Q. The diameter of the objective of the telescope is $0.1\ metre$ and wavelength of light is $6000\ \mathring{\mathrm{A}}$. Its resolving power would be approximately
A) $7.32 \times 10^{- 6}rad$
B) $1.36 \times 10^{6}rad$
C) $7.32 \times 10^{- 5}rad$
D) $1.36 \times 10^{5}rad$
Explanation
(d) Resolving power $= \frac{d}{1.22\ \lambda} = \frac{0.1}{1.22 \times 6000 \times 10^{- 10}}$
$\equiv 1.36 \times 10^{5}radian$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.