Question

Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star:
\[
  A.{\text{ }}305 \times {10^{ - 9}}{\text{radian}} \\
  B.{\text{ }}152.5 \times {10^{ - 9}}{\text{radian}} \\
  C.{\text{ }}610 \times {10^{ - 9}}{\text{radian}} \\
  D.{\text{ }}457.5 \times {10^{ - 9}}{\text{radian}} \\
 \]

Answer 1

- Hint – In this question, the diameter and wavelength of telescope objective is given, so in order to calculate the resolution of the telescope we will apply the formula of limit of resolution of telescope which is mentioned in the solution. Most importantly in order to find the answer we will take care of the units of the parameters.
Formula used- $\theta = \dfrac{{1.22\lambda }}{D}$

Complete step-by-step solution -
Given that
Diameter $ = 200cm = 200 \times {10^{ - 2}}m$
Wavelength $500nm = 500 \times {10^{ - 9}}m$
We know limit of resolution of telescope is given as:
$\theta = \dfrac{{1.22\lambda }}{D}$
Here,
$\theta $ is the resolution of the telescope.
$\lambda $ is the wavelength of light.
$D$ is the diameter of the objective of a telescope.
Substitute the given values in above formula we have
$
  \because \theta = \dfrac{{1.22\lambda }}{D} \\
   \Rightarrow \theta = \dfrac{{1.22 \times 500 \times {{10}^{ - 9}}}}{{200 \times {{10}^{ - 2}}}} \\
   \Rightarrow \theta = 305 \times {10^{ - 9}}{\text{radian}} \\
 $
Hence the limit of resolution of telescope is $305 \times {10^{ - 9}}{\text{radian}}$
So, the correct answer is option A.

Additional information- The resolution limit (or solving power) is a function of the objective lens' ability to distinguish information present in the object in the picture adjacent to it. It's the distance in the picture between two points that is just being resolved in the shot. Ultimately, the resolving power of an optical device is reduced by aperture diffraction. And an optical device can not produce a flawless picture of a point.

Note- The resolution limit (or resolving power) is a function of the objective lens' ability to distinguish information contained in the object in the image adjacent to it. It's the distance in the picture between two points that is just being resolved in the shot. Ultimately, the resolving power of an optical system is limited by aperture diffraction. So an optical system cannot create a perfect picture of a point.