Question

A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.

Answer 1

Answer:

The smallest diameter is $D =122 \ m$

Explanation:

From the question we are told that

       The resolution of the telescope is $\theta = 0.00350 \ arc \ second$

           The wavelength is  $\lambda = 1.70 \mu m = 1.70 *10^{-6} \ m$

From the question we are told that

        $1 arc \ sec = \frac{1}{3600^o}$

So      $0.00350 \ arc \ second = x$

Therefore

             $x = 0.00350 * \frac{1}{3600 }$

              $x = ( 9.722*10^{-7} )^o$

Now  $1^o = \frac{\pi}{180}$

   So  $(9.722*10^{-7})^o = \theta$

  =>    $\theta = (9.722*10^{-7}) * \frac{\pi}{180}$

           $\theta = 1.69*10^{-8} rad$

The smallest diameter is mathematically represented  as

          $D = \frac{1.22 \lambda }{\theta }$

substituting values

           $D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8} }$

           $D =122 \ m$