Q.Describe what is Rayleigh's criterion for resolution. Explain it for a telescope and a microscope.

[!TLDR] Rayleigh's criterion: two objects are just resolved when the first minimum of one's diffraction pattern coincides with the central maximum of the other's; for a telescope this gives $\theta_{min}=1.22\lambda/D$, and for a microscope $a_{min}=0.61\lambda/\text{N.A.}$ (self-luminous case). [!ANSWER] T …

Lord Rayleigh's criterion states that two objects are considered JUST RESOLVED by an optical instrument when the FIRST MINIMUM of one object's own diffraction pattern coincides exactly with the CENTRAL MAXIMUM of the other object's diffraction pattern (and, symmetrically, vice versa). If the two objects are closer together than this, their two central diffraction maxima overlap too heavily, and the resultant (summed) intensity shows a single smooth peak with no discernible dip -- the objects are NOT resolved. At exactly the Rayleigh separation, the resultant intensity shows a small but clearly noticeable dip between two peaks -- the borderline, just-resolved case. Separated further still, the dip becomes deep and unambiguous -- the objects are clearly, comfortably resolved.

For a TELESCOPE, viewing two distant point objects (effectively stars) with angular separation $\theta$ through an objective of aperture (diameter) D: because the objects are so far away, only their angular -- not linear -- separation matters. Since stars are self-luminous point sources, applying Abbe's theory of Fraunhofer diffraction by a circular aperture (giving Airy-disc diffraction patterns, a bright central disc surrounded by concentric rings) with Rayleigh's criterion, the path difference between extreme rays at the first dark ring works out to $1.22\lambda$, and the resulting geometry gives the LIMIT OF RESOLUTION directly as $\theta_{min}=\dfrac{1.22\lambda}{D}$. The RESOLVING POWER of the telescope is the reciprocal of this: $R=\dfrac{1}{\theta_{min}}=\dfrac{D}{1.22\lambda}$ -- so a LARGER objective aperture D, or a SHORTER observing wavelength, both improve a telescope's resolving power. …

State Rayleigh's criterion (first minimum of one pattern at the central maximum of the other), then apply it via Abbe's Airy-disc first-dark-ring path difference ($1.22\lambda$) separately to the telescope's angular geometry (aperture D) and the microsco …

Mixing up the telescope's angular resolving-power formula ($\theta_{min}=1.22\lambda/D$, using the FULL aperture diameter D) with the microscope's linear one ($a_{min}=0.61\lambda/\text{N.A.}$, using numerical aperture, itself defined via a HALF-angle $\alpha$), since the …