Optics Using Lenses
Figure 1313 shows a converging lens. A light ray incident on the lens bends twice, once when entering the lens (usually glass) and once when leaving it, both times bending toward the axis. You should work this out by tracing the rays in an exaggerated diagram. The lens is designed so that parallel light rays on the left converge to a point on the right. The distance from the lens to the point of convergence is the focal length f. Light rays from the right will also focus to a point after a distance f.
Figure 1313
Figure 1314
If the distance from the lens to the object is d_{0} the distance from the lens to the image is d_{i} and the focal length of the lens is f. then we have

Raytracing method for a converging lens
 Draw the lens, the object being observed, and both focuses.
 Draw a ray parallel to the principal axis and passing through the lens. Bend the ray to go through the opposite focus.
 Draw a ray passing through the object and the focus on the same side of the lens. This ray becomes parallel to the principal axis when it passes through the lens.
The intersecting point is the location of the image. If the rays do not intersect on the side opposite the object, extend them backwards until they do.
Raytracing method for a diverging lens
 Draw the lens, object, and focus on the same side as the object.
 Draw a ray parallel to the principal axis. After it passes through the lens it bends up. as if it came from the focus. Extend the ray backwards.
 Draw a ray going through the vertex (center) of the lens and passing straight through.
 The intersection of the extended ray in step 2 and the ray step 3 gives the location of the image.
 Draw a ray diagram.
 What is the magnification of the image?
Figure 1315 shows the ray diagram.
Figure 1315
In this example, we had to extend the rays backwards in step 4 in order to find the image. For this reason the image is virtual, meaning light rays are not actually coming from the position from which they seem to come.
If the light rays pass through the point from which they seem to come, then the image is said to be real. We can get the magnification by first calculating the exact position of the image:
Then
The negative sign for d_{i }means the image is on the same side as the object. The positive sign for m means the image is upright (not inverted).
 What is the focal length of the lens?
 Is the image upright or inverted?
We calculate the focus as follows:
Figure 1316 shows the appropriate diagram.
Figure 1316
We draw the horizontal line on the left of the lens in step 2. The other line, you will notice, takes the near focus into account. The image is where the lines meet, inverted and very small.
Because the image is located where light rays actually converge, the image is real. Figure 1316 is a physics diagram. In an actual eye, there are many rays which converge to a point, and none of them need to be the two which we have drawn here. See Figure 1317 for a more realistic diagram.
Figure 1317
 Where is the image of the candle when viewed through the lens?
 What is the size of the image?
 Is the image inverted or upright?
 Is the image real or virtual?
(You should try this solution yourself before you read about it. Then work it out with the book.) Figure 1318 shows the ray diagram.
Figure 1318
We had to extend a ray backwards in step 2. That means the image is virtual. Also the image is upright. The location of the image can be gotten from the equation:
where the negative result indicates the image is behind the lens, which we knew from the diagram. The magnification is given by
where the positive result indicates the image is upright. That gives the magnification. The size is then (0.333) (9.0 cm) = 3.0 cm.
It is a general rule that if only one lens or mirror is involved in a problem, then the image is either both real and inverted or both virtual and upright. It will never be real and upright, for example.