A pin of length 2.00 cm is positioned perpendicular to the...

Given height of the object, $h_o = 2.00 \text{ cm}$

Height of the inverted image, $h_i = -1.00 \text{ cm}$

Distance between the object and the image, $D = 40.0 \text{ cm}$

Magnification of the lens is given by $m = \dfrac{v}{u} = \dfrac{h_i}{h_o}$

$\dfrac{v}{u} = \dfrac{-1.00}{2.00} = -0.5 \Rightarrow v = -0.5u$

For a real image, the object and image are on opposite sides of the lens. The distance between them is $v - u = 40.0 \text{ cm}$.

Substituting $v = -0.5u$:

$-0.5u - u = 40.0$

$-1.5u = 40.0 \Rightarrow u = -\dfrac{80}{3} \approx -26.7 \text{ cm}$

The distance of the lens from the pin is $|u| = 26.7 \text{ cm}$.

The image distance is $v = -0.5 \times \left(-\dfrac{80}{3}\right) = \dfrac{40}{3} \text{ cm}$.

Using the lens formula, $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$:

$\dfrac{1}{f} = \dfrac{3}{40} - \left(-\dfrac{3}{80}\right)<br><br>\dfrac{1}{f} = \dfrac{6}{80} + \dfrac{3}{80} = \dfrac{9}{80}<br><br>$f = $\dfrac{80}{9} \approx$8.89 \text{ cm}$<br><br>The focal length of the lens is$8.89 \text{ cm} and its distance from the pin is $26.7 \text{ cm}$.