A pin of length 2.0 cm lies along the principal axis of a...

Given:
Length of the pin, $L_o = 2.0 \text{ cm}$
Focal length of the converging lens, $f = +6 \text{ cm}$
Distance of the centre of the pin from the lens $= 11 \text{ cm}$

Since the pin lies along the principal axis, its ends are located at distances of $11 - 1 = 10 \text{ cm}$ and $11 + 1 = 12 \text{ cm}$ from the optical centre of the lens.
Thus, the object coordinates for the two ends are $u_1 = -10 \text{ cm}$ and $u_2 = -12 \text{ cm}$.

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

For the first end ($u_1 = -10 \text{ cm}$):
$v_1 = \dfrac{-10 \times 6}{-10 + 6} = \dfrac{-60}{-4} = 15 \text{ cm}$

For the second end ($u_2 = -12 \text{ cm}$):
$v_2 = \dfrac{-12 \times 6}{-12 + 6} = \dfrac{-72}{-6} = 12 \text{ cm}$

The size of the image is the distance between the images of the two ends:
$L_i = |v_2 - v_1| = |12 - 15| = 3 \text{ cm}$