Question

A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification

Answer 1

Answer:

The magnification is  $m = 12$

Explanation:

From the question  we are told that

   The object distance is $u = 36.2 \ cm$

     The focal length is  $v = 39.5 \ cm$

From the lens equation we have that

         $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

=>     $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

substituting values

       $\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}$

       $\frac{1}{v} = -0.0023$

=>   $v = \frac{1}{0.0023}$

=>   $v =-433.3 \ cm$

The magnification is mathematically represented as

         $m =- \frac{v}{u}$

substituting values

        $m =- \frac{-433.3}{36.2}$

         $m = 12$