question_answer

When the object is self-luminous, the resolving power of a microscope is given by the expression

A)            \[\frac{2\mu \sin \theta }{1.22\,\lambda }\]              

B)            \[\frac{\mu \sin \theta }{\lambda }\]

C)            \[\frac{2\mu \cos \theta }{1.22\ \lambda }\]             

D)            \[\frac{2\mu }{\lambda }\]