Answer:
The smallest diameter is [tex]D =122 \ m[/tex]
Explanation:
From the question we are told that
The resolution of the telescope is [tex]\theta = 0.00350 \ arc \ second[/tex]
The wavelength is [tex]\lambda = 1.70 \mu m = 1.70 *10^{-6} \ m[/tex]
From the question we are told that
[tex]1 arc \ sec = \frac{1}{3600^o}[/tex]
So [tex]0.00350 \ arc \ second = x[/tex]
Therefore
[tex]x = 0.00350 * \frac{1}{3600 }[/tex]
[tex]x = ( 9.722*10^{-7} )^o[/tex]
Now [tex]1^o = \frac{\pi}{180}[/tex]
So [tex](9.722*10^{-7})^o = \theta[/tex]
=> [tex]\theta = (9.722*10^{-7}) * \frac{\pi}{180}[/tex]
[tex]\theta = 1.69*10^{-8} rad[/tex]
The smallest diameter is mathematically represented as
[tex]D = \frac{1.22 \lambda }{\theta }[/tex]
substituting values
[tex]D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8} }[/tex]
[tex]D =122 \ m[/tex]
