Question

The objective of a microscope has magnifying power of 20×, a numerical aperture of 0.4, a working distance (WD) of 4 mm (~f ). For a wavelength of light of 450 nm, what is the resolving power of this objective? What is the diameter of its lens?

Answer 1

The resolving power is given by

$r=\frac{0.61\lambda }{na}$

Given $\lambda = 450\times 10^{-9}m$ , $na=0.4$

0.61 X 450 X 10-9 -m 0.4

$r=686.25nm$

The diameter D will be given by

$D=2\times f\times sin(na)$

$D=2\times WD\times sin(na)$

$D=2\times 4mm\times sin(0.4)=3.1153mm$