We now look at the practical combination of thin lenses which yields optical instruments such as the one shown in the accompanying photo of Caribbean science student Carson King, who won top price at a Science Exhibition with his self-designed and constructed 2" aperture refracting telescope (which he converted to an astro -camera to take photos of the Moon.)

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 in analyzing a thin 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 blog 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.

Problems:

(1) 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.

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

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

b) A rule for observing extended astronomical objects, such as planets or nebulae, is that the telescope magnification should not exceed 60x per inch of objective aperture.

If the astronomical telescope of this problem is used to observe the planet Jupiter,is the condition met or not? If not, what focal length eyepiece is needed to get the

*maximum*angular magnification?

3) The objective of a telescope has a focal length F = 30 in. When it is used for an object at a great distance, then the distance between the objective and eyepiece is 32 in. What is the angular magnification?

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