- ALength of telescope
- BFocal length of objective
- CDiameter of the objective
- DFocal length of eyepiece
Explanation:
Resolving power of telescope $\text{R}=\frac{1}{\Delta\theta}=\frac{\text{a}}{1.22\lambda}$
where, $\Delta\theta$ is angular separation between two objects.
a is the diameter of the objective.
$\lambda$ is wavelength of light.
So, clearly resolving power of a telescope depends on diameter of the objective.
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