The focal length of the objective of a microscope is \( f_o \) and that of the eye piece is \( f_e \). The aperture of objective is \( d_o \) and that of eye piece is \( d_e \). Which of the following relation is correct?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( f_o \lt f_e,\ d_o \lt d_e \)
In a compound microscope the objective forms a magnified real image of a tiny object kept just beyond its focus, and the eyepiece acts as a magnifier for that image.
Focal lengths: the magnification of the microscope is large when both focal lengths are small, and the objective is made with the smaller one, so \( f_o \lt f_e \) (for example \( f_o = 1\,\mathrm{cm} \) and \( f_e = 2\,\mathrm{cm} \)).
Apertures: the object is very close to the objective and is tiny, so the objective collects light through a small aperture. The eyepiece has to take in the whole cone of light that spreads out from the first image and pass it to the eye, so it has the larger aperture, \( d_o \lt d_e \).