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modified: modernOpticsProblemSet1.tex

new file:   modernOpticsProblemSet1Fig1aTake2.png
new file:   modernOpticsProblemSet1Fig1f.png
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451 notes/blogit/modernOpticsProblemSet1.tex
 @@ -28,137 +28,220 @@ \makeanswer{modernOptics:problemSet1:1}{ \begin{enumerate} -\item[(a)] My interpretation of this problem is that are given the transfer matrix from class - -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1030} -M = -\begin{bmatrix} -1 & 0 \\ --1/f & 1 -\end{bmatrix}, -\end{dmath} +\item[(a)] -describing the input/output response of a thin lens in the paraxial approximation. +% FIXME: uncomment for final notes compilation. The blundering way has some value I think. +%My interpretation of this problem is that are given the transfer matrix from class +% +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1030} +%M = +%\begin{bmatrix} +%1 & 0 \\ +%-1/f & 1 +%\end{bmatrix}, +%\end{dmath} +% +%describing the input/output response of a thin lens in the paraxial approximation. +% +%We wish to relate this to the geometries of the lens system illustrated in figure (\ref{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1a}) that we expect this describes. +% +%\imageFigure{modernOpticsProblemSet1Fig1a}{Thin paraxial lens with image in input and output conjugate planes}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1a}{0.3} +% +%At the point $(0,a)$ with a horizontal input ray, our transfer matrix gives us +% +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1050} +%\begin{bmatrix} +%a \\ +%-\beta +%\end{bmatrix} +%= +%\begin{bmatrix} +%1 & 0 \\ +%-1/f & 1 +%\end{bmatrix} +%\begin{bmatrix} +%a \\ +%0 +%\end{bmatrix}. +%\end{dmath} +% +%The angular portion is +% +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1070} +%-\beta = -\frac{a}{f} + 0, +%\end{dmath} +% +%or +% +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1090} +%\beta = \frac{a}{f}. +%\end{dmath} +% +%Similarily, when the output is horizontal at the point $(0, -b)$ we have +% +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1110} +%\begin{bmatrix} +%-b \\ +%0 +%\end{bmatrix} +%= +%\begin{bmatrix} +%1 & 0 \\ +%-1/f & 1 +%\end{bmatrix} +%\begin{bmatrix} +%-b \\ +%-\theta +%\end{bmatrix}. +%\end{dmath} +% +%The angular component of this product is +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1130} +%0 = \frac{b}{f} - \theta +%\end{dmath} +% +%or +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1150} +%\theta = \frac{b}{f}. +%\end{dmath} +% +%Because this describes a paraxial system, these angles approximate the tangents, so we can recast \ref{eqn:modernOpticsProblemSet1:1090} and \ref{eqn:modernOpticsProblemSet1:1150} as +% +%\begin{subequations} +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1170} +%\tan \beta = \frac{a}{f} +%\end{dmath} +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1190} +%\tan \theta = \frac{b}{f}. +%\end{dmath} +%\end{subequations} +% +%Looking back to the figure, we see that we have $b/n = \tan \theta$, so by \ref{eqn:modernOpticsProblemSet1:1190} $n = f$, the focal length. Similarily $a/o = \tan \beta$, so by \ref{eqn:modernOpticsProblemSet1:1170}, we have $o = f$, also the focal length. The geometry of the figure associated with image formation is now fully determined by the transfer matrix, and we are free to extract some additional relations. In particular note that we have +% +%\begin{subequations} +%\label{eqn:modernOpticsProblemSet1:1210a} +%\label{eqn:modernOpticsProblemSet1:1210} +%\tan \theta = \frac{a}{x} = \frac{b}{f} = \frac{a + b}{s} +% +%\label{eqn:modernOpticsProblemSet1:1230} +%\tan \beta = \frac{b}{x'} = \frac{a}{f} = \frac{a + b}{s'} +% +%\end{subequations} +% +%We have respectively +% +%\begin{subequations} +%\label{eqn:modernOpticsProblemSet1:1250a} +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1250} +%\frac{b}{a} = \frac{f}{x} +%\end{dmath} +%\begin{dmath}\label{eqn:modernOpticsProblemSet1:1270} +%\frac{b}{a} = \frac{x'}{f} +%\end{dmath} +%\end{subequations} +% +%but we've seen that $x = s - f$ and $x' = s' - f$, so we have +% +%\label{eqn:modernOpticsProblemSet1:1290} +%\frac{b}{a} = \frac{f}{s - f} = \frac{s' -f}{f}. +% +% +%Rearranging we have +% +%\label{eqn:modernOpticsProblemSet1:1310} +%f^2 = (s' - f)(s - f) = s s' - f s - s' f + f^2, +% +% +%or +% +%\label{eqn:modernOpticsProblemSet1:1330} +%s s' = f s + f s'. +% +% +%Dividing through by $f s s'$ we have +% +%\label{eqn:modernOpticsProblemSet1:1350} +%\inv{f} = \inv{s'} + \inv{s}, +% +% +%as expected. -We wish to relate this to the geometries of the lens system illustrated in figure (\ref{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1a}) that we expect this describes. +Our system and the associated transfer matrices labels is illustrated in figure (\ref{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1aTake2}). -\imageFigure{modernOpticsProblemSet1Fig1a}{Thin paraxial lens with image in input and output conjugate planes}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1a}{0.3} +\imageFigure{modernOpticsProblemSet1Fig1aTake2}{Input and output conjugate planes for paraxial thin lens}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1aTake2}{0.4} -At the point $(0,a)$ with a horizontal input ray, our transfer matrix gives us +The system tranfer matrix, a composition of a free propagation matrix, then a thin lens paraxial matrix, and one more free propagation matrix is -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1050} +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1790} +M += M_3 M_2 M_1 += \begin{bmatrix} -a \\ --\beta +1 & s' \\ +0 1 \end{bmatrix} -= \begin{bmatrix} 1 & 0 \\ -1/f & 1 \end{bmatrix} \begin{bmatrix} -a \\ -0 -\end{bmatrix}. -\end{dmath} - -The angular portion is - -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1070} --\beta = -\frac{a}{f} + 0, -\end{dmath} - -or - -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1090} -\beta = \frac{a}{f}. -\end{dmath} - -Similarily, when the output is horizontal at the point $(0, -b)$ we have - -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1110} -\begin{bmatrix} --b \\ -0 +1 & s \\ +0 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ +1 - s'/f & s' \\ -1/f & 1 \end{bmatrix} \begin{bmatrix} --b \\ --\theta +1 & s \\ +0 1 +\end{bmatrix} += +\begin{bmatrix} +1 - s'/f & s + s' - s s'/f \\ +-1/f & -s/f + 1 \end{bmatrix}. \end{dmath} -The angular component of this product is -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1130} -0 = \frac{b}{f} - \theta -\end{dmath} - -or -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1150} -\theta = \frac{b}{f}. -\end{dmath} - -Because this describes a paraxial system, these angles approximate the tangents, so we can recast \ref{eqn:modernOpticsProblemSet1:1090} and \ref{eqn:modernOpticsProblemSet1:1150} as - -\begin{subequations} -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1170} -\tan \beta = \frac{a}{f} -\end{dmath} -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1190} -\tan \theta = \frac{b}{f}. -\end{dmath} -\end{subequations} - -Looking back to the figure, we see that we have $b/n = \tan \theta$, so by \ref{eqn:modernOpticsProblemSet1:1190} $n = f$, the focal length. Similarily $a/o = \tan \beta$, so by \ref{eqn:modernOpticsProblemSet1:1170}, we have $o = f$, also the focal length. The geometry of the figure associated with image formation is now fully determined by the transfer matrix, and we are free to extract some additional relations. In particular note that we have - -\begin{subequations} -\label{eqn:modernOpticsProblemSet1:1210a} -\begin{equation}\label{eqn:modernOpticsProblemSet1:1210} -\tan \theta = \frac{a}{x} = \frac{b}{f} = \frac{a + b}{s} -\end{equation} -\begin{equation}\label{eqn:modernOpticsProblemSet1:1230} -\tan \beta = \frac{b}{x'} = \frac{a}{f} = \frac{a + b}{s'} -\end{equation} -\end{subequations} - -We have respectively +Consider ray $(B)$ from the figure, where we have -\begin{subequations} -\label{eqn:modernOpticsProblemSet1:1250a} -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1250} -\frac{b}{a} = \frac{f}{x} -\end{dmath} -\begin{dmath}\label{eqn:modernOpticsProblemSet1:1270} -\frac{b}{a} = \frac{x'}{f} +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1810} +\begin{bmatrix} +0 \\ +\alpha +\end{bmatrix} +\rightarrow +\alpha +\begin{bmatrix} +s + s' - s s'/f \\ +-s/f + 1 +\end{bmatrix} += +\begin{bmatrix} +0 \\ +\alpha' +\end{bmatrix}. \end{dmath} -\end{subequations} -but we've seen that $x = s - f$ and $x' = s' - f$, so we have +With -\begin{equation}\label{eqn:modernOpticsProblemSet1:1290} -\frac{b}{a} = \frac{f}{s - f} = \frac{s' -f}{f}. +\begin{equation}\label{eqn:modernOpticsProblemSet1:1830} +y = y' = \alpha ( s + s' - s s'/f ), \end{equation} -Rearranging we have - -\begin{equation}\label{eqn:modernOpticsProblemSet1:1310} -f^2 = (s' - f)(s - f) = s s' - f s - s' f + f^2, -\end{equation} - -or +we must have for all $\alpha$ -\begin{equation}\label{eqn:modernOpticsProblemSet1:1330} -s s' = f s + f s'. +\begin{equation}\label{eqn:modernOpticsProblemSet1:1850} +s + s' = \frac{s s'}{f}. \end{equation} -Dividing through by $f s s'$ we have +Dividing through by $s s'$ we have -\begin{equation}\label{eqn:modernOpticsProblemSet1:1350} -\inv{f} = \inv{s'} + \inv{s}, +\begin{equation}\label{eqn:modernOpticsProblemSet1:1870} +\boxed{ +\inv{s'} + \inv{s} = \inv{f}, +} \end{equation} as expected. @@ -214,7 +297,7 @@ Let's consider the system as the compound action of three transfer matrices as illustrated in figure (\ref{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig2b1}). -\imageFigure{modernOpticsProblemSet1Fig2b1}{2b1: FIXME: CAPTION}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig2b1}{0.2} +\imageFigure{modernOpticsProblemSet1Fig2b1}{Newton's form, an image with magnification}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig2b1}{0.4} Compounding the transfer matrices we have @@ -305,17 +388,81 @@ \item[(c)] -Here we refer to figure +Here we refer to figure (\ref{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1c}). +\imageFigure{modernOpticsProblemSet1Fig1c}{Position distribution at the focus of a lens}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1c}{0.3} + +Our system transfer matrix is + +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1610} +M += M_3 M_2 M_1 += +\begin{bmatrix} +1 & f \\ +0 & 1 +\end{bmatrix} +\begin{bmatrix} +1 & 0 \\ +-1/f & 1 +\end{bmatrix} +\begin{bmatrix} +1 & x + f \\ +0 & 1 +\end{bmatrix} += +\begin{bmatrix} +0 & f \\ +-1/f & 1 +\end{bmatrix} +\begin{bmatrix} +1 & x + f \\ +0 & 1 +\end{bmatrix} += +\begin{bmatrix} +0 & f \\ +-1/f & -x/f +\end{bmatrix}. +\end{dmath} + +A ray is transformed according to +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1630} +\begin{bmatrix} +y \\ +\theta +\end{bmatrix} +\rightarrow +\begin{bmatrix} +0 & f \\ +-1/f & -x/f +\end{bmatrix} +\begin{bmatrix} +y \\ +\theta +\end{bmatrix} += +\begin{bmatrix} +f \theta \\ +-\inv{f} ( y - x \theta ) +\end{bmatrix}. +\end{dmath} + +In particular +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1650} +y' = f \theta, +\end{dmath} + +demonstrating the claim that at the focus, the position is an angular distribution of the incident beam. This is clearly independent of $x$ so the input plane position is irrelavant. \item[(d)] Consider figure (\ref{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1d}). -\imageFigure{modernOpticsProblemSet1Fig1d}{1d: FIXME: CAPTION}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1d}{0.2} +\imageFigure{modernOpticsProblemSet1Fig1d}{Two identical lenses separated by twice focus}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1d}{0.2} The transfer matrix $M = M_5 M_4 M_3 M_2 M_1$ for the system is -\begin{dmath}\label{eqn:modernOpticsProblemSet1:n} +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1670} M = M_5 M_4 M_3 M_2 M_1 = @@ -370,7 +517,7 @@ Consider any ray from the source going towards the lens along the horizontal. We have -\begin{dmath}\label{eqn:modernOpticsProblemSet1:n} +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1690} \begin{bmatrix} y \\ 0 @@ -384,7 +531,7 @@ The ratio of the output to the input height to be -\begin{dmath}\label{eqn:modernOpticsProblemSet1:n} +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1710} \frac{y'}{y} = -1, \end{dmath} @@ -395,6 +542,98 @@ This is actually demonstrated above. \item[(f)] + +Here we consider figure (\ref{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1f}). + +\imageFigure{modernOpticsProblemSet1Fig1f}{Cat's eye. Lens with mirror behind at focus}{fig:modernOpticsProblemSet1:modernOpticsProblemSet1Fig1f}{0.4} + +Our system transfer matrix is + +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1730} +M = +M_7 +M_6 +M_5 +M_4 +M_3 +M_2 +M_1 += +\begin{bmatrix} +1 & s' \\ +0 & 1 +\end{bmatrix} +\begin{bmatrix} +1 & 0 \\ +-1/f & 1 +\end{bmatrix} +\begin{bmatrix} +1 & f \\ +0 & 1 +\end{bmatrix} +\begin{bmatrix} +1 & 0 \\ +0 & 1 +\end{bmatrix} +\begin{bmatrix} +1 & f \\ +0 & 1 +\end{bmatrix} +\begin{bmatrix} +1 & 0 \\ +-1/f & 1 +\end{bmatrix} +\begin{bmatrix} +1 & s \\ +0 & 1 +\end{bmatrix} += +\begin{bmatrix} +1 - s'/f & s' \\ +-1/f & 1 +\end{bmatrix} +\begin{bmatrix} +1 & 2 f \\ +0 & 1 +\end{bmatrix} +\begin{bmatrix} +1 & s \\ +-1/f & -s/f + 1 +\end{bmatrix} += +\begin{bmatrix} +1 - s'/f & 2f - s' \\ +-1/f & -1 +\end{bmatrix} +\begin{bmatrix} +1 & s \\ +-1/f & -s/f + 1 +\end{bmatrix} += +\begin{bmatrix} +-1 & 2 f - s' -s \\ +0 & -1 +\end{bmatrix} +\end{dmath} + +We see that the angle of the output light is unchanged except for sign, so we have no scattering in the paraxial limit. Observe that if the emitter is positioned at $s = f$ we have + +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1750} +M = +\begin{bmatrix} +-1 & f - s' \\ +0 & -1 +\end{bmatrix} +\end{dmath} + +so + +\begin{dmath}\label{eqn:modernOpticsProblemSet1:1770} +y' = -y + (f -s') \alpha. +\end{dmath} + +If the observation is also made at the focus, then the image magnification is unity. The image is magnified (negatively) for any position $\Abs{s'} > f$ without any angular distortion (other than the uniform sign change). + \end{enumerate} }
BIN  notes/blogit/modernOpticsProblemSet1Fig1aTake2.png