Lens equation:

$\overline{)\frac{\mathbf{1}}{{\mathit{s}}_{\mathit{o}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathit{s}}_{\mathit{i}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathit{f}}}$

Magnification:

$\overline{){\mathbf{m}}{\mathbf{=}}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{o}}}{\mathbf{=}}{\mathbf{-}}\frac{{\mathbf{s}}_{\mathbf{i}}}{{\mathbf{s}}_{\mathbf{o}}}}$

**A.**

We first determine the position of the image from the first lens.

1/s_{i1} = 1/f - 1/s_{o1}

1/s_{i1} = 1/9 - 1/36 = 1/12

s_{i1} = 12 cm

The figure shows a combination of two identical lenses. (Intro 1 figure)

A. Find the position of the final image of the 2.0-cm-tall object.

B. Find the size of the final image of the 2.0-cm-tall object.

C. Find the orientation of the final image of the 2.0-cm-tall object.

A. erect

B. inverted

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