Explain the resolving power of optical instruments and explain the resolving power of telescopes.

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(N/A) Resolving Power: The ability of an optical instrument to produce distinctly separate images of two closely placed objects is called its resolving power.
Due to the phenomenon of diffraction in optical instruments,it is difficult to distinguish between objects and their images that are very close to each other.
The angular resolution of a telescope is determined by its objective lens. Increasing the magnification through the eyepiece does not improve the resolution; the eyepiece only magnifies the image already formed by the objective.
When a parallel beam of light is incident on a convex lens,the lens focuses the beam. However,due to diffraction,the light is focused into a finite circular area (the Airy disk) instead of a single point. This is shown in the figure.
Dark and bright rings are obtained around the central bright region,known as Airy's rings.
If $f$ is the focal length of the lens and $2a$ is the diameter of the circular aperture,the linear width of the central maximum is given by $\frac{1.22 \lambda f}{2a}$.
Therefore,the radius of the central maximum is:
$r_{0} \approx \frac{1.22 \lambda f}{2a} = \frac{0.61 \lambda f}{a}$
Using the relation $r_{0} = f \Delta \theta$,we get:
$f \Delta \theta \approx \frac{0.61 \lambda f}{a}$
$\therefore \Delta \theta \approx \frac{0.61 \lambda}{a}$
Here,$\Delta \theta$ is the minimum angle required to see two images as just separated,which defines the angular resolution of the telescope.
Thus,when the diameter of the objective $(2a)$ is larger,$\Delta \theta$ becomes smaller. This means that for a telescope,a larger aperture $a$ results in a higher resolving power.

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