The focal length of the objective of a microscope if f_o and that of the eye piece if f_e.The aperture of objective is d_o and that of eye piece is d_e.Which of the following relation is correct?

The correct option is:
$f_o<f_e,d_o<d_e$

Step-by-Step Explanation:

In a compound microscope, the objective lens and the eyepiece (ocular lens) serve different, specific purposes:
1. Objective Lens: The objective lens is positioned close to the object being viewed. To obtain high magnification and to focus on extremely small, close objects, it must have a very short focal length. Additionally, because the specimen is very small, a small aperture (diameter) is sufficient to gather light from it. Therefore, the objective lens has a small focal length (_fo) and a small aperture (_do).
2. Eyepiece: The eyepiece acts as a magnifying glass to enlarge the intermediate real image formed by the objective lens. To allow the viewer to comfortably observe the magnified image and to gather enough light from the intermediate image, the eyepiece needs a larger aperture. It also has a larger focal length compared to the objective lens. Therefore, the eyepiece has a larger focal length (_fe) and a larger aperture (_de) than the objective lens.

Comparing these parameters:
- For focal lengths:
$f_o<f_e$
- For apertures:
$d_o<d_e$

This confirms that the correct relationship is $f_o<f_e,d_o<d_e$.