Question

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

• A. $\frac{2\mu \sin \left(\theta \right)}{1.22\lambda }$
• B. $\frac{\mu \sin \left(\theta \right)}{\lambda }$
• C. $\frac{2\mu \cos \left(\theta \right)}{1.22\lambda }$
• D. $\frac{2\mu }{\lambda }$

Answer 1

The correct option is A $\frac{2\mu \sin \left(\theta \right)}{1.22\lambda }$

For self luminous body, resolving power is given by $\frac{2\mu \sin \left(\theta \right)}{1.22\lambda }$