Question

The focal length of the objective of a compound microscope is $f_0$ and its distance from the eyepiece is $L.$ The object is placed at a distance $u$ from the objective. For proper working of the instrument,

• A. $L<u$
• B. $L>u$
• C. $f_0<L<2f_0$
• D. $L>2f_0$

Answer 1

Answer: (B), (D)

The correct options are
B $L>u$
D $L>2f_0$

In the perfect condition of a compound microscope, the object is placed between $F_0-2F_0$, so that an inverted real image is formed, which will act as the object for the eye-piece. The final image thus produced is a virtual image, and for that, the real image produced by the first lens must be in the range of $F_e$ and the optical center.

So, $L>F_o+F_e$

For a compound microscope, $F_e>F_o$ and thus, we can write that $L>2F_o$

Since the object is placed between $F_0-2F_0$, then $L>u$

Therefore, option (B) and (D) are correct.