Explain the resolving power of microscope.

Explanation

Image of a point like object by an objective of 3 microscope is shown in figure.
Let diameter of the lens be D and its focal length be f. As object distance is kept greater than that of f. Let an image distance be v.
The angular width of central maximum due to the effect of diffraction is `theta=(1.22lamda)/(D)`
Linear width of central maximum `=vtheta`
`:.vtheta=((1.22lamda)/(D))v" "......(1)`
If image of two point like objects are at a separation less than vo, then `vtheta`, it will be seen as a mixed single object.
If minimum distance for two images of two point like objects found to be resolved is `d_(m)`
`:.d_(m)=((1.22lamda)/(D))(v)/(m)" "......(2)`
[ `:.m=(v)/(f)` magnification ]
`:.d_(m)=((1.22lamda)/(D))f" "[:.(v)/(m)=f]`
From figure `(2)/(f)=tanbeta`
`:.(D)/(f)=2tanbeta" "......(3)`
From equation (2) and (3),
`d_(m)=(1.22lamda)/(2tanbeta)`
If `beta` is too small and is in radian then `tanbeta~~sinbeta`
`:.d_(m)=(1.22lamda)/(2sinbeta)" "......(4)`
If medium ofrefractive index"n" is placed between on object and objective then from equation (4) `:.d_(m)=(1.22lamda)/(2nsinbeta)" "......(5)`
Where `nsinbeta` is called numerical aperture and `(1)/(d_(m))` is called resolving power of microscope.
If refractive index "n" of medium is increased, resolving power of microscope can be increased.
Value of `sinbeta` is not more than 1, so resolving power (R.P.) of microscope is inversely proportional to the wavelength `lamda`.