Đáp án:

$f=18cm$

Giải thích các bước giải:

ảnh thu được đều ngược chiều với vật

=> cả 2 vị trí đều cho ảnh thật 

${{d}_{2}}={{d}_{1}}-6(1);d{{'}_{2}}=d{{'}_{1}}+27cm;{{A}_{2}}{{B}_{2}}=2{{A}_{1}}{{B}_{1}}$

ta có:

$\left\{ \begin{align}
  & \dfrac{{{A}_{1}}{{B}_{1}}}{AB}=-\dfrac{d{{'}_{1}}}{{{d}_{1}}} \\ 
 & \dfrac{{{A}_{2}}{{B}_{2}}}{AB}=-\dfrac{d{{'}_{2}}}{{{d}_{2}}} \\ 
\end{align} \right.$

$\begin{align}
  & \Rightarrow \dfrac{{{A}_{2}}{{B}_{2}}}{{{A}_{1}}{{B}_{1}}}=\dfrac{d{{'}_{2}}}{{{d}_{2}}}.\dfrac{{{d}_{1}}}{d{{'}_{1}}} \\ 
 & \Leftrightarrow 2=\dfrac{(d{{'}_{1}}+27).{{d}_{1}}}{({{d}_{1}}-6).d{{'}_{1}}} \\ 
 & \Rightarrow 2.({{d}_{1}}.d{{'}_{1}}-6.d{{'}_{1}})=(d{{'}_{1}}+27).d(2) \\ 
\end{align}$

mà ta có:

$\begin{align}
  & \dfrac{1}{f}=\dfrac{1}{{{d}_{1}}}+\dfrac{1}{d{{'}_{1}}}=\dfrac{1}{{{d}_{2}}}+\dfrac{1}{d{{'}_{2}}} \\ 
 & \Leftrightarrow \dfrac{1}{{{d}_{1}}}+\dfrac{1}{d{{'}_{1}}}=\dfrac{1}{{{d}_{1}}-6}+\dfrac{1}{d{{'}_{1}}+27} \\ 
 & \Rightarrow \dfrac{{{d}_{1}}-6-{{d}_{1}}}{{{d}_{1}}({{d}_{1}}-6)}=\dfrac{d_{1}^{\prime }-d_{1}^{\prime }-27}{d_{1}^{\prime }(d_{1}^{\prime }+27)} \\ 
 & \Rightarrow 6d{{\prime }_{1}}(d{{\prime }_{1}}+27)=27{{d}_{1}}({{d}_{1}}-6)(3) \\ 
\end{align}$

từ (2) và (3) ta có:

$\dfrac{(2)}{(3)}\Leftrightarrow \dfrac{2d_{1}^{\prime }}{27{{d}_{1}}}=\dfrac{{{d}_{1}}}{6d_{1}^{\prime }}$

$\Rightarrow 27d_{1}^{2}=12d_{1}^{'2}\Rightarrow d{{'}_{1}}=\frac{3}{2}{{d}_{1}}(4)\text{ }\!\!~\!\!\text{ }$

thay (4) vào (2): 

$\begin{align}
  & 6.\dfrac{3}{2}.{{d}_{1}}.(\dfrac{3}{2}{{d}_{1}}+27)=27{{d}_{1}}.({{d}_{1}}-6) \\ 
 & \Leftrightarrow \dfrac{3}{2}{{d}_{1}}+27=3.({{d}_{1}}-6) \\ 
 & \Rightarrow {{d}_{1}}=30cm;d{{'}_{1}}=\dfrac{3}{2}{{d}_{1}}=\dfrac{3}{2}.30=45cm \\ 
\end{align}$

tiêu cự:

$\begin{align}
  & \dfrac{1}{f}=\dfrac{1}{{{d}_{1}}}+\dfrac{1}{d{{'}_{1}}} \\ 
 & =\dfrac{1}{30}+\dfrac{1}{45}=\dfrac{1}{18} \\ 
 & \to f=18cm \\ 
\end{align}$