ch02/figs/hw-near-focal-point-photo.jpg ch02/figs/hw-near-focal-point.svg @@ -1778,5 +1778,27 @@ the maximum occurs close to, but not exactly at, $f_\zu{o}$. %%graph%% resonance func=1/sqrt(3*(x**2-1)**2+x**2) xlo=0 xhi=3 ylo=0 yhi=1.1 with=lines xtic_spacing=1 ytic_spacing=.5 format=eps \smallfig{resonance}{The function of problem \ref{hw:resonance}, with $a=3$, $b=1$, and $f_\zu{o}=1$.}% +\widefig[t]{hw-near-focal-point}{Problem \ref{hw:near-focal-point}. A set of light rays is emitted from +the tip of the glamorous movie star's nose on the film, and reunited to form a spot on the screen which is the image of the +same point on his nose. The distances have been distorted for +clarity. The distance $y$ represents the entire length of the theater from front to back.} +% +\begin{hwwithsoln}{near-focal-point} +In a movie theater, the image on the screen is formed by a lens in the projector, and originates from one of the frames on the strip of celluloid film (or, +in the newer digital projection systems, from a liquid crystal chip). +Let the distance from the film to the lens be $x$, and let the distance from the lens to the screen be $y$. The projectionist +needs to adjust $x$ so that it is properly matched with $y$, or else the image will be out of focus. There is therefore +a fixed relationship between $x$ and $y$, and this relationship is of the form +\begin{equation*} + \frac{1}{x}+\frac{1}{y} = \frac{1}{f} \qquad , +\end{equation*} +where $f$ is a property of the lens, called its focal length. A stronger lens has a shorter focal length. +Since the theater is large, and the projector is relatively small, $x$ is much less than $y$. +We can see from the equation that if $y$ is sufficiently large, the left-hand side of the equation +is dominated by the $1/x$ term, and we have $x \approx f$. Since the $1/y$ term doesn't completely +vanish, we must have $x$ slightly greater than $f$, so that the $1/x$ term is slightly less than +$1/f$. Let $x=f+\der x$, and approximate $\der x$ as being infinitesimally small. +Find a simple expression for $y$ in terms of $f$ and $\der x$. +\end{hwwithsoln} \end{hwsection} ch02/ch02.tex @@ -614,6 +614,20 @@ the square root in the original expression for $s$. Then by the chain rule, We looked for the place where $\der I/\der x$ was zero, but $\der s/\der f$ could also be zero if one of the other factors was zero. This is what happens at $f=0$, where $\der x/\der f=0$. +\hwsolnhdr{near-focal-point} + +\begin{align*} + y &= \left(\frac{1}{f}-\frac{1}{x}\right)^{-1} \\ + &= \left(\frac{1}{f}-\frac{1}{f+\der x}\right)^{-1} \\ + &= f\left(1-\frac{1}{1+\der x/f}\right)^{-1} \\ +\intertext{Applying the geometric series $1/(1+r)=1+r+r^2+\ldots$,} + y &\approx f\left(1-\left(1-\frac{\der x}{f}\right)\right)^{-1} \\ + &= \frac{f^2}{\der x} +\end{align*} + +As checks on our result, we note that the units work out correctly (meters squared divided by +meters give meters), and that the result is indeed large, since we divide by the small quantity $\der x$. + \beginsolutions{6} \hwsolnhdr{sequence-weierstrass} ch99/hwans.tex 4504e8368e3b692854bb9686ea4e6914f4e87df6 Ben Crowell githubcrowell09@lightandmatter.com http://github.com/bcrowell/calculus/commit/9413dc273ec4aa166bb564d7eafc628af7e35310 9413dc273ec4aa166bb564d7eafc628af7e35310 2009-04-19T16:09:41-07:00 2009-04-19T16:09:41-07:00 add move projector example 46168e3a6a6d4044854e0adaf7dd458c981af4d1 Ben Crowell githubcrowell09@lightandmatter.com