145 lines
5.4 KiB
TeX
145 lines
5.4 KiB
TeX
\section{Physical Principals}
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The following physical basics are necessary for understanding the experiment:
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\subsection{Lenses}
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Light is defined as an electromagnetic wave. But for the purposes of this experiment it is to be interpreted as a ray
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that is being emitted by a light source, which can be broken or reflected. When parallel rays of light hit a convex lens they get broken
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in such a way that they gathers at the focal point F of said lense (see $\ref{A}$). G represents the height of hte object which is being
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viewed through the lense, g is the distance between the object and the lense. B is the image of the object and b similarly the distance between
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the lense and the image.
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\label{A}
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\begin{figure}[h!]
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\centering
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\includegraphics[angle=270,scale=0.5]{Beam path of a concex lense.jpg}
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\caption{beam path of a convex lense}
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\end{figure}
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The relationship between B and G or b and g is described in $\ref{1}$
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\label{1}
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\begin{align*}
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\frac{B}{G} & = \frac{b}{g} \tag{1}
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\end{align*}
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\subsection{Magnification}
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b is the width of the eye and g is the distance between the object and the eye, which should be
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\begin{equation}
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a_{0} = 25 cm
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\end{equation}
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for maximum resolution. Magnification means that the image which reaches the back of the eye is enlarged through the use
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of an optical intrument. The proportion of B and G can be calcuated thus:
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\label{2}
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\begin{align*}
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\gamma & = \frac{tan(\alpha_{2})}{tan(\alpha_{1})} \tag{1}
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\end{align*}
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\label{B}
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\begin{figure}[h!]
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\centering
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\includegraphics[angle=270,scale=0.5]{Effects of a lense on the eye (1).jpg}
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\caption{beam path of a ray of light through eye with (right) and without (left) lense}
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\end{figure}
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A magnifying glass relies on the principle that when the object is placed between the focal point F and the convex lense a virtual
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image is created, which is further back and larger than the object. Since the image is virtual it cannot be projected onto a screen,
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but it has the effect of an enlarged real image being presented to the observer.
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The magnification, based on the equation $\ref{2}$, can be calculated thus:
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\label{3}
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\begin{align*}
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\gamma_{magnifying} & = \frac{tan(\alpha-{2})}{tan(\alpha_{1})} \\
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& = \frac{G/f}{G/a_{0}} \\
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& = \frac{a_{0}}{f} \tag{3}
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\end{align*}
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\subsection{Microscope}
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\label{C}
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\begin{figure}[h!]
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\centering
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\includegraphics[angle=270,scale=0.5]{Beam path microscope.jpg}
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\caption{beam path of ray of light through a microscope}
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\end{figure}
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A microscope constitutes of two convex lenses: an objecte and the ocular. The obeject is placed between the focal length and twice the
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focal length of the objective. This creates a real image between the objective and the ocular, which should be in betweem the ocular and
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its focal point $F_{ok}$. The real image $B_{Z}$ is then magnified following the same principle as that of the magnifying glass, which to
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an enlarged image being presented to the viewer throgh the ocular.
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In this case, the magnification of two lenses is calculated using the equation:
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\label{4}
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\begin{equation}
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\gamma_{mik} = \gamma_{ob} \cdot \gamma_{ok} \tag{4}
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\end{equation}
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The theoretical magnification of the obejective can be calculated using the tube length t and the distance between the ocular and the viewers
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eye ($a_{0}$).
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\label{5}
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\begin{equation}
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\gamma_{ob} = \frac{t}{f_{ob}} \tag{5}
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\end{equation}
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Based on the equation $\ref{3}$ the calculation of the magnification of the ocular is apparent:
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\label{3.1}
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\begin{align*}
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\gamma_{ok} &= \frac{a_{0}}{g} \\
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&= \frac{a_0}{f_{ok}} \tag{3.1}
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\end{align*}
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For $\gamma_{ok}$ at the distance of $a_{0}$ ($\gamma_{ok}(a_{0})$) the equation $\ref{6}$ has to be taken into account:
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\label{6}
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\begin{align*}
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\frac{1}{f} &= \frac{1}{g} + \frac{1}{b} \tag{6} \\
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\frac{1}{g} &= \frac{1}{f} + \frac{1}{a_{0}}
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\end{align*}
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The final formula for $\gamma_{ok}$ is then as follows:
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\label{3.2}
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\begin{equation}
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\gamma_{ok}(a_{0}) = [\frac{a_0}{f_{ok}} + 1]\tag{3.2}
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\end{equation}
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\subsection{Resolution}
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The resolution of an optical is defined as the minimum distance of two points that can still be observed as separate.
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If the distance is smaller than this, diffraction takes place and the wo points interfere with each other. Light can be
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discribed as radial waves that add together to become wave front. If an obstacle is small enough, the radial waves of
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a wave front rearrange into a new wave front, which goes around the obstacle. If a grid, whose openings are about the same
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size as the wavelenght of the wave, were to be used as the obstacle, the ampltiudes of the waves would either add up
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(constructive interference) or cancel eachother out (destructive interference). If the two point were to interfere with each
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other, then an interference pattern would be visible, rather than a clear image.
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The angle, at which the light reaches the eye $\epsilon$ can be calculated using the equation:
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\label{7}
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\begin{equation}
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tan(\epsilon) = [\frac{B/2}{f_{ok}} + 1]\tag{7}
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\end{equation}
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B being the measured resolution limit (the difference between the smallest pinhole-size at which the grid is still visible and
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the next).
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The minimum distance $d_{min}$ is then calculated thus, A being the numerical aperture:
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\label{8}
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\begin{align*}
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d_{min} &= \frac{\lambda}{A} + \frac{1}{b} \tag{8} \\
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A &= n \cdot sin(\epsilon) \tag{8.1}
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\end{align*}
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n is the refraction index and $\lambda$ refers to the wavelength of the light being used.
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\newpage
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