A point object O is situated on the principal axis of a convex lens...

Using the lens formula to find the position of the image:

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

Substituting $u = -40 \text{ cm}$ and $f = 20 \text{ cm}$:

$\dfrac{1}{v} - \dfrac{1}{-40} = \dfrac{1}{20}$

$\dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{40} = \dfrac{1}{40}$

$v = 40 \text{ cm}$

The image is formed at a distance of $40 \text{ cm}$ to the right of the lens. The rays from the object pass through the lens aperture, which has a radius of $R = \dfrac{10}{2} = 5 \text{ cm}$, and converge at the image point.

After converging at the image, the rays diverge. The eye is positioned $60 \text{ cm}$ to the right of the lens, which is at a distance of $60 - 40 = 20 \text{ cm}$ from the image.

Let the radius of the diverging beam at the position of the eye be $r$. Using similar triangles formed by the image point, the lens aperture, and the cross-section of the beam at the eye:

$\dfrac{r}{20} = \dfrac{5}{40}$

$r = 20 \times \dfrac{5}{40} = 2.5 \text{ cm}$

For the image to be visible, the eye must be within this diverging beam of light. Therefore, the maximum distance $h$ below the principal axis is $2.5 \text{ cm}$.