Q. 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
A) $(1.22\lambda)/a$
B) $(1.22a)/\lambda$
C) $\lambda m/(1.22a)$
D) $a/(1.22\lambda)$
Explanation
The correct answer is Option D: $a/(1.22\lambda)$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.