Q. 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) $\frac{2\mu}{\ \lambda}$
Explanation
The correct answer is Option A: $\frac{2\mu\sin\theta}{1.22\ \lambda}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.