Thin lenses

An 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 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.

Consider first a converging lense, $f > 0$. If the object is moved far to the left, the incoming rays are nearly parallel. In this case, $1/x_o$ negligible in the thin lens equation, $-\frac{1}{x_o}+\frac{1}{x_i}=\frac{1}{f}$, 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 parallel and $x_i=\infty$. If , 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.