Coupon Accepted Successfully!


Optics Using Lenses

Figure 13-13 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.

..\art 13 jpg\figure 13-tm.jpg

Figure 13-13

A diverging lens has the property that parallel light rays incident from the left spread apart after going through the lens, as if they were coming from a point source a distance –f from the lens. Figure 13-14 shows such a lens. For diverging lenses, the focal length is negative.

..\art 13 jpg\figure 13-tn.jpg

Figure 13-14

For lens problems, there are two formulas and a ray-tracing method. It is probably worth your while to learn both. The formulas are better for calculating numbers, but ray tracing is better at answering qualitative questions. The method of ray tracing will sound confusing at first, but it will become clearer as you work through the examples.
If the distance from the lens to the object is d0 the distance from the lens to the image is di and the focal length of the lens is f. then we have
is the magnification of the image. If d, is greater than zero, this indicates that the image is on the other side of the lens from the object.

Ray-tracing method for a converging lens

  1. Draw the lens, the object being observed, and both focuses.
  2. Draw a ray parallel to the principal axis and passing through the lens. Bend the ray to go through the opposite focus.
  3. 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.



Ray-tracing method for a diverging lens

  1. Draw the lens, object, and focus on the same side as the object.
  2. 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.
  3. Draw a ray going through the vertex (center) of the lens and passing straight through.
  4. The intersection of the extended ray in step 2 and the ray step 3 gives the location of the image.
A boy uses a magnifying glass (converging lens with focal length 0.03 m) to observe a bug which is 0.02 m from the lens.
  1. Draw a ray diagram.
  2. What is the magnification of the image?

Figure 13-15 shows the ray diagram.


..\art 13 jpg\figure 13-to.jpg

Figure 13-15


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:




The negative sign for di means the image is on the same side as the object. The positive sign for m means the image is upright (not inverted).

The eye contains a lens whose focal length can be adjusted. A candle (2 cm long) sits 0.1 m from the lens of the eye, and the image is focused on the retina. Assume the length of the eye from front to back is 2.5 cm.
  1. What is the focal length of the lens?
  2. Is the image upright or inverted?

We calculate the focus as follows:



Figure 13-16 shows the appropriate diagram.


..\art 13 jpg\figure 13-tp.jpg

Figure 13-16


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 13-16 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 13-17 for a more realistic diagram.


..\art 13 jpg\figure 13-tq.jpg

Figure 13-17

The diverging lens on a pair of glasses has a focal length of 3.0 m. A candle is 9.0 cm tall and 6.0 m away.
  1. Where is the image of the candle when viewed through the lens?
  2. What is the size of the image?
  3. Is the image inverted or upright?
  4. 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 13-18 shows the ray diagram.


..\art 13 jpg\figure 13-tr.jpg

Figure 13-18


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.

Test Your Skills Now!
Take a Quiz now
Reviewer Name