确定光的折射点

manthanein

如图所示，光线经A点，跨过分界的直线CD，折射后经过B点。AC垂直于CD，BD垂直于CD。设AC=a，BD=b，CD=m，光线在折射前的速度为v1，折射后的速度为v2。E为折射点。
如何确定E的位置？（比如说，求出DE或CE）

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倪举鹏

解方程不就得了  问题太明显

manthanein

我知道解方程，可真的解起来呢？

KeyTo9_Fans

$|DE|=argmin_x\{\sqrt{a^2+(m-x)^2}/v_1+\sqrt{b^2+x^2}/v_2\}$

函数求导，

导数$=0$，

解方程。

人工太麻烦，

数学工具用起来

manthanein

求导我知道，人工解这个方程够麻烦的，哪位能帮帮忙呢

manthanein

毕竟我没有什么神通广大的数学工具……

cn8888

这个问题很简单!

数学星空

对\(f(x)=\frac{\sqrt{a^2+(m-x)^2}}{v_1}+\frac{\sqrt{b^2+x^2}}{v_2}=t\)......................(1)

关于\(x\)求导：

得到\(\frac{a-x}{v_1(\sqrt{(m-x)^2+a^2})}= \frac{x}{v_2(\sqrt{x^2+b^2})}=k\)..................(2)

若记

\(v_1=\frac{1}{n_1},v_2=\frac{1}{n_2}\)

\(\frac{m-x}{\sqrt{(m-x)^2+a^2}}=\sin(\theta_1),\frac{x}{\sqrt{x^2+b^2}}=\sin(\theta_2)\)

则有光折射公式：\(\frac{n_1}{\sin(\theta_1)}=\frac{n_2}{\sin(\theta_2)}\).............(3)

注意：这里的\(\theta_1=\frac{\pi}{2}-\angle AEC,\theta_2=\frac{\pi}{2}-\angle BED\)与物理课本上的入射角与反射角是一致的

关于(1)(2)我们得到：

\(-a^2k^2v_1^2-k^2m^2v_1^2+2k^2mv_1^2x-k^2v_1^2x^2+m^2-2mx+x^2=0\)..................(4)

\(-b^2k^2v_2^2-k^2v_2^2x^2+x^2=0\)........................................................................................(5)

\(ktv_1^2v_2^2-mv_2^2-v_1^2x+v_2^2x=0\)...............................................................................(6)


对上面三个方程消元求解得到：

\(b^2m^2v_2^2-2b^2mv_2^2x+(-a^2v_1^2+b^2v_2^2-m^2v_1^2+m^2v_2^2)x^2+2m(v_1-v_2)(v_1+v_2)x^3-(v_1-v_2)(v_1+v_2)x^4=0\).........(7)


\(m^4-2m^2(a^2v_1^2+b^2v_2^2+m^2v_1^2+m^2v_2^2)k^2+(a^4v_1^4-2a^2b^2v_1^2v_2^2+2a^2m^2v_1^4+4a^2m^2v_1^2v_2^2+b^4v_2^4+4b^2m^2v_1^2v_2^2+2b^2m^2v_2^4+m^4v_1^4+4m^4v_1^2v_2^2+m^4v_2^4)k^4-2v_1^2v_2^2(a^4v_1^2-a^2b^2v_1^2-a^2b^2v_2^2+2a^2m^2v_1^2+a^2m^2v_2^2+b^4v_2^2+b^2m^2v_1^2+2b^2m^2v_2^2+m^4v_1^2+m^4v_2^2)k^6+v_1^4v_2^4(a^2+2ab+b^2+m^2)(a^2-2ab+b^2+m^2)k^8=0\)....................(8)


\((a^2+2ab+b^2+m^2)(a^2-2ab+b^2+m^2)(a^2v_1^2v_2^2-a^2v_2^4-b^2v_1^4+b^2v_1^2v_2^2+m^2v_1^2v_2^2)^2-2v_1^2v_2^2(a^6v_1^4v_2^2-3a^6v_1^2v_2^4+2a^6v_2^6+a^4b^2v_1^6-2a^4b^2v_1^4v_2^2+4a^4b^2v_1^2v_2^4-3a^4b^2v_2^6+3a^4m^2v_1^4v_2^2-5a^4m^2v_1^2v_2^4+3a^4m^2v_2^6-3a^2b^4v_1^6+4a^2b^4v_1^4v_2^2-2a^2b^4v_1^2v_2^4+a^2b^4v_2^6+2a^2b^2m^2v_1^6-3a^2b^2m^2v_1^4v_2^2-3a^2b^2m^2v_1^2v_2^4+2a^2b^2m^2v_2^6+3a^2m^4v_1^4v_2^2-a^2m^4v_1^2v_2^4+a^2m^4v_2^6+2b^6v_1^6-3b^6v_1^4v_2^2+b^6v_1^2v_2^4+3b^4m^2v_1^6-5b^4m^2v_1^4v_2^2+3b^4m^2v_1^2v_2^4+b^2m^4v_1^6-b^2m^4v_1^4v_2^2+3b^2m^4v_1^2v_2^4+m^6v_1^4v_2^2+m^6v_1^2v_2^4)t^2+v_1^4v_2^4(a^4v_1^4-6a^4v_1^2v_2^2+6a^4v_2^4-6a^2b^2v_1^4+10a^2b^2v_1^2v_2^2-6a^2b^2v_2^4+2a^2m^2v_1^4-2a^2m^2v_1^2v_2^2+6a^2m^2v_2^4+6b^4v_1^4-6b^4v_1^2v_2^2+b^4v_2^4+6b^2m^2v_1^4-2b^2m^2v_1^2v_2^2+2b^2m^2v_2^4+m^4v_1^4+4m^4v_1^2v_2^2+m^4v_2^4)t^4+2v_1^6v_2^6(a^2v_1^2-2a^2v_2^2-2b^2v_1^2+b^2v_2^2-m^2v_1^2-m^2v_2^2)t^6+t^8v_1^8v_2^8=0\)....................(9)

manthanein

原理我是懂的。计算看上去够麻烦，不过还是谢谢了。

gxqcn

好像小时候，学光学的时候，也有楼主同样的疑问。