a/ \(n\sin i=\sin r\Leftrightarrow n\sin i=\cos i\left(r=90^0-i\right)\)

\(\Rightarrow\tan i=\dfrac{1}{n}=\dfrac{1}{\sqrt{3}}\Rightarrow i=\dfrac{\pi}{6}\)

b/ \(i\ge i_{gh};\sin i_{gh}=\dfrac{1}{n}\Leftrightarrow\sin45=\dfrac{1}{n}\Rightarrow n=\sqrt{2}\)

\(\dfrac{1}{f}=\dfrac{1}{d_1}+\dfrac{1}{d_1'};\left\{{}\begin{matrix}d_2=d_1-2\\d_2'=d_1'+b\end{matrix}\right.;\dfrac{A"B"}{A'B'}=\dfrac{5}{3}\Leftrightarrow\dfrac{d_2'.d_1}{d_1'.d_2}=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{\left(d_1'+b\right).d_1}{d_1'.\left(d_1-2\right)}=\dfrac{5}{3}\)\(\Leftrightarrow\dfrac{20\left(d_1'+b\right)}{d_1'\left(20-2\right)}=\dfrac{5}{3}\)

\(\dfrac{A"B"}{A'B'}=\dfrac{5}{3}\Leftrightarrow\dfrac{k_2}{k_1}=\dfrac{f-d_1}{f-d_2}=\dfrac{5}{3}\Leftrightarrow\dfrac{f-20}{f-20+2}=\dfrac{5}{3}\Rightarrow f=....\)

\(\Rightarrow d_1'=\dfrac{fd_1}{d_1-f}=...;\dfrac{20\left(d_1'+b\right)}{18d_1'}=\dfrac{5}{3}\Rightarrow b=...\)

Bai 1:

\(\omega=\sqrt{\dfrac{k}{m}}\Rightarrow m=\dfrac{k}{\omega^2};v=x'=-\omega A\sin\left(\omega t+\varphi\right)\)

\(W_d=\dfrac{1}{2}mv^2=\dfrac{1}{2}\dfrac{k}{\omega^2}.\omega^2A^2.\sin^2\left(\omega t+\varphi\right)=\dfrac{1}{4}kA^2\left[1-\cos\left(2\omega t+2\varphi\right)\right]\)

\(\Rightarrow W_d=\dfrac{1}{4}.\omega^2.m.A^2\left[1-\cos\left(2\omega t+2\varphi\right)\right]=\dfrac{1}{4}.100.0,2.4\left[1-\cos\left(20t\right)\right]=20\left[1-\cos\left(20t\right)\right]\)

Bai 2:

\(W_t=\dfrac{1}{2}kx^2=m\omega^2A^2.\dfrac{\cos\left(2\omega t+\varphi\right)+1}{4}=\dfrac{1}{4}m\omega^2.A^2\left[1+\cos\left(2\omega t+2\varphi\right)\right]\)

\(\Rightarrow W_t=\dfrac{1}{4}.0,1.100.36.\left[1+\cos\left(2.10t\right)\right]=90.\left[1+\cos20t\right]\)

\(\sin i=n\sin r\Leftrightarrow\sin45^0=1,4.\sin r\Rightarrow r=...\)

\(\Rightarrow r'=A-r;n\sin r'=\sin i'\Leftrightarrow n\sin\left(A-r\right)=\sin i'\Rightarrow i'=....\)

\(D=i+i'-A=....\)

\(OC_c=40cm\)

a/ Các vật cách mắt gần nhất 25cm, nghĩa là ảnh tạo thành hiện ở cực cận

\(\Rightarrow\dfrac{1}{f}=D=\dfrac{1}{d}-\dfrac{1}{OC_C}=\dfrac{1}{0,25}-\dfrac{1}{0,4}=1,5\)

b/\(\dfrac{1}{f}=D=\dfrac{1}{d}-\dfrac{1}{OC_C}\Leftrightarrow1=\dfrac{1}{d}-\dfrac{1}{0,4}\Rightarrow d=\dfrac{2}{7}\left(m\right)\)

\(\dfrac{1}{f}=\dfrac{1}{d}+\dfrac{1}{d'}\Leftrightarrow-\dfrac{1}{20}=\dfrac{1}{d}+\dfrac{1}{d'}\)

\(d-\left|d'\right|=10\Leftrightarrow d+d'=10\left(d'< 0\right)\)

\(\Rightarrow d=\dfrac{d'f}{d'-f}=\dfrac{-20.\left(10-d\right)}{10-d+20}=\dfrac{20d-200}{30-d}\Leftrightarrow30d-d^2=20d-200\Leftrightarrow d=...\left(cm\right)\)

Công thức là xong thôi mà?

\(\sin i=n\sin r\Leftrightarrow\sin60^0=1,5\sin r\Rightarrow r=...\)

Vâng, em làm được rồi ạ. Em cảm mơn anh nhaa. Lát anh ngủ ngon nhé :3

À vâng em hiểu rồi ạ. Câu này thì sao anh nhỉ? Vẫn là xét đạo hàm tại x0 ạ