the prism are parallel. (Enter all correct choices in
alphabetical order, e.g., B, CD, or ABE).
\begin{quote}
A) Incorrect since the rays are refracted by different amounts within the
medium.\\
B) Correct.\\
C) Incorrect since red light is refracted less than violet light; thus it
would be on top.\\
D) Incorrect. Not sure why this is incorrect.\\
E) Incorrect since the light is refracted at every boundary where it is
incident with a nonzero angle (it would have been refracted while exiting
the prism).\\
CAPA is only looking for the correct answers. In this case, B.
\end{quote}
10) In the above figures $f$ is located at the focal point of the lens. For
each statement enter T or F.
\begin{figure}[h]
\begin{center}
\includegraphics[width=4.6674in]{capa13-10.png}
\end{center}
\end{figure}
\begin{quote}
QUESTION:\\
A) An object placed between $f$ and the lens in Fig. $a$ results in an image
on the right side of the lens.\\
B) The lens in Fig. $b$ is a converging lens.\\
C) An object placed to the left of $f$ and the lens in Fig. $a$ produces a
virtual image.\\
D) An object placed to the left of $f$ in Fig. $b$ results in an image on
the right side of the lens.\\
E) An object placed to the left of $f$ and the lens in Fig. $b$ produces a
virtual image.\\
F) An object placed to the left of $f$ in Fig. $a$ results in an image on
the right side of the lens.\\
G) An object placed between $f$ and the lens in Fig. $a$ produces a real
image.\\
ANSWER:\\
Playing with the JAVA lens applet will help answer these questions.\\
A) False, an object placed here will produce a virtual image.
Not enough ``bending power'' to make a real image. Consequently, the
virtual image is formed, upright, on the left side of the lens (with the
object).\\
B) False, a lens with this shape is a diverging lens. Rays will bend
toward the thicker part of a lens, thus causing them to diverge from a point
in this image.\\
C) False, this position will produce a real image.\\
D) False, a diverging lens cannot produce a real image.\\
E) True, see (D)\\
F) True, if the image is further away than one focal length, a real image is
formed, inverted, on the opposite side of the lens.\\
G) False, see (A)\\
CAPA is looking for true/false answers to these questions. Thus, for this
problem, FFFFTTF
\end{quote}
11) Starting with a real object, which of the following statements are TRUE
about the image formed by a single lens? (Enter all correct choices in
alphabetical order, e.g. B, CD, or ABE).
\begin{quote}
QUESTION:\\
A) A diverging lens always produces a virtual, upright, reduced image.\\
B) A converging lens can never produce a virtual, upright, reduced image.\\
C) A converging lens cannot produce a real, inverted reduced image.\\
D) For a converging lens an object has to be placed between the focal length
and the lens in order to form a virtual image.\\
E) A converging lens can produce a virtual, upright, enlarged image.\\
F) A diverging lens can produce a real, inverted, reduced image.\\
ANSWER:\\
Playing with the JAVA lens applet will help answer these questions.\\
A) True.\\
B) True, the virtual, upright images it forms are always magnified.\\
C) False, an object placed further than $2f$ away from the lens will be a
real, inverted, and reduced image.\\
D) True, see (10A).\\
E) True, this is the only type of virtual image it can produce.\\
F) False, a diverging lens cannot produce a real image.\\
CAPA is looking for the letters of the correct answers, in this case, ABDE
\end{quote}
12) How much is the light ray displaced after passing through the $5.30\ cm$
thick sheet of transparent material ($n=1.33$) with an incident angle of
$\theta = 34.0^{\circ}$.
\begin{figure}[h]
\begin{center}
\includegraphics[width=6.1021in]{capa13-12.png}
\end{center}
\end{figure}
\begin{quote}
$d$ = Given\\
$n_{1}$ = Given (Air)\\
$n_{2}$ = Given\\
$\theta _{1}$ = Given\\
Starting with Snell's Law,
\[
n_{1}\sin \theta _{1}=n_{2}\sin \theta _{2}
\]
Thus,
\[
\theta _{2}=\sin ^{-1}\left( \frac{n_{1}}{n_{2}}\sin \theta _{1}\right)
=\sin ^{-1}\left( \frac{1}{n_{2}}\sin \theta _{1}\right)
\]
The hypotenuse, $h$, of this first triangle can be found by
\begin{eqnarray*}
\cos \theta _{2} & = & \frac{d}{h}\\
h & = & \frac{d}{\cos \theta _{2}}
\end{eqnarray*}
The angle between the refracted ray and the original ray is then
\[
\alpha =\theta _{1}-\theta _{2}
\]
Looking at the second triangle, we can see that
\begin{eqnarray*}
\sin \alpha & = & \frac{x}{h}\\
x & = & h\sin \alpha
\end{eqnarray*}
\end{quote}
\end{document}