$A$ thin convex lens of focal length $25 \, cm$ is cut into two pieces $0.5 \, cm$ above the principal axis. The top part is placed at $(0,0)$ and an object is placed at $(-50 \, cm, 0)$. Find the coordinates of the image.

(D) The focal length of the lens remains $f = 25 \, cm$ after cutting.
The principal axis of the new lens passes through its optical center at $(0,0)$.
The object is placed at $(-50 \, cm, 0)$. Since the lens is cut $0.5 \, cm$ above the original principal axis,the new principal axis is $0.5 \, cm$ below the original one.
Relative to the new principal axis,the object is at a height of $y_o = 0.5 \, cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{25} = \frac{1}{v} - \frac{1}{-50} \implies \frac{1}{v} = \frac{1}{25} - \frac{1}{50} = \frac{1}{50} \implies v = 50 \, cm$.
Using magnification $m = \frac{v}{u} = \frac{y_i}{y_o}$:
$m = \frac{50}{-50} = -1$.
Thus,$y_i = m \cdot y_o = -1 \cdot 0.5 \, cm = -0.5 \, cm$.
The image coordinates relative to the new principal axis are $(50 \, cm, -0.5 \, cm)$.
Since the new principal axis is at $y = -0.5 \, cm$ relative to the original axis,the absolute $y$-coordinate is $-0.5 + (-0.5) = -1.0 \, cm$.
Therefore,the coordinates of the image are $(50 \, cm, -1.0 \, cm)$.