Lens maker equation:

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

Magnification;

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

**a)**

For the converging lens:

1/60 + 1/s_{i} = 1/40

1/s_{i} = 1/40 - 1/60 = 1/120

s_{i} = 120 cm

This image distance is the object distance for the diverging lens, which is now 40cm in front of the lens. This is because the lenses are 160cm apart.

Therefore:

1/140 + 1/s_{i} = 1/-40

A converging lens with a focal length of 40 cm and a diverging lens with a focal length of -40 cm are 160 cm apart. A 1.0-cm-tall object is 60 cm in front of the converging lens.

a) Calculate the image position in front of the second lens

b) Calculate the image height.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

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What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Maric's class at UH.