The realm of geometric optics is one of the most basic sub-disciplines of physics with the most general applications - from magnifying glasses, to binoculars to microscopes and telescopes as well as a host of other instruments. Their design begins with the use of thin lenses - both convex and concave- to effect diverging and converging rays.

The
theoretical results for ray tracing in a bi-convex lens are depicted in the graphic below, with the
Object O positioned on the *left side of the lens*, which has a radius of
curvature R2, and center of curvature C2, and an image I on the right side of
the lens.

The right side of the lens has a radius of curvature R1 and
center of curvature C1. The object distance is then s, and the image distance
is s'. These can then be used to form the thin lens formula and the
"*lens makers equation*".

1/s + 1/ s' = 1/f

The lens maker's formula is:

1/f
= (n - 1) [1/R1 - 1/R2]

Where n is the refractive index.

It is the practical combination of such thin lenses which yields optical instruments, such as the telescope. Such a lens combination was the basis for the simple refracting telescope designed by Barbadian physics student Carson King:

This 2-inch (50mm) aperture instrument was also used to take photos of the Moon and no surprise it won first prize at the 1978 Barbados Science Exhibition. The typical thin lens combo
is shown in the illustrated example, asking '*Where is the final image?'*

The key step is to use the known focal lengths (f1 = 10 cm and f2 = 20 cm) and then perform the working as shown.

The trick is to find the image distance s1' for the first lens, then having done that find the object distance s2 of the second lens. One can also obtain the total magnification (lateral) using the magnification formula. We look at the procedure then complete the solution for the converging lens system shown.*Procedure for analyzing a thin (convex) lens combination*:

1) The image of the first lens (L1) is calculated as if the 2nd lens (L2) is not present.

2) The image of the first lens is treated as the object of the 2nd lens.

3) If the image of the first lens lies to the right of the 2nd lens, the image is treated as a virtual object for the 2nd lens (that is, s is negative). Refer again to the sign rules in the previous section on lenses.

4) The image of the 2nd lens is the final image of the system.

*Application*:

Using the thin lens eqn. for lens L1:

1/s1 + 1/s1' = 1/15 + 1/s1' = 1/10 cm

therefore:

**s1' = 30 cm**

e.g.: 1/ s1' = 1/15 - 1/10 = 5/150 or s1' = 150 cm/ 5

And for the 2nd lens:

1/s2 + 1/s2' = 1/f2

® 1/ (-10 cm) + 1/s2' = 1/20 cm

or 1/s2' = 1/20 cm + 1/ 10 cm = 30 / 200 cm

^{-1}or s2' = 200/30 = (20/3) cm

Thus, the final image lies (20/3) cm

**.**

*to the right of the 2nd lens*The lateral magnification for each lens is defined as before (see, e.g. solutions to previous problems):

M1 = (-s1'/ s1) = - (30 cm/ 15 cm) = -2

M2 = (-s2'/s2) = -(20/3)cm/ -10 cm = 2/3

Then the total magnification of the lens system is:

M1 M2 = (-2)(2/3) = -4/3

So it is: real, inverted and enlarged by 4/3 times over the object.

In the case of the refracting telescope, such as shown in the photo, the magnifying power is defined by:

m = F/f(e)

where F is the focal length of the objective (the main or front lens) and f(e) is the focal length of the eyepiece. Hence, to get a large magnification one needs F to be fairly large and f(e) to be small.

*:*

__Thin Diverging Lens__

Figure 2 above shows a typical ray construction as it would be applied to a diverging lens and in many ways is similar to the converging lens except that now the object is formed on the

*same side of the lens*as the image and is

*larger than the image*. To summarize the numbered rays:

(1) Is drawn from the top of the object (O) to the optical axis of the lens.

(2) Is drawn from the top of O through the center of the lens and continued as a straight line

(3) Is drawn from the top of I (image) to the lens optical axis, thence to the opposite focus (F)

(4) is drawn starting from the first focus (F) on the left side of the lens, intersecting the top of I then on toward the optical axis at upper part of lens.

*Example
Problem (1a)*:

The
focal length of a diverging lens is 30 cm, and the object is 40 cm away. Find
the image distance.

*Solution*:

We have: f = - 30 cm

s = 40 cm

The thin lens equation for diverging lenses is:

1/s + 1/ s' = - 1/f

and in this case, we will write:

1/40 + 1/ s' = - 1/30

Then:

1/s' = -1/f -1/s = - 1/30 - 1/40

1/s' = - 7/120

s' = -120/7 cm = -17.1 cm

According to our sign conventions (Fig. 1, bottom) s' is negative if the image
is in front of the lens. Since s' = -17.1 cm it must be located 17.1 cm in
front of the lens conforming with what we see in the ray diagram of Fig. 2

*Example Problem (1b)*:

Using
a graphical construction find the image formed by a diverging lens with a focal
length of 30 cm when the object is 60 cm from the lens.

*Solution*: The diagram shown below
provides the scale solution.

* Suggested Problems*:

1) The objective lens of
an astronomical telescope has a focal length of 6 ft. The eyepiece has a focal
length of 2 inches.

Find the angular magnification that the telescope will produce when used for
distant objects.

(2)
A telephoto lens consists of a converging lens of focal length 6 cm placed 4 cm
in front of a diverging lens of focal length (-2.5 cm).

a) Do a graphical construction of the system showing where the image would be.

b) Compare the size of the image formed by this combination with the size of
the image that would be formed by the positive lens alone.

** See Also**:

Basic Physics (Light and Optics)

*And:*

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