MCQ
When the object is self-luminous, the resolving power of a microscope is given by the expression
- ✓$\frac{{2\mu \sin \theta }}{{1.22\,\lambda }}$
- B$\frac{{\mu \sin \theta }}{\lambda }$
- C$\frac{{2\mu \cos \theta }}{{1.22\;\lambda }}$
- D$\frac{{2\mu }}{\lambda }$