Một kính hiển vi với vật kính có tiêu cự 0,4cm, thị kính có tiêu cự f2=4cm

* Hướng dẫn giải

Chọn D

+ Sơ đồ tạo ảnh: 

$\begin{array}{l}\underset{d_1}{\underbracket{AB}}\stackrel{O_1}{\to }\underset{\mathcal{l}=f_1+\delta +f_2}{\underbracket{\underset{d_1^{/}{d}_2}{\underbracket{A_1{B}_1}}}}\stackrel{O_2}{\to }\underset{0}{\underbracket{\underset{d_2^{/}{d}_M=\left[O{C}_C;O{C}_V\right]}{\underbracket{A_2{B}_2}}}}\\ \stackrel{Mat}{\to }V\end{array}$

$\begin{array}{l}{d}_1=0,41\Rightarrow d_1^{/}=\frac{d_1{f}_1}{d_1-f_1}=16,4\\ \Rightarrow d_2=\mathcal{l}-d_1^{/}=3,6\Rightarrow d_2^{/}=\frac{d_2{f}_2}{d_2-f_2}=-36\end{array}$

$\Rightarrow d_M=-d_2^{/}=36\in \left[O{C}_C;O{C}_V\right]$

=>Mắt nhìn thấy ảnh.

$$\begin{gathered} \Rightarrow \left|k\right|=\left|k_1{k}_2\right|=\left|\frac{d_1^{/}{d}_2^{/}}{d_1{d}_2}\right|=\left|\frac{16,4.36}{0,41.3,6}\right|=400 \\ \left\{\begin{array}{l}\alpha =\approx \tan \alpha =\frac{A_2{B}_2}{d_M}=\frac{\left|k\right|AB}{d_M}=\\ 400.\frac{0,01}{36}=\frac{1}{9}\left(rad/s\right)\\ G=\frac{\alpha }{{\alpha }_0}\approx \frac{\tan \alpha }{\tan {\alpha }_0}=\frac{\frac{A_2{B}_2}{d_M}}{\frac{AB}{O{C}_C}}=\left|k\right|\frac{O{C}_C}{d_M}=\\ 400.\frac{27}{36}=300\end{array}\right. \end{gathered}$$