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$\frac{2 \mu}{\lambda}$