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

For the first convex lens, the incident light beam is parallel to the principal axis. Therefore, the image is formed at its focal point.

Using the lens formula for the first lens:
$u_1 = -\infty$
$f_1 = +10 \text{ cm}$
$\Rightarrow v_1 = +10 \text{ cm}$

This real image formed by the first lens acts as an object for the second lens. Since the distance between the lenses is $15 \text{ cm}$, the object distance for the second lens is:
$u_2 = -(15 \text{ cm} - 10 \text{ cm}) = -5 \text{ cm}$

Now, applying the lens formula for the second lens:
$\dfrac{1}{v_2} - \dfrac{1}{u_2} = \dfrac{1}{f_2}$

Substitute $u_2 = -5 \text{ cm}$ and $f_2 = +10 \text{ cm}$:
$\dfrac{1}{v_2} - \dfrac{1}{-5} = \dfrac{1}{10}$

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

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

$\Rightarrow v_2 = -10 \text{ cm}$

The negative sign indicates that the final image is virtual and is formed $10 \text{ cm}$ in front of the second lens. Since the rays appear to originate from a virtual point before the second lens, the light beam diverges as it exits the lens system.