 | Physik M 513.805 (511.015) |
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| Physik M 513.805 (511.015) |
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Thin lensesAn object $o$ is a distance $x_o$ in front of a thin lens at $x=0$. The object is $y_o$ from the optical axis. A lens is thin when it is in much thinner than the focal length $f$. The image $i$ is found at a distance $x_i$ on the other side of the lens. The relationships between the object position $(x_o,y_o)$ and the image position $(x_i,y_i)$ are, \[ \begin{equation} x_i=\frac{fx_o}{f+x_o}\qquad \text{and} \qquad y_i=\frac{fy_o}{f+x_o}. \end{equation} \]Often the equation for $x_i$ is rewritten as, \[ \begin{equation} -\frac{1}{x_o}+\frac{1}{x_i}= \frac{1}{f}. \end{equation} \]This is called the thin lens equation. The magnification is, \[ \begin{equation} \frac{y_i}{y_o}=\frac{f}{f+x_o}. \end{equation} \]In the drawing, light rays leave the object at $x_o$. These rays are bent by the lens. The image $x_i$ is the point where the bent rays come together. The blue dots at at the focal length of the lens. $D$ is the optical power of the lens measured in dioptre, $D=1/f$ where $f$ is measured in meters. If the object is moved far to the left, the incoming rays are nearly parallel. In this case, $1/x_o$ negligible in the equation and $x_i \approx f$. Then move the object to the right and notice that the image moves to the right. When $-x_o=f$ the lens bends the outgoing rays so that they are paralleland $x_i=\infty$. If the object is closer to the lens than $f$, the lens bends the rays so that they diverge on the right. The diverging rays on the right meet at a point that is on the left of the lens. This is a virtual image since the light rays do not pass through the image point. The focal length is positive for a converging lens and is negative for a diverging lens. Try both signs for the focal length. |