resolving power of telescope

resolving power of telescope

1. What Resolving Power Means for Telescope

Resolving power = reciprocal of minimum angular separation $θ_{min}$ that telescope can just resolve (Rayleigh). Larger aperture ⇒ smaller $θ_{min}$ ⇒ higher resolving power.

Often quoted as resolving power $R_P = D/(1.22λ)$ (angular) or simply $θ_{min}$.

For telescope objective diameter $D$, focal length $f_0$, eyepiece discriminates linear separation $y = f_0 θ_{min}$ at focal plane.

2. Why It Matters

Telescope's job is to see faint close stars/galaxies — light-gathering ∝$D²$ and resolution ∝$D$ both improve with aperture. That's why large telescopes (Keck 10 m, ELT 39 m) built.

3. Worked Examples

(a) Telescope D=0.1 m, λ=550 nm: θ_min=6.7 µrad ⇒ 1.4" arcsec; two stars 1" apart unresolved with this scope, need D≈0.14 m ✓  (θ_min ∝1/D)

(b) Human eye D≈2 mm? Actually 3 mm ⇒ θ≈1'. That's eye's limit; telescope magnifies but cannot beat diffraction — magnification without resolution is empty.

(c) Radio telescope D=100 m, λ=21 cm (H line): θ_min=1.22·0.21/100=0.0026 rad≈0.15° (≈9') — much worse than optical despite large D, because λ large.

(d) Need to resolve binary stars 0.5" apart in visible 550 nm: D=1.22λ/θ =1.22·5.5e-7/(0.5·4.85e-6)≈0.28 m ⇒ 11 inch scope.

(e) Resolving power vs magnifying power: magnifying power = f0/fe, independent. Useful magnification ≈ R·eye resolution (~1') ⇒ ~D(mm)·0.5×? Rule of thumb max useful ~2×D(mm).

4. Relation to Microscope (contrast)

• Telescope: angular separation.
• Microscope: linear separation $y_{min}=0.61λ/NA$ (next levels).

5. Exam Points

• State $θ_{min}=1.22λ/D$ and $R= D/(1.22λ)$ — 5 marks, very frequent.
• Compute $θ_{min}$ for given $D,λ$ — 5 marks, very frequent.
• Explain why larger $D$ improves both brightness and resolution — 3-5 marks, frequent.
• Distinguish resolving vs magnifying power — 3 marks.