Two convex lenses, each having a focal length of 10 cm , are...

For the first lens, the object is at infinity, so $u_1 = -\infty$ and $f_1 = +10 \text{ cm}$.

Using the lens formula:

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

$\dfrac{1}{v_1} - \dfrac{1}{-\infty} = \dfrac{1}{10} \Rightarrow v_1 = +10 \text{ cm}$

The image formed by the first lens acts as a real object for the second lens. Since the distance between the two lenses is $15 \text{ cm}$, the object distance for the second lens is:

$u_2 = -(15 - 10) = -5 \text{ cm}$

For the second lens, $f_2 = +10 \text{ cm}$. Applying the lens formula again:

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

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

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

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

$v_2 = -10 \text{ cm}$

The negative sign indicates that the final virtual image is formed $10 \text{ cm}$ in front of the second lens.

The distance of this image from the first lens is $15 \text{ cm} - 10 \text{ cm} = 5 \text{ cm}$. Thus, the final image is located $5 \text{ cm}$ from the first lens towards the second lens.