A pin of length 2.00cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00cm is formed at a distance of 40.0cm from the pin. Find the focal length of the lens and its distance from the pin.
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Given that,
(-u) + v = 40cm = distance between object and image
$h_0 = 2cm, h_i = 1cm$
Since $\frac{\text{h}_\text{i}}{\text{h}_0}=\frac{\text{v}}{-\text{u}}=$ magnification
$\Rightarrow \frac{1}{2}=\frac{\text{v}}{-\text{u}}\Rightarrow\text{u}=-2\text{v} \ ...(1)$
Now, $\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}+\frac{1}{2\text{v}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{3}{2\text{v}}=\frac{1}{\text{f}}\Rightarrow\text{f}=\frac{2\text{v}}{3} \ ...(2)$
Again, $(-\text{u})+\text{v}=40$
$\Rightarrow3\text{v}=40\Rightarrow\text{v}=\frac{40}{3}\text{cm}$
$\therefore \ \text{f}=\frac{2\times40}{3\times3}=8.89\text{cm}=$ focal length
From eqn. (1) and (2)
$u = -2v = -3f = -3(8.89) = 26.7cm =$ object distance.
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