Answer
(a)
$$\phi\approx \left(9 \times 10^{-6}\right)^{\circ}$$
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(b)
$$\theta\approx 4 \mathrm{\ m}$$
The largest optical telescopes with single mirrors are about $8\ m$ in diameter.
Work Step by Step
(a)
$$100 \operatorname{ly}\left(\frac{1 \mathrm{pc}}{3.26 \mathrm{ly}}\right)=30.67 \mathrm{pc}=\frac{1}{\phi^{\prime \prime}} \rightarrow$$
$$\phi=\left(\frac{1}{30.67}\right)^{\prime \prime}\left(\frac{1^{\prime}}{60^{\prime \prime}}\right)\left(\frac{1^{\circ}}{60^{\prime}}\right)$$
$$=\left(9.06 \times 10^{-6}\right)^{\circ} \approx \left(9 \times 10^{-6}\right)^{\circ}$$
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(b)
We use the Rayleigh criterion, Eq. $25–7$.
We choose a wavelength of $550\ nm$, in the middle of the visible range
$$\theta=\frac{1.22 \lambda}{D} \rightarrow D=\frac{1.22 \lambda}{\theta} $$
$$= \frac{1.22\left(550 \times 10^{-9} \mathrm{m}\right)}{\left[\left(9.06 \times 10^{-6}\right)^{\circ}\right]\left(\pi \operatorname{rad} / 180^{\circ}\right)}=4.24 \mathrm{m} \approx 4 \mathrm{m}$$
The largest optical telescopes with single mirrors are about $8\ m$ in diameter.
