A.5,8 cm.B.6,8 cm.C.12 cm.D.8,5 cm.

duongpham565656

2 năm trước

A.5,8 cm.
B.6,8 cm.
C.12 cm.
D.8,5 cm.

Loga Hóa Học lớp 12

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nguyenlanhuong02122006

Đáp án đúng: A

Phương pháp giải:

Công thức định luật khúc xạ ánh sáng: \(\sin i = n\sin r\) Công thức lượng giác: \(\dfrac{1}{{{{\sin }^2}x}} = 1 + \dfrac{1}{{{{\tan }^2}x}}\) Giải chi tiết:Từ hình vẽ ta thấy:\(\begin{array}{l}\sin i = \dfrac{{A'I}}{{AI}} = \dfrac{{CB}}{{AB}} = \dfrac{{40}}{{\sqrt {{{30}^2} + {{40}^2}} }} = 0,8\\ \Rightarrow \dfrac{1}{{{{\sin }^2}i}} = 1 + \dfrac{1}{{{{\tan }^2}i}} \Rightarrow \dfrac{1}{{0,{8^2}}} = 1 + \dfrac{1}{{{{\tan }^2}i}} \Rightarrow \tan i = \dfrac{4}{3}\end{array}\) Áp dụng công thức định luật khúc xạ ánh sáng, ta có:\(\begin{array}{l}\sin i = n\sin r \Rightarrow \sin r = \dfrac{{\sin i}}{n} = \dfrac{{0,8}}{{\dfrac{4}{3}}} = 0,6\\ \Rightarrow \dfrac{1}{{{{\sin }^2}r}} = 1 + \dfrac{1}{{{{\tan }^2}r}} \Rightarrow \dfrac{1}{{0,{6^2}}} = 1 + \dfrac{1}{{{{\tan }^2}r}} \Rightarrow \tan r = 0,75\end{array}\) Lại có: \(h = II' = \dfrac{2}{3}AC = \dfrac{2}{3}.30 = 20\,\,\left( {cm} \right)\) Ta có:\(\begin{array}{l}\left\{ \begin{array}{l}I'B = h.\tan i\\I'D = h.\tan r\end{array} \right. \Rightarrow d = I'B - I'D = h\left( {\tan i - \tan r} \right)\\ \Rightarrow d = 20.\left( {\dfrac{4}{3} - 0,75} \right) \approx 5,8\,\,\left( {cm} \right)\end{array}\)

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