Question

A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 1.5 m away. He then wants to take a picture of an object that is 30 cm away.
How far must the lens move to focus on this second object?

Answer 1

Answer:

The distance is $z = 0.008 \ m$

Explanation:

From the question we are told that

   The  focal length is  $f = 50 \ mm = 50*10^{-3} \ m$

   

Generally the lens equation is mathematically represented as  

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

At  image  distance  u =  1.5 m

       $\frac{1}{1.5} + \frac{1}{v} = \frac{1}{50 *10^{-3}}$

=>      $\frac{1}{50 *10^{-3}} - \frac{1}{1.5} = \frac{1}{v}$

=>$v = 0.052 \ m$

At  image  distance  $u = 30\ cm = 0.30 \ m$

        $\frac{1}{0.3} + \frac{1}{v_1} = \frac{1}{50 *10^{-3}}$

=>     $\frac{1}{50 *10^{-3}} - \frac{1}{0.30 } = \frac{1}{v_1}$

=>    $v_1 = 0.06 \ m$

The distance the lens need to move is evaluate as

   $z = |v - v_1|$

   $z = |0.052 - 0.06|$

   $z = 0.009 \ m$