MCQOPTIONS
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| 1. |
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. | \[\frac{2\mu }{\lambda }\] |
| Answer» B. \[\frac{\mu \sin \theta }{\lambda }\] | |