Đáp án:

a) \(\dfrac{1}{f} = \dfrac{1}{d} + \dfrac{1}{{d'}}\)

b) \(20\sqrt 2 \left( {cm} \right)\)

Giải thích các bước giải:

a) Ta có: 

\(\begin{array}{l}
\Delta OAB \sim \Delta OA'B' \Rightarrow \dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{d}{{d'}}\\
\Delta F'OI \sim \Delta F'A'B' \Rightarrow \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{F'A'}}\\
 \Rightarrow \dfrac{{AB}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{f}{{d' - f}}\\
 \Rightarrow \dfrac{d}{{d'}} = \dfrac{f}{{d' - f}} \Rightarrow d.d' - df = d'f\\
 \Rightarrow \dfrac{1}{f} - \dfrac{1}{{d'}} = \dfrac{1}{d} \Rightarrow \dfrac{1}{f} = \dfrac{1}{d} + \dfrac{1}{{d'}}
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
\dfrac{1}{f} = \dfrac{1}{{{d_A}}} + \dfrac{1}{{{d_A}'}} \Rightarrow \dfrac{1}{{20}} = \dfrac{1}{{40}} + \dfrac{1}{{{d_A}'}} \Rightarrow {d_A}' = 40cm\\
\dfrac{{{h_A}'}}{{{h_A}}} = \dfrac{{{d_A}'}}{{{d_A}}} \Rightarrow \dfrac{{{h_A}'}}{{20}} = \dfrac{{40}}{{40}} \Rightarrow {h_A}' = 20cm\\
\dfrac{1}{f} = \dfrac{1}{{{d_B}}} + \dfrac{1}{{{d_B}'}} \Rightarrow \dfrac{1}{{20}} = \dfrac{1}{{30}} + \dfrac{1}{{{d_B}'}} \Rightarrow {d_B}' = 60cm\\
\dfrac{{{h_B}'}}{{{h_B}}} = \dfrac{{{d_B}'}}{{{d_B}}} \Rightarrow \dfrac{{{h_B}'}}{{20}} = \dfrac{{60}}{{30}} \Rightarrow {h_B}' = 40cm\\
h = \sqrt {{{\left( {{d_A}' - {d_B}'} \right)}^2} + {{\left( {{h_A}' - {h_B}'} \right)}^2}} \\
 \Rightarrow h = \sqrt {{{\left( {40 - 60} \right)}^2} + {{\left( {20 - 40} \right)}^2}}  = 20\sqrt 2 \left( {cm} \right)
\end{array}\)