Ray Optics and Optical Instruments

Question: The diameter of the objective of a telescope is $a$, its magnifying power is $m$ and wavelength of light is $\lambda .$The resolving power of the telescope is

1) $(1.22\,\lambda )/a$

2) $(1.22\,a)/\lambda $

3) $\lambda m/(1.22a)$

4) $a/(1.22\lambda \,)$

Solution:

Explanation: No Explanation

Microscope and Telescope

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