Lens maker equation:

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

Image distance:

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

Magnification:

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

You are using a converging lens with a focal length of 10 cm what will the image distance be and what will the magnification of that image be if you place an object:

a) 30 cm to the left of the lens.

b) 20 cm to the left of the lens.

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