At a distance of 15 cm from a convex lens of focal length 10 cm , a...

For the first lens, using the lens formula:

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

Given $u_1 = -15 \text{ cm}$ and $f_1 = +10 \text{ cm}$.

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

$\dfrac{1}{v_1} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3 - 2}{30} = \dfrac{1}{30}$

$v_1 = +30 \text{ cm}$

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

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

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

Given $f_2 = +5 \text{ cm}$.

Using the lens formula:

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

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

$\dfrac{1}{v_2} = \dfrac{1}{5} - \dfrac{1}{10} = \dfrac{2 - 1}{10} = \dfrac{1}{10}$

$v_2 = +10 \text{ cm}$

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

Total magnification of the combination is $m = m_1 \times m_2 = (-2) \times (-1) = 2$.

The size of the final image is $h_i = |m| \times h_o = 2 \times 5 \text{ mm} = 10 \text{ mm}$.