Ray Optics and Optical Instruments

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

1) $\frac{2\mu \sin \theta }{1.22\,\lambda }$

2) $\frac{\mu \sin \theta }{\lambda }$

3) $\frac{2\mu \cos \theta }{1.22\ \lambda }$

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

Solution:

Explanation: No Explanation

Microscope and Telescope

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