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:The distance is [tex]z = 0.008 \ m[/tex]Explanation:From the question we are told that    The  focal length is  [tex]f = 50 \ mm = 50*10^{-3} \ m[/tex]    Generally the lens equation is mathematically represented as        [tex]\frac{1}{u} + \frac{1}{v} = \frac{1}{f}[/tex]At  image  distance  u =  1.5 m        [tex]\frac{1}{1.5} + \frac{1}{v} = \frac{1}{50 *10^{-3}}[/tex]=>      [tex]\frac{1}{50 *10^{-3}} - \frac{1}{1.5} = \frac{1}{v}[/tex]=>[tex]v = 0.052 \ m[/tex]At  image  distance  [tex]u = 30\ cm = 0.30 \ m[/tex]         [tex]\frac{1}{0.3} + \frac{1}{v_1} = \frac{1}{50 *10^{-3}}[/tex]=>     [tex]\frac{1}{50 *10^{-3}} - \frac{1}{0.30 } = \frac{1}{v_1}[/tex]=>    [tex]v_1 = 0.06 \ m[/tex]The distance the lens need to move is evaluate as    [tex]z = |v - v_1|[/tex]    [tex]z = |0.052 - 0.06|[/tex]    [tex]z = 0.009 \ m[/tex]