Câu 1:
a) Hình vẽ
b) Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{4}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' + OF'}} = \dfrac{6}{{OA' + 6}}\\
\Rightarrow \dfrac{4}{{OA'}} = \dfrac{6}{{OA' + 6}} \Rightarrow OA' = 12cm\\
\Rightarrow \dfrac{{AB}}{{A'B'}} = \dfrac{1}{3} \Rightarrow A'B' = 6cm
\end{array}\)
Câu 2:
a) Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{30}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{{20}}{{OA' - 20}}\\
\Rightarrow \dfrac{{30}}{{OA'}} = \dfrac{{20}}{{OA' - 20}} \Rightarrow OA' = 60cm\\
\Rightarrow \dfrac{{AB}}{{A'B'}} = \dfrac{1}{2} \Rightarrow A'B' = 6cm
\end{array}\)
b) Di chuyển AB lại gần thấu kính 20 cm thì OA = 10, vật cho ảnh ảo qua thấu kính.
Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{10}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' + OF'}} = \dfrac{{20}}{{OA' + 20}}\\
\Rightarrow \dfrac{{10}}{{OA'}} = \dfrac{{20}}{{OA' + 20}} \Rightarrow OA' = 20cm\\
\Rightarrow \dfrac{{AB}}{{A'B'}} = \dfrac{1}{2} \Rightarrow A'B' = 6cm
\end{array}\)