If an optical telescope focusing light of wavelength 550 nm has a perfectly ground mirror, what would the minimum mirror diameter have to be so that the telescope could resolve a Jupiter-size planet around our nearest star, Alpha Centauri, which is about 4.3 lightyears from earth? (Consult Appendix F.)

The minimum diameter of the mirror to resolve the planet is 197.35 m .

According to Rayleigh, two images can just be resolved if the center of one's diffraction pattern coincides with the first minimum of the diffraction pattern of the other. The relation between the angular separation of the two images to be resolved $\mathbf{\theta }$, the wavelength of light used $\mathbf{\lambda }$and the diameter of the aperture D is given as-

$\mathbf{sin\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{D}}$

If $\mathit{\theta }$is very small,

$$\begin{gathered} \mathbf{sin\theta }\mathbf{\approx }\mathbf{\theta } \\ \mathbf{\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{D}} \end{gathered}$$

For the given values, the angular separation for resolving the planet situated at a distance from earth and has a diameter is;

$$\begin{gathered} \mathrm{\theta }=\frac{\mathrm{d}}{\mathrm{L}} \\ =\frac{1.4\times {10}^8\mathrm{m}}{4.3\times \left(9.46\times {10}^{15}\mathrm{m}\right)} \\ =3.4\times {10}^{-9}\mathrm{rad} \end{gathered}$$

Therefore, the minimum mirror diameter should be -

$$\begin{gathered} \mathrm{D}=1.22\frac{\mathrm{\lambda }}{\mathrm{\theta }} \\ =1.22\times \frac{550\times {10}^{-9}\mathrm{m}}{3.4\times {10}^{-9}\mathrm{rad}} \\ =197.35\mathrm{m} \end{gathered}$$

Hence, the minimum mirror diameter is 197.35 m , and ground-based optical telescopes are unable to resolve the planet in Alpha Centauri.