(N/A) Suppose the object has a height $h$. The maximum angle it can subtend and be clearly visible (without a lens) is when it is at the near point,i.e.,at a distance $D$.
$\tan \theta_{0} = \frac{h}{D}$
For small angles,
$\tan \theta_{0} \approx \theta_{0} \implies \theta_{0} = \frac{h}{D} \quad \dots (1)$
Now,if the object is placed at the focus $f$ of the convex lens,the image is formed at infinity. The angle subtended at the eye by the image is $\theta_{i}$,where
$\tan \theta_{i} = \frac{h}{f}$
For small angles,
$\tan \theta_{i} \approx \theta_{i} \implies \theta_{i} = \frac{h}{f} \quad \dots (2)$
The angular magnification $m$ is defined as the ratio of the angle subtended by the image to the angle subtended by the object at the near point:
$m = \frac{\theta_{i}}{\theta_{0}} = \frac{h/f}{h/D} = \frac{D}{f}$
Thus,the magnification for the image formed at infinity is $m = \frac{D}{f}$.