Đáp án:
\(\begin{array}{l}
17.B. - 15cm\\
\frac{1}{f} = \dfrac{1}{d} + \dfrac{1}{{d'}} = \dfrac{1}{{30}} + \dfrac{1}{{ - 10}} = \dfrac{1}{{ - 15}}\\
\Rightarrow f = - 15cm\\
18.d = 25cm\\
\dfrac{{d'}}{d} = \dfrac{{h'}}{h} = 4 \Rightarrow d' = 4d\\
\dfrac{1}{f} = \dfrac{1}{d} + \dfrac{1}{{d'}} = \dfrac{1}{d} + \dfrac{1}{{4d}} = \dfrac{5}{{4d}}\\
\Rightarrow d = \dfrac{{5f}}{4} = \dfrac{{5.20}}{4} = 25cm\\
19.D.12cm\\
\dfrac{{d'}}{d} = \dfrac{{h'}}{h} = \dfrac{3}{2} \Rightarrow d' = 1,5d = 1,5.20 = 30cm\\
\dfrac{1}{f} = \dfrac{1}{d} + \dfrac{1}{{d'}} = \dfrac{1}{{20}} + \dfrac{1}{{30}} = \dfrac{1}{{12}}\\
\Rightarrow f = 12cm\\
20.Bd = 80cm,d' = - 40cm\\
|\dfrac{{d'}}{d}| = \dfrac{{h'}}{h} = \dfrac{2}{4} \Rightarrow |d'| = 0,5d \Rightarrow d' = - 0,5d(d' < 0)\\
d - |d'| = 40 \Rightarrow d - 0,5d = 40\\
\Rightarrow d = 80cm \Rightarrow d' = - 0,5d = 40cm\\
21.B.16cm\\
DK:f < d < 2f \Rightarrow 12 < d < 24\\
d + d' = 64 \Rightarrow d' = 64 - d\\
\dfrac{1}{f} = \dfrac{1}{d} + \dfrac{1}{{d'}} \Rightarrow \dfrac{1}{{12}} = \dfrac{1}{d} + \dfrac{1}{{64 - d}}\\
\Rightarrow d = 16cm\\
22.D\\
23.D\\
24.{i_{gh}} = 48,59^\circ \\
\sin {i_{gh}} = \dfrac{{{n_2}}}{{{n_1}}} = \dfrac{1}{{\dfrac{4}{3}}} = 0,75 \Rightarrow {i_{gh}} = 48,59^\circ \\
25.A\\
26.D\\
27.B\\
D = \dfrac{1}{f} \Rightarrow f = \dfrac{1}{D} = \dfrac{1}{{ - 4}} = - 0,25m = - 25cm\\
\dfrac{1}{f} = \dfrac{1}{d} + \dfrac{1}{{d'}} \Rightarrow \dfrac{1}{{ - 25}} = \dfrac{1}{{25}} + \dfrac{1}{{d'}}\\
\Rightarrow d' = - 12,5cm\\
|\dfrac{{d'}}{d}| = \dfrac{{h'}}{h} \Rightarrow h' = |\dfrac{{d'}}{d}|h = |\dfrac{{ - 12,5}}{{25}}|h = 0,5h\\
28.D\\
{e_c} = - L|\dfrac{{\Delta i}}{{\Delta t}}|\\
29.B.\alpha = 0^\circ \\
\phi = NBS\cos \alpha \Rightarrow \cos \alpha = \dfrac{\phi }{{NBS}} = \dfrac{{{{10}^{ - 6}}}}{{1.0,{{05}^2}{{.4.10}^{ - 4}}}} = 1\\
\Rightarrow \alpha = 0^\circ \\
30.D.10V\\
{e_c} = \dfrac{{\Delta \phi }}{{\Delta t}} = \dfrac{{1,6 - 0,6}}{{0,1}} = 10V
\end{array}\)
