A point object is placed on the optic axis of a convex lens of focal length $f$ at a distance of $2 f$ to the left of it. The diameter of the lens is $d$. An eye is placed at a distance of $3 f$ to the right of the lens and a distance $h$ below the optic axis. The maximum value of $h$ to see the image is

The object is placed at $u = -2f$. Using the lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$, we have $\dfrac{1}{v} - \dfrac{1}{-2f} = \dfrac{1}{f}$, which gives $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{2f} = \dfrac{1}{2f}$. Thus, $v = 2f$. The image is formed at a distance of $2f$ to the right of the lens.

The eye is placed at a distance of $3f$ to the right of the lens. Let the optic axis be the x-axis. The image is formed at $(2f, 0)$. The eye is at $(3f, -h)$.

The light rays from the image at $(2f, 0)$ must pass through the lens of diameter $d$ to reach the eye. The lens extends from $y = -d/2$ to $y = d/2$ at $x = 0$.

Consider the ray passing through the edge of the lens at $(0, d/2)$ and the image at $(2f, 0)$. The equation of this line is $y - 0 = m(x - 2f)$. Since it passes through $(0, d/2)$, we have $d/2 = m(0 - 2f)$, so $m = -d/(4f)$.

The ray equation is $y = -\dfrac{d}{4f}(x - 2f)$. At the eye position $x = 3f$, the height $y$ is $y = -\dfrac{d}{4f}(3f - 2f) = -\dfrac{d}{4f}(f) = -d/4$.

The eye is at a distance $h$ below the axis, so $h = |y| = d/4$. Any point further than $d/4$ below the axis will not receive light from the image formed by the lens.

Answer: $\dfrac{d}{4}$