Explain the resolving power of a microscope.

(N/A) The image of a point-like object formed by the objective lens of a microscope is shown in the figure.
Let the diameter of the lens be $D$ and its focal length be $f$. The object distance is kept greater than $f$. Let the image distance be $v$.
The angular width of the central maximum due to the effect of diffraction is $\theta = \frac{1.22 \lambda}{D}$.
The linear width of the central maximum is $v \theta$.
$\therefore v \theta = \left( \frac{1.22 \lambda}{D} \right) v \quad \dots (1)$
If the images of two point-like objects are at a separation less than $v \theta$,they will be seen as a single merged object.
Let the minimum distance for two images of two point-like objects to be resolved be $d_m$.
$\therefore d_m = \left( \frac{1.22 \lambda}{D} \right) \frac{v}{m}$,where $m = \frac{v}{f}$ is the magnification.
Substituting $m = \frac{v}{f}$,we get $d_m = \left( \frac{1.22 \lambda}{D} \right) f \quad \dots (2)$
From the figure,$\frac{D/2}{f} = \tan \beta$.
$\therefore \frac{D}{f} = 2 \tan \beta \quad \dots (3)$
Substituting equation $(3)$ into equation $(2)$:
$d_m = \frac{1.22 \lambda}{2 \tan \beta}$
If $\beta$ is very small and is in radians,then $\tan \beta \approx \sin \beta$.
$\therefore d_m = \frac{1.22 \lambda}{2 \sin \beta}$