A microscope has objective of aperture 8mm and focal length `2.5 cm`. Estimate its resolving power. Given `lambda= 5500 dotA`.

Explanation

We assume that the object is placed a little beyond its focal distance, say 2.5 cm.

`tan alpha = (4mm)/(2.5cm) = (0.4)/(2.5) = 0.16`
Since `sin alpha` small , `sin alpha ~~ tan alpha = 0.16`
Thus ,` Delta x = (1.22 lamda)/(2mu sin alpha) = (1.22 (5500 xx 10^(-10) m))/(2 xx 1 xx 0.16) = 2 xx 10^(-6) cm`