__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:

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

**Part A**

We're given:

f_{1} = 50 cm

f_{2} = - 48 cm

h_{o} = 2.4 cm

s_{o} = 70 cm

For the conversing lens using the lens equation:

1/50 = 1/s_{i} + 1/70

A converging lens with a focal length of 50 cm and a diverging lens with a focal length of -48 cm are 223 cm apart. A 2.4-cm-tall object is 70 cm in front of the converging lens.

Part A

Calculate the distance between the final image and the diverging lens.

Part B

Calculate the image height.

Frequently Asked Questions

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

Our tutors have indicated that to solve this problem you will need to apply the Thin Lens And Lens Maker Equations concept. You can view video lessons to learn Thin Lens And Lens Maker Equations. Or if you need more Thin Lens And Lens Maker Equations practice, you can also practice Thin Lens And Lens Maker Equations practice problems.