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\begin{center}
{\Large {Solution Derivations for Capa \#14\\[20pt]} }
\end{center}
1) An image of the moon is focused onto a screen using a converging lens of
focal length ($f = 34.8\ cm$). The diameter of the moon is
$3.48 \times 10^6\ m$, and its mean distance from the earth is
$3.85 \times 10^8\ m$. What is the diameter of the moon's image?
\begin{quote}
$f$ = Given\\
$h$ = Given (diameter of moon, height of object)\\
$l$ = Given (distance to object)\\
The magnification equation can be used here:
\[
\frac{h^{\prime }}{h}=-\frac{l^{\prime }}{l}
\]
We are looking for $h^{\prime }$, the height (diameter) of the image on the
screen. Since the moon is so far away, we can treat the light rays as
being parallel. Thus, the image distance $l^{\prime }$ is the focal length
of the lens.
\[
h^{\prime }=-\frac{hf}{l}
\]
\end{quote}
2) A $0.54\ cm$ high object is placed $8.5\ cm$ in front of a diverging lens
whose focal length is $-7.5\ cm$. What is the height of the image?
\begin{quote}
$h$ = Given\\
$l$ = Given\\
$f$ = Given\\
We can't use the magnification equation immediately because we don't know
the height of the image or the distance to it. But, the lens equation also
relates these quantities. Thus, we can start with it and solve for the
first unknown, the image distance.
\begin{eqnarray*}
\frac{1}{l}+\frac{1}{l^{\prime }} & = & \frac{1}{f}\\
\frac{1}{l^{\prime }} & = & \frac{1}{f}-\frac{1}{l}\\
l^{\prime } & = & \frac{1}{\frac{1}{f}-\frac{1}{l}}
\end{eqnarray*}
Now, we can use the magnification equation.
\begin{eqnarray*}
\frac{h^{\prime }}{h} & = & -\frac{l^{\prime }}{l}\\
h^{\prime } & = & -\frac{hl^{\prime }}{l}\\
& = & -\frac{h}{l}\frac{1}{\frac{1}{f}-\frac{1}{l}}\\
& = & -\frac{h}{l\left( \frac{1}{f}-\frac{1}{l}\right) }
\end{eqnarray*}
\end{quote}
3) A magnifying glass uses a converging lens with a focal length of $15.5\ cm$.
It produces a virtual and upright image that is 2.7 times larger than the
object. How far is the object from the lens?
\begin{quote}
$f$ = Given\\
$M$ = Given\\
$l$ = ?\\
The magnification equation in another form is
\[
M=-\frac{l^{\prime }}{l}
\]
Thus,
\[
l^{\prime }=-lM
\]
Now, using the lens equation, the object distance can be solved for
\begin{eqnarray*}
\frac{1}{f} & = & \frac{1}{l}+\frac{1}{l^{\prime }}=\frac{1}{l}-\frac{1}{lM}=\frac{M}{lM}-\frac{1}{lM}\\
\frac{1}{f} & = & \frac{M-1}{lM}\\
lM & = & \left( M-1\right) f\\
l & = & \frac{\left( M-1\right) f}{M}
\end{eqnarray*}
\end{quote}
4) What is the image distance? (Think carefully about whether the answer is
positive or negative.)
\begin{quote}
The image distance formula was a necessary step for the final formula in \#3. It is
\[
l^{\prime }=-lM
\]
Since $l$ was calculated in \#3, simple plug this in. Note the signs, the
algebra will take care of it.
\end{quote}
5) In the 7 diagrams below, the solid arrow represents the object and the
dashed arrow the image. The rectangle shows the position of an SINGLE OPTICAL
ELEMENT. Match each diagram with the appropriate optical element. (If the
first corresponds to B, and the next 6 to C, enter BCCCCCC.)
\begin{quote}
Use the thin lens Java applet.
\end{quote}
6) Which of the following statements are correct? (All of the statements
concern real objects. Give ALL correct answers, i.e., C, AE, DEF, etc.)
\begin{quote}
QUESTION:\\
A) When an object is placed between a concave mirror and its focal point, the
image is virtual.\\
B) The image produced by a convex mirror is always closer to the mirror than
it would be in a plane mirror for the same object distance.\\
C) A concave mirror always forms a virtual image of a real object.\\
D) A concave mirror always forms an enlarged real image of a real object.\\
E) The virtual image formed by a convex mirror is always enlarged.\\
F) A virtual image formed by a concave mirror is always enlarged.\\
G) A convex mirror never forms a real image of a real object.\\
ANSWER:\\
Use the thin lens Java applet.
\end{quote}
7) An apple is placed $12.6\ cm$ in front of a diverging lens with a focal
length of magnitude $22.0\ cm$. What is the image distance $i$ of the image
of the apple through this lens?
\begin{quote}
$l$ = Given\\
$f$ = Magnitude given, value is negative by convention\\
$l^{\prime }$ = ?\\
The lens equation will work here.
\begin{eqnarray*}
\frac{1}{l}+\frac{1}{l^{\prime }} & = & \frac{1}{f}\\
l^{\prime } & = & \frac{1}{\frac{1}{f}-\frac{1}{l}}
\end{eqnarray*}
\end{quote}
8) What is the magnification of the image of the apple?
\begin{quote}
The magnification is given by
\[
M=\left| \frac{l^{\prime }}{l}\right|
\]
\end{quote}
9) Which of the following statements are true? (If, `C' and `D' are true and
the rest are false, answer `FFTT'.) The image of the apple is... (Note: ``in
front of'' means same side as the object; ``behind'' means the other side.)
\begin{quote}
QUESTION:\\
A) In front of the lens.\\
B) Virtual.\\
C) Upright.\\
D) Larger in size.\\
ANSWER:\\
Due to the potential variability of this question, only remarks can be made.\\
A) For a diverging lens, the image will be in front of the lens. For a
converging lens, the image will behind the lens for a real image, in front
of the lens for a virtual image.\\
B) A diverging lens will always produce a virtual image. A converging
lens will produce a real image unless the object is within the focal length.\\
C) A diverging lens' image will always be upright. A converging lens'
image will be inverted if it's real, upright if it's virtual.\\
D) A diverging lens' image will always be reduced. A converging lens'
image will be reduced if the object is greater than 2f away, otherwise it
will be enlarged. This results since the object and image are the same
size at 2f.
\end{quote}
10) {\bf NOTE:} We suggest you use ray diagrams to qualitatively understand
these questions. A candle $8.40\ cm$ high is placed in front of a thin
converging lens of focal length $32.0\ cm$. What is the image distance $i$
when the object is placed $94.0\ cm$ in front of the same lens?
\begin{quote}
$h$ = Given\\
$f$ = Given\\
$l$ = Given\\
$l^{\prime }$ = ?\\
A ray diagram will help with these problems. But, use the lens equation to
work out mathematically what will happen.
\begin{eqnarray*}
\frac{1}{l}+\frac{1}{l^{\prime }} & = & \frac{1}{f}\\
l^{\prime } & = & \frac{1}{\frac{1}{f}-\frac{1}{l}}
\end{eqnarray*}
\end{quote}
11) What is the size of the image? (Note: an inverted image will have a
`negative' size.)
\begin{quote}
The magnification equation will work here.
\begin{eqnarray*}
\frac{h^{\prime }}{h} & = & -\frac{l^{\prime }}{l}\\
h^{\prime } & = & -\frac{hl^{\prime }}{l}
\end{eqnarray*}
\end{quote}
12) Is the image real(R) or virtual(V); upright(U) or inverted(I); larger(L)
or smaller(S) or unchanged(UC); in front of the lens(F) or behind the lens(B)?
Answer these questions in the order they are posed. (for example, if the
image is real, inverted, larger and behind the lens, then enter `RILB'.)
\begin{quote}
This is where the ray diagram will help.
\end{quote}
13) A 79 year old man has a near point of $107\ cm$. What power lens (in
$diopters$) does the man need in order to be able to see an object clearly at
the normal near point of $25\ cm$? Assume that the person will get contact
lenses, that is neglect the two $cm$ that glasses would be away from the eye.
Recall that diopters is $1/f$, where $f$ is the focal length in meters.
(Enter answer (in units of diopters) without units.)
\begin{quote}
Near Point = $l^{\prime }$ = Given\\
Normal Near\ Point = $l$ = Given\\
The measurement of diopters is simply the ratio
\[
p=\frac{1}{f}
\]
where $f$ is the focal length of the lens. The object of correcting the
vision is to take an object at the normal near point and produce the image
where the person's near point is. Thus, they will be able to focus. To
do this, the object distance, $l$, is the normal near point. The image
will be produced at the person's near point, labelled $l^{\prime }$ for the
image distance. Thus,
\[
p=\frac{1}{f}=\frac{1}{l}+\frac{1}{l^{\prime }}
\]
\end{quote}
14) A 187x astronomical telescope is adjusted for a relaxed eye when the two
lenses are $1.49\ m$ apart. What is the focal length of the eyepiece?
\begin{quote}
$m$ = Given\\
$d$ = Given (separation of lenses)\\
The angular magnification for telescopes has the following definition
\[
m=\frac{f_{o}}{f_{e}}
\]
where $f_{o}$ is the objective lens focal length and $f_{e}$ is the eyepiece
focal length. Since we are trying to solve for the eyepiece focal length,
it will be easier to solve this equation for $f_{o}$.
\begin{equation}
f_{o}=mf_{e} \label{eqn:f_naught}
\end{equation}
Presumably in a telescope, the observer is looking at an object very far
away. Thus, the object distance is effectively infinity (i.e.
$l_{o}=\infty $). The first lens in a telescope will take the
distant object and place the image at the focal point of the lens (since the
rays from the object will be effectively parallel, so $l_{o}^{\prime }=f_{o}$
). This image at the focal point will be the object for the eyepiece.
The object will be at the focal point of the eyepiece lens (i.e. $l_{e}=f_{e}$
). This way, the rays will emerge parallel from the eyepiece (which will
by themselves produce an image at infinity) but will be focused by the eye
into a point. Thus, the focal length of the objective lens and the focal
length of the eyepiece will be equal to the total separation between the
lenses. So,
\[
f_{e}+f_{o}=d
\]
But from the above reasoning and with equation (\ref{eqn:f_naught}), this becomes
\begin{eqnarray*}
f_{e} + mf_{e} & = & d\\
f_{e}\left( 1 + m \right) & = & d\\
f_{e} & = & \frac{d}{\left( 1+m\right) }
\end{eqnarray*}
\end{quote}
15) Two narrow slits are illuminated by a laser with a wavelength of
$538\ nm$. The interference pattern on a a screen located $x = 5.20\ m$ away
shows that the second-order bright fringe is located $y = 9.40\ cm$ away from
the central bright fringe. Calculate the distance between the two slits.
\begin{quote}
$\lambda $ = Given\\
$x$ = Given\\
$y$ = Given\\
$m$ = Given\\
Since this fringe is located a distance $x$ away from the slit and a
distance $y$ from the center, a triangle is formed. The angle that the ray
producing this fringe makes with the central ray is then
\[
\theta =\tan ^{-1}\left( \frac{y}{x}\right)
\]
From the book, the additional distance the second ray must travel before
hitting the same point on the screen is $d\sin \theta $. When this
distance is equal to an integer number of wavelengths, constructive
interference results in a bright ``fringe.'' The order of the fringe is
the number of additional wavelengths the second ray travelled from the
first. Zero order is the center fringe, first order is the next, etc.
Thus,
\begin{eqnarray*}
d\sin \theta & = & m\lambda\\
d & = & \frac{m\lambda }{\sin \theta }
\end{eqnarray*}
\end{quote}
\end{document}