Đáp án:
$\begin{align}
& {{d}_{1}}=30cm \\
& d{{'}_{1}}=-7,5cm \\
\end{align}$
Giải thích các bước giải:
$f=-10cm;{{A}_{1}}{{B}_{1}};{{d}_{2}}={{d}_{1}}-15cm;d{{'}_{2}}=d{{'}_{1}}+1,5cm$
Ta có:
$\begin{align}
& \dfrac{1}{f}=\dfrac{1}{{{d}_{1}}}+\dfrac{1}{d{{'}_{1}}}\Rightarrow d{{'}_{1}}=\dfrac{f.{{d}_{1}}}{{{d}_{1}}-f}=\dfrac{-10.{{d}_{1}}}{{{d}_{1}}+10} \\
& \dfrac{1}{f}=\dfrac{1}{{{d}_{2}}}+\dfrac{1}{d{{'}_{2}}}\Rightarrow d{{'}_{2}}=\dfrac{f.{{d}_{2}}}{{{d}_{2}}-f}=\dfrac{-10.{{d}_{2}}}{{{d}_{2}}+10} \\
\end{align}$
với đề bài:
$\begin{align}
& d{{'}_{2}}=d{{'}_{1}}+1,5 \\
& \Leftrightarrow \dfrac{-10.{{d}_{2}}}{-10-{{d}_{2}}}=\frac{-10.{{d}_{1}}}{{{d}_{1}}+10}+1,5 \\
& \Leftrightarrow \dfrac{-10.({{d}_{1}}-15)}{({{d}_{1}}-15)+10}=\dfrac{-10.{{d}_{1}}}{{{d}_{1}}+10}+1,5 \\
& \Rightarrow {{d}_{1}}=30cm \\
\end{align}$
ảnh ảo
$d{{'}_{1}}=\dfrac{-10.30}{30+10}=-7,5cm$
