If $$f_{0} = 5\ cm, \lambda = 600{A^{0}}, a = 1\ cm$$ for a microscope, then what will be its resolving power.
A
$$10.9\times 10^{3} / m$$
B
$$10.9\times 10^{5} / m$$
C
$$10.9\times 10^{4} / m$$
D
$$11.9\times 10^{5} / m$$
Correct option is B. $$10.9\times 10^{5} / m$$
$$R.P. = \dfrac {2\mu \sin \theta}{1.22\lambda}$$
$$\mu = 1$$
$$\tan \theta \simeq \sin \theta = \dfrac {a}{f} = \dfrac {1}{5} = 0.2\Rightarrow R.D. = \dfrac {2\times 1\times 0.2}{1.22\times 6\times 10^{-7}} = \dfrac {4\times 10^{6}}{3.66}$$
$$RP. = 10.9\times 10^{5} / m$$.