\input{labpreamble}
\begin{document}
\subsubsection*{PHSX 101L\hfill Experiment 13\hfill Name:\underline{\hspace*{2in}}}

\underline{Title:} Lenses and Image Formation

\underline{Object:} (1) To measure the focal length of a converging 
lens. (2) To use this lens to form real images and to compare the 
measured position of each image with theory.  (3) Also use the lens to 
observe a virtual image, estimate is magnifying power, and compare 
with theory.

\underline{Theory:} The focal point of a converging lens is defined as 
that point where a beam of incoming parallel light rays will be 
brought to a focus.  For a diverging lens, a beam of parallel light 
rays will be spread out such that it will appear to diverge from a 
focal point located in front of the lens.

Since lenses can bend light rays as stated above, it is therefore
possible to form images using lenses.  The location and nature of an 
image formed by a lens can be predicted either from a ray tracing 
diagram or by means of a lens equation.

To locate the image by diagram, three rays (known as the principal 
rays) are drawn:
\vspace{-.2in}
\begin{enumerate}
\item \underline{A parallel ray} is drawn from the top of the object 
parallel to the lens axis.  When this ray passes through the lens 
it is bent so as to pass through the primary focal point.

\item \underline{A central ray} is drawn from the top of the object toward the 
center of the lens.  It passes through the lens undeviated.

\item \underline{A focal point ray} is drawn from the top of the object through 
the secondary focal point.  When it passes though the lens it will 
emerge traveling parallel to the lens axis.

\end{enumerate}

The point at which these three rays converge gives the location and 
nature of the image.  (The three principal rays are not the only rays 
coming from the object, but they are sufficient to determine the 
location of the image.)


The image location and nature can also be found using the thin-lens 
equation, which relates together the object distance $x_{o}$, the 
image distance $x_{i}$, and the lens focal length $f$:
\begin{equation}\label{thinlens}
\frac{1}{x_{o}} + \frac{1}{x_{i}} = \frac{1}{f}.
\end{equation}

If the image is on the opposite side of the lens from the object,
$x_{i}$ is considered to be positive (real image); if on the same side, then
negative (virtual image).  Also, $f$ is positive for a converging lens and negative
for a diverging lens.

The magnification of the image is calculated from the ratio of the 
image distance to the object distance:
\begin{equation}\label{mag}
M=-\frac{x_{i}}{x_{o}}
\end{equation}
$M$ will be positive if the image is erect and negative if the image is 
inverted.

\newpage The \emph{near point} of the eye is defined to be the closest
distance for which the eye can accommodate to focus light on the
retina.  It increases with age, but averages about 25~cm for
20-year-olds.

When a converging lens is used as a magnifying glass, the lens is held 
next to the eye and the object to be viewed is placed just inside the 
focal point of the lens.  The image will be a virtual image, 
upright, and further away from the lens than the object, and will be 
larger than the object.  The \emph{magnifying power} of such a lens is 
defined as the ratio of the angular size of the object when viewed 
through the lens ({\it i.e.,} the image) compared to the angular size of the object when 
viewed at the eye's near point without the lens.  In the small-angle 
approximation, this is equal to the ratio of the eye's near point to the 
focal length of the lens.
\begin{equation}\label{angmag}
M=\frac{\theta_{\rm image,\ lens}}{\theta_{\rm near\ point,\ no\ lens}} \approx 
\frac{25~\rm cm}{f}
\end{equation}

\underline{Procedure:} 
\vspace{-.25in}
\begin{enumerate}\setlength{\itemsep}{1.5pt}
\item Measure the focal length of your converging lens by placing a 
lighted object on one side of the lens, and a viewing screen on the 
other.  Adjust the positions of the lens and screen until the object 
and the focused image are the same distance from the lens.  Under these 
conditions the focal length will be one-half the object distance (or 
one-half the image distance since they are the same in this case).  
Measure $f$ carefully for your upcoming predictions to be 
good.

\item\label{calculate} Use equation \ref{thinlens} to calculate the
image distance when $x_{o} = 50.0$~cm and when $x_{o} = 25.0$~cm.

\item Position a lighted object on the optical bench at each of the 
distances specified in procedure \ref{calculate}.  Locate the image on a 
viewing screen and measure the image distance in each case.  Compare 
your measured results with the results predicted in procedure 
\ref{calculate}.  

\item Find the percent difference for each case.

\item Use your calculated $x_{i}$ and $x_{o}$ (from procedure
\ref{calculate}) and equation \ref{mag} to calculate the image
magnification when $x_{o} = 50.0$~cm and when $x_{o} = 25.0$~cm.

\item Use a ruler to measure the height of the object and image; 
record the measured magnification.

\item Find the percent difference for each case.

\item Predict (\emph{pre} means before you do the experiment) what
will happen to the image if you use a piece of opaque paper to cover
the bottom half of the lens when $x_{o} = 25.0$~cm, and also if you
cover the lens with an opaque paper with a 2~cm diameter hole cut in
the center.  Record your predictions.  Then do the experiment and
record what really happened.

\item\label{ten} Calculate the image distance when $x_{o} = 10.0$~cm.  What will be 
the nature of this image?

\item Position the lighted object on the optical bench at $x_{o} =
10.0$~cm.  Can you locate the image on a viewing screen?  Explain your
observations.  How did you eventually see this image?

\item Place an object, such as your ruler, at a distance just less 
than the focal length from the lens.  Place your eye close to the 
lens and look through the lens at the object.  Also look at the object 
without the lens at your near point (25 cm if you are young) from 
your eye.  Estimate the size of the image compared to the size of the 
object.  How do your observations compare with the calculated 
magnifying power (given by equation \ref{angmag}) of this lens?

\end{enumerate}

\pagebreak
\underline{Results:} 
\vspace{-.25in}
\begin{enumerate}
\item Focal length of the lens $f$:\hfill \underline{\hspace{1.2in}}
\vspace{.3in}

\hfill$x_{o} = 50.0$~cm\hspace{.7in}$x_{o} = 25.0$~cm\hspace{.2in}
\vspace{.1in}

\item Predicted image distance $x_{i}$:
\hfill \underline{\hspace{1.2in}}\hspace{.5in}\underline{\hspace{1.2in}}
\vspace{.1in}

\item Measured image distance $x_{i}$:
\hfill \underline{\hspace{1.2in}}\hspace{.5in}\underline{\hspace{1.2in}}
\vspace{.1in}

\item Percent difference:
\hfill \underline{\hspace{1.2in}}\hspace{.5in}\underline{\hspace{1.2in}}
\vspace{.1in}

\item Predicted magnification of image: 
\hfill \underline{\hspace{1.2in}}\hspace{.5in}\underline{\hspace{1.2in}}
\vspace{.1in}

\item Measured magnification of image: 
\hfill \underline{\hspace{1.2in}}\hspace{.5in}\underline{\hspace{1.2in}}
\vspace{.1in}

\item Percent difference:
\hfill \underline{\hspace{1.2in}}\hspace{.5in}\underline{\hspace{1.2in}}
\vspace{.1in}

\item Prediction:
\vspace{.4in}

What really happened:
\vspace{.4in}

\item Predicted image distance $x_{i}$ when $x_{o} = 
10.0$ cm:\hfill \underline{\hspace{1.2in}}
\vspace{.1in}

Nature of the image (real or virtual):\hfill \underline{\hspace{1.2in}}
\vspace{.1in}

\item Explain observations and how you located the image:
\vspace{1.in}

\item Estimated size of image compared to size of object:
\hfill\underline{\hspace{1.2in}}
\vspace{.1in}

Calculated magnifying power:\hfill \underline{\hspace{1.2in}}
\vspace{.1in}

\end{enumerate}

\vspace{-.2in}
\underline{Conclusions:}
\vspace{.4in}

\newpage
\underline{Practice:} 
For each of the lenses shown below, use a straight edge to draw the 
three principal rays and locate and label the image.  Measure the 
image distance with a ruler.  Indicate virtual 
or real.

\vspace{.2in}
\begin{picture}(590,200)
\put(0,100){\line(1,0){590}} % horizontal axis
\put(300,0){\line(0,1){200}} % vertical axis
\qbezier(300,0)(304,2)(310,18) % start of curved lens profile
\qbezier(300,0)(296,2)(290,18) % start of curved lens profile
\qbezier(300,200)(304,198)(310,182) % start of curved lens profile
\qbezier(300,200)(296,198)(290,182) % start of curved lens profile
\put(318,1){\shortstack[l]{converging\\lens}}
\put(200,100){\circle*{3}} % left focus
\put(196,85){\shortstack{$f$}}
\put(400,100){\circle*{3}} % right focus
\put(396,85){\shortstack{$f$}}
\put(50,150){\line(1,-5){10}}  % make the triangular object
\put(50,150){\line(-1,-5){10}}
\put(50,85){\makebox(0,0){object}}
\put(5,5){\makebox(0,0)[l]{$x_{i} = 
$\underline{\hspace{.6in}}cm\hspace{.4in}Real or virtual?}}
\end{picture}
\noindent

\vspace{.2in}
\begin{picture}(590,200)
\put(0,100){\line(1,0){590}} % horizontal axis
\put(300,0){\line(0,1){200}} % vertical axis
\qbezier(300,0)(304,2)(310,18) % start of curved lens profile
\qbezier(300,0)(296,2)(290,18) % start of curved lens profile
\qbezier(300,200)(304,198)(310,182) % start of curved lens profile
\qbezier(300,200)(296,198)(290,182) % start of curved lens profile
\put(318,1){\shortstack[l]{converging\\lens}}
\put(150,100){\circle*{3}} % left focus
\put(146,85){\shortstack{$f$}}
\put(450,100){\circle*{3}} % right focus
\put(446,85){\shortstack{$f$}}
\put(210,125){\line(1,-5){5}}  % make the triangular object
\put(210,125){\line(-1,-5){5}}
\put(210,85){\makebox(0,0){object}}
\put(5,5){\makebox(0,0)[l]{$x_{i} = 
$\underline{\hspace{.6in}}cm\hspace{.4in}Real or virtual?}}
\end{picture}
\noindent

\vspace{.2in}
\begin{picture}(590,200)
\put(0,100){\line(1,0){590}} % horizontal axis
\put(300,0){\line(0,1){200}} % vertical axis
\put(280,0){\line(1,0){40}}  % lens edge
\put(280,200){\line(1,0){40}} %lens edge
\qbezier(280,0)(285,2)(286,18) % start of curved lens profile
\qbezier(320,0)(315,2)(314,18) % start of curved lens profile
\qbezier(320,200)(315,198)(314,182) % start of curved lens profile
\qbezier(280,200)(285,198)(286,182) % start of curved lens profile
\put(330,1){\shortstack[l]{diverging\\lens}}
\put(150,100){\circle*{3}} % left focus
\put(146,85){\shortstack{$f$}}
\put(450,100){\circle*{3}} % right focus
\put(446,85){\shortstack{$f$}}
\put(100,150){\line(1,-5){10}}  % make the triangular object
\put(100,150){\line(-1,-5){10}}
\put(100,85){\makebox(0,0){object}}
\put(5,5){\makebox(0,0)[l]{$x_{i} = $\underline{\hspace{.6in}}cm\hspace{.4in}Real or virtual?}}
\end{picture}
\noindent


\end{document}
