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$\frac{2 \mu}{\lambda}$
✓
Answer
Correct option: A.
$\frac{2 \mu \sin \theta}{1.22 \lambda}$
(a) $\frac{2 \mu \sin \theta}{1.22 \lambda}$
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