A 5 mm high pin is situated at a distance of 15 cm from a convex...

For the first lens:

$\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}$

$\dfrac{1}{v_1} - \dfrac{1}{-15} = \dfrac{1}{10}$

$\dfrac{1}{v_1} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{1}{30} \Rightarrow v_1 = 30 \text{ cm}$

Magnification of the first lens, $m_1 = \dfrac{v_1}{u_1} = \dfrac{30}{-15} = -2$

The image formed by the first lens acts as the object for the second lens. The distance between the lenses is $40 \text{ cm}$.

Object distance for the second lens, $u_2 = -(40 - 30) = -10 \text{ cm}$

For the second lens:

$\dfrac{1}{v_2} - \dfrac{1}{u_2} = \dfrac{1}{f_2}$

$\dfrac{1}{v_2} - \dfrac{1}{-10} = \dfrac{1}{5}$

$\dfrac{1}{v_2} = \dfrac{1}{5} - \dfrac{1}{10} = \dfrac{1}{10} \Rightarrow v_2 = 10 \text{ cm}$

Since $v_2$ is positive, the final image is real.

Magnification of the second lens, $m_2 = \dfrac{v_2}{u_2} = \dfrac{10}{-10} = -1$

Total magnification, $m = m_1 \times m_2 = (-2) \times (-1) = 2$

Since the total magnification is positive, the final image is erect with respect to the original object.

Thus, the final image is erect and real.