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, d_{o} is the distance
from the object to the lens, and is counted positive for real objects (which is
all we shall ever consider). Similarly, d_{i} 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 d_{o} and d_{i} 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 d_{i}
and d_{o} 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 h_{o} and h_{i} 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 d_{i} or d_{o}
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" (d_{i}
=
¥)
and eq. (1) implies that d_{o} =
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 (d_{o}
=
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 f_{A},
the focal length of lens A. Is
this consistent with the measure of f_{A}
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 f_{B}.
Is this value consistent with the measure of f_{B} 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 (f_{red})
is a little different from that for blue light (f_{blue}).

**Fig. 3.** Using
a collimated beam to measure f_{B}

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
f_{red} and f_{blue}
for lens A. Compare these
values and discuss which color of light must be traveling slower.

PRE-LAB:

- 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.
- 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?