Experiment 3     



Lenses are the basis of many modern optical instruments, such as microscopes, telescopes, cameras, and eyeglasses. In this experiment, you will work with two thin lenses. You will measure their focal lengths in several ways and study the magnification and chromatic aberration produced by one of them. Finally, you will use both of them to make a simple astronomical telescope.

The main properties of any thin lens are summarized by the thin lens equation:


In this equation, f is the focal length of the lens and is positive for converging lenses (which you will be using in this experiment) but negative for diverging lenses. As shown in Fig. 1, do is the distance from the object to the lens, and is counted positive for real objects (which is all we shall ever consider). Similarly, di is the distance from the image to the lens and is counted positive for real images (as in Fig. 1) but negative for virtual images. (A virtual image is created when the rays of light from a point on the object are still diverging once they pass through the lens.  These exit rays will appear as if they are coming from a point behind the lens.)

Fig.1. The distances do and di that appear in the lens equation, illustrated for the case of a real image produced by a converging lens. The two points labeled F are the focal points of the lens.

The first part of this lab is to find the focal lengths of lenses A and B by measuring values of di and do and using eq. (1) to calculate f.  For each lens, make measurements of do, ho, di, and hi at 3 different distances between the object and the lens.  Here, the image is real and can be located on a frosted glass screen.  Using the lens equation above, find f for each lens.   As you bring the object closer and closer to the lens, at what distance from the lens is it no longer possible to form an focused image on your screen.  Why are the rays of light no longer focusing to an image?  If you wanted to keep the same object-to-lens distance, you could swap the lens out for a new one.  What would be different about the new lens and why would you now be able to create a real image?

Use your measurements of the sizes of ho and hi of the object and image and compute the magnification of the object for each measurement you made:




Referring to Fig. 1 and using the fact that with similar triangles (all the angles the same) the ratio of the sides are also the same (e.g., di/hi=do/ho), you can see that M is also equal to


Using your measurements, check the agreement between these two expressions for M.  


The second part of this experiment is to measure f for all three lenses by arranging that either di or do is infinite. First, using a point source of light, adjust the position of lens A so that it produces a parallel, or collimated, beam of light as in Fig. 2. In this case the image is at "infinity" (di = ¥) and eq. (1) implies that do = f;  

Fig. 2. When a converging lens produces a parallel beam of light we say that the light is collimated. This happens when the source is at the lens's focal point (do = f). You can check that the light is collimated by seeing if the diameter D of the beam at points far from the lens is equal to the diameter of the lens itself.

That is, the distance from source to lens should now equal fA, the focal length of lens A.  Is this consistent with the measure of fA from part 1?

Now place lens B in the collimated beam produced by lens A as in Fig. 3. Since the light rays approaches B traveling parallel, the object distance for lens B is infinite and the image distance should now be the focal length of the lens in question. In this way, you can get another value for fB.  Is this value consistent with the measure of fB from part 1?

Still using the point source of light, you can study the chromatic aberration of lens A. Chromatic aberration (slight differences in the lens focusing properties for different colors or frequencies of light) occurs in all but the most expensive lenses.  Basically what is happening is that the glass bends light of different wavelengths by slightly different amounts which is equivalent to saying the light of different frequencies travels at slightly different speeds through glass (remember the concrete/sand interface… it’s like saying different frequencies of light are slowed down more by the sand than other frequencies and thus bend more).  This phenomenon is put to good use in a prism when one wants to disperse the different colors of white light.  It is also the cause for a rainbow.  For a given lens, therefore, the focal length for red light (fred) is a little different from that for blue light (fblue).

Fig. 3. Using a collimated beam to measure fB

When one uses white light (as we do in this experiment) there is no single position at which all colors are brought to a sharp focus. So far in this experiment, you have been using an image which is a compromise among the slightly different positions for different colors. As the last part of the experiment, try to locate the best position (the most in focus) for the red end of the spectrum and then for the blue end, so to measure rough values for fred and fblue for lens A.  Compare these values and discuss which color of light must be traveling slower.




  1. I have a candle that is 60 cm away from a lens.  I find that the lens creates a real image of the candle 20 cm away from the lens.  Use the lens equation to determine the focal length of the lens.
  2. I have a point source of light and a lens, and I am trying to create a collimated beam of light exiting the lens (as in Figure 2) (that is all rays parallel, not converging or diverging).  I use a piece of paper to look at the light exiting the lens (so I move the piece of paper from just behind the lens to far from the lens and look at how the size of the light beam changes).  If I find that the circle of light is continually growing bigger as it exits the lens, should I move the point source further away from or closer to the lens in order to get a collimated beam?