Question

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

• A. $\frac{\left(1.22\lambda \right)}{a}$
• B. $\frac{1.22a}{\lambda }$
• C. $\lambda \left(1.22a\right)$
• D. $\frac{a}{1.22\lambda }$

Answer 1

The correct option is D $\frac{a}{1.22\lambda }$

Resolving power of telescope:
$R=\frac{1}{\theta }=\frac{a}{1.22\lambda }$

where $\theta$ is angular resolution, a is diameter of the objective and $\lambda$ is wavelength of light.