Đáp án:
$\begin{align}
& 2)n=1,52 \\
& 4)f=9cm \\
\end{align}$
Giải thích các bước giải:
Câu 2:
$i={{40}^{0}};180-(i'+r)={{115}^{0}}$
ta có:
$\left\{ \begin{align}
& i=i'={{40}^{0}} \\
& 180-(i'+r)={{115}^{0}} \\
\end{align} \right.$
góc khúc xạ:
$\begin{align}
& 180-i-r={{115}^{0}} \\
& \Leftrightarrow r=180-40-115={{25}^{0}} \\
\end{align}$
chiết suất:
$\begin{align}
& \operatorname{s}\text{ini=n}\text{.sinr} \\
& \Rightarrow n=\dfrac{\sin 40}{\sin 25}=1,52 \\
\end{align}$
Câu 4:
$\begin{align}
& AB=2cm;{{A}_{1}}{{B}_{1}}=6cm; \\
& {{d}_{2}}={{d}_{1}}+1,5cm;{{A}_{2}}{{B}_{2}}=4cm \\
\end{align}$
ta có:
$\begin{align}
& \frac{{{A}_{1}}{{B}_{1}}}{AB}=\frac{d{{'}_{1}}}{{{d}_{1}}}\Leftrightarrow \frac{6}{2}=\frac{d{{'}_{1}}}{{{d}_{1}}}\Rightarrow d{{'}_{1}}=3{{d}_{1}} \\
& \frac{{{A}_{2}}{{B}_{2}}}{AB}=\frac{d{{'}_{2}}}{{{d}_{2}}}\Leftrightarrow \frac{4}{2}=\frac{d{{'}_{2}}}{{{d}_{2}}}\Rightarrow d{{'}_{2}}=2{{d}_{2}} \\
\end{align}$
mà:
$\begin{align}
& \dfrac{1}{f}=\dfrac{1}{d{}_{1}}+\dfrac{1}{d{{'}_{1}}}=\dfrac{1}{{{d}_{2}}}+\dfrac{1}{d{{'}_{2}}} \\
& \Leftrightarrow \dfrac{1}{{{d}_{1}}}+\dfrac{1}{3{{d}_{1}}}=\dfrac{1}{{{d}_{2}}}+\dfrac{1}{2{{d}_{2}}} \\
& \Leftrightarrow \dfrac{4}{3{{d}_{1}}}=\dfrac{3}{2.({{d}_{1}}+1,5)} \\
& \Rightarrow {{d}_{1}}=12cm \\
\end{align}$
tiêu cự:
$\begin{align}
& \frac{1}{f}=\dfrac{1}{d{}_{1}}+\dfrac{1}{d{{'}_{1}}} \\
& \Leftrightarrow \dfrac{1}{f}=\dfrac{1}{12}+\dfrac{1}{3.12} \\
& \Rightarrow f=9cm \\
\end{align}$
