Q. The diameter of objective of a telescope is 1 m. its resolving limit for the light of wavelength $4538\ Å,$ will be
A) $5.54\ \times \ 10^{- 7}\ rad$
B) $2.54\ \times \ 10^{- 4}\ rad$
C) $6.54\ \times \ 10^{- 7}\ rad$
D) None of the above
Explanation
(a) Here $d = 1\ m,\ \lambda = 4538\ Å$
= 4.538$\times 10^{- 7}m$
Resolving limit $\theta = \frac{1.22\lambda}{d}$
$= \frac{1.22 \times 4.538 \times 10\ \bar{}⁷}{1}$
= 5.54$\times 10^{- 7}$ rad
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.