An optical instrument employed for angular magnification contains...

The focal length of the objective lens is $f_o = \dfrac{100}{P_o} = \dfrac{100}{25} = 4 \text{ cm}$.

The focal length of the eyepiece is $f_e = \dfrac{100}{P_e} = \dfrac{100}{20} = 5 \text{ cm}$.

Since both the objective and the eyepiece have small focal lengths, the instrument is a compound microscope. In a telescope, the objective lens has a much larger focal length compared to the eyepiece.

For a telescope in normal adjustment (least strained eye), the tube length would be $L = f_o + f_e = 4 + 5 = 9 \text{ cm}$, which contradicts the given tube length of $25 \text{ cm}$.

For a microscope in normal adjustment, the tube length is $L = v_o + f_e$.

$25 = v_o + 5 \Rightarrow v_o = 20 \text{ cm}$.

Since $v_o > f_o$, the objective forms a real image, which is consistent with the principle of a compound microscope.