Câu 1: 

Góc tới là:

\(i = {90^o} - {60^o} = {30^o}\)

Câu 2: 

a) Do \({N_2} > {N_1} \Rightarrow \) đó là máy tăng thế.

b) Ta có: 

\(\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{N_1}}}{{{N_2}}} \Rightarrow \dfrac{{500}}{{4000}} = \dfrac{{400}}{{{U_2}}} \Rightarrow {U_2} = 3200V\)

c) Công suất hao phí là:

\({P_{hp}} = {\left( {\dfrac{P}{{{U_2}}}} \right)^2}R = {\left( {\dfrac{{{{10}^6}}}{{3200}}} \right)^2}.40 = 3906250W\)

d) Để công suất hao phí giảm 1 nửa thì HĐT cần tăng \(\sqrt 2 \) lần:

\({U_2}' = {U_2}\sqrt 2  = 3200\sqrt 2 V\)

Câu 3: 

Ta có: 

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} \Rightarrow \dfrac{1}{4} = \dfrac{{24}}{{OA'}} \Rightarrow OA' = 96cm\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' + OF'}} = \dfrac{f}{{f + 96}}\\
 \Rightarrow \dfrac{1}{4} = \dfrac{f}{{f + 96}} \Rightarrow f = 32cm
\end{array}\)

Câu 4: 

Ta có: 

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{20}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{{16}}{{OA' - 16}}\\
 \Rightarrow \dfrac{{20}}{{OA'}} = \dfrac{{16}}{{OA' - 16}} \Rightarrow OA' = 80cm
\end{array}\)

Câu 6: 

Ta có: 

\(\begin{array}{l}
f = \dfrac{{25}}{G} = \dfrac{{25}}{{2,5}} = 10cm\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{1}{5}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' + OF'}} = \dfrac{{10}}{{OA' + 10}}\\
 \Rightarrow \dfrac{1}{5} = \dfrac{{10}}{{OA' + 10}} \Rightarrow OA' = 40cm\\
 \Rightarrow OA = 8cm
\end{array}\)