计算专题3 凸透镜的二阶像散

本文最后更新于 2024年7月31日 晚上

计算专题3 凸透镜的二阶像散

前言

这一篇专题旨在计算我在竞赛阶段算过三遍却没有一次算对的问题,之前CPhO-S某张试卷中的“凸透镜的像散”. 这是对之前的自我的一次挑战,也是为了14号开始去某物理竞赛机构打工而进行的一次计算能力的复健. 同时,毕竟8月中下旬就要开学,在此之前我还是希望自己的能力,至少是计算能力,要能够维持一定的水准.

希望这次的计算专题能够达到我想要的效果.

问题描述

如图所示,一个凸透镜,两面球面半径分别为r1r_1r2r_2,光线平行主光轴入射,距离主光轴rr. 凸透镜的像散是指边缘光汇聚点与傍轴光汇聚点(即焦点)之间的一段距离Δ\Delta,本题所求即为在二阶近似下的Δ\Delta值(保留O(r2)O(r^2)阶项,rfr\ll fff为焦距).

计算

角度设定和初步运算

如图所示,设定所用到的各个角度.

\begin{aligned}\\\\ \end{aligned}

设第一次入射角为ii,折射角为jj,第二球面的法线在光线入射处与主光轴夹角为α\alpha,则第二次的入射角为ij+αi-j+\alpha,折射角为β\beta. 设最终出射光线与主光轴的交点距离光心xx,又因为可做薄透镜近似,xx实际上是交点与第二球面顶点之间的距离.

由几何关系,易知:

sini=rr1,sinα=rr2,x=rtan(βα)\begin{aligned} \sin i=\frac{r}{r_1}\,,\quad\sin\alpha=\frac{r}{r_2}\,,\quad x=\frac{r}{\tan(\beta-\alpha)}\\\\ \end{aligned}

又有折射定律

nsinj=sini,nsin(ij+α)=sinβ\begin{aligned} n\sin j=\sin i\,,\quad n\sin(i-j+\alpha)=\sin\beta\\\\ \end{aligned}

所以

x=rcos(βα)sin(βα)=rcosβcosα+sinβsinαsinβcosαcosβsinα=rcosβ1r2r22+nsin(ij+α)rr2nsin(ij+α)1r2r22cosβrr2Δ=fx=frcosβ1r2r22+nsin(ij+α)rr2nsin(ij+α)1r2r22cosβrr2=1(n1)(1r1+1r2)rcosβ1r2r22+nsin(ij+α)rr2nsin(ij+α)1r2r22cosβrr2\begin{aligned} x&=r\cdot\frac{\cos(\beta-\alpha)}{\sin(\beta-\alpha)}=r\cdot\frac{\cos\beta\cos\alpha+\sin\beta\sin\alpha}{\sin\beta\cos\alpha-\cos\beta\sin\alpha}\\\\ &=r\cdot\frac{\cos\beta\cdot\sqrt{1-\frac{r^2}{r_2^2}}+n\sin(i-j+\alpha)\cdot\frac{r}{r_2}}{n\sin(i-j+\alpha)\cdot\sqrt{1-\frac{r^2}{r_2^2}}-\cos\beta\cdot\frac{r}{r_2}}\\\\ \Longrightarrow&\quad\Delta=f-x\\\\&=f-r\cdot\frac{\cos\beta\cdot\sqrt{1-\frac{r^2}{r_2^2}}+n\sin(i-j+\alpha)\cdot\frac{r}{r_2}}{n\sin(i-j+\alpha)\cdot\sqrt{1-\frac{r^2}{r_2^2}}-\cos\beta\cdot\frac{r}{r_2}}\\\\&=\frac{1}{(n-1)(\frac{1}{r_1}+\frac{1}{r_2})}-r\cdot\frac{\cos\beta\cdot\sqrt{1-\frac{r^2}{r_2^2}}+n\sin(i-j+\alpha)\cdot\frac{r}{r_2}}{n\sin(i-j+\alpha)\cdot\sqrt{1-\frac{r^2}{r_2^2}}-\cos\beta\cdot\frac{r}{r_2}}\\\\ \end{aligned}

至此,我们得到了像散的初步表达式,之后进行较为细致的小量处理.

深入计算与小量处理

目前还有cosβ\cos\betasin(ij+α)\sin(i-j+\alpha)两个量没有具体表达式,先对这两个量进行化简.

sin(ij+α)=sini(cosαcosj+sinαsinj)+cosi(sinαcosjcosαsinj)=rr1(1r2r221r2n2r12+rr2rnr1)+1r2r12(rr21r2n2r121r2r22rnr1)\begin{aligned} \sin(i-j+\alpha)&=\sin i(\cos \alpha\cos j+\sin\alpha\sin j)+\cos i(\sin\alpha\cos j-\cos\alpha\sin j)\\\\&=\frac{r}{r_1}(\sqrt{1-\frac{r^2}{r_2^2}}\cdot\sqrt{1-\frac{r^2}{n^2r^2_1}}+\frac{r}{r_2}\cdot\frac{r}{nr_1})+\\\\&\quad\,\sqrt{1-\frac{r^2}{r_1^2}}\cdot(\frac{r}{r_2}\cdot\sqrt{1-\frac{r^2}{n^2r^2_1}}-\sqrt{1-\frac{r^2}{r_2^2}}\cdot\frac{r}{nr_1})\\\\ \end{aligned}

注意到,sin(ij+α)\sin(i-j+\alpha)的最低阶项就是一阶项,所以xx表达式里分母中最低阶的项就是一阶项,再与前面的rr相乘后分母变为零阶. 又因为整个xx一定是正常量级,所以分子最低阶必须是零阶. 而问题要求最终结果保留到二阶量,所以分子、与rr相乘后的分母都需要保留到二阶项,这就要求sin(ij+α)\sin(i-j+\alpha)cosβ\cos\beta在前期计算时必须保留三阶项.

继续:

sin(ij+α)rr1(1r22r22r22n2r12+r2nr1r2)+(1r22r12)(rr2r32n2r12r2rnr1+r32nr1r22)rr1r32r1r22r32n2r13+r3nr12r2+rr2r32n2r12r2rnr1+r32nr1r22r32r12r2+r32nr13=(1r1+1r21nr1)r+(12r1r2212n2r13+1nr12r212n2r12r2+12nr1r2212r12r2+12nr13)r3=(1r1+1r21nr1)r+12nr1(n1)(1nr11r2)(1r1+1r2)\begin{aligned} \sin(i-j+\alpha)&\approx\frac{r}{r_1}(1-\frac{r^2}{2r_2^2}-\frac{r^2}{2n^2r^2_1}+\frac{r^2}{nr_1r_2})+\\\\&\quad\,(1-\frac{r^2}{2r^2_1})(\frac{r}{r_2}-\frac{r^3}{2n^2r_1^2r_2}-\frac{r}{nr_1}+\frac{r^3}{2nr_1r_2^2})\\\\&\approx\frac{r}{r_1}-\frac{r^3}{2r_1r^2_2}-\frac{r^3}{2n^2r_1^3}+\frac{r^3}{nr^2_1r_2}+\\\\&\quad\,\frac{r}{r_2}-\frac{r^3}{2n^2r_1^2r_2}-\frac{r}{nr_1}+\frac{r^3}{2nr_1r_2^2}-\frac{r^3}{2r_1^2r_2}+\frac{r^3}{2nr_1^3}\\\\&=(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})r+(-\frac{1}{2r_1r_2^2}-\frac{1}{2n^2r_1^3}+\\\\&\quad\,\frac{1}{nr_1^2r_2}-\frac{1}{2n^2r_1^2r_2}+\frac{1}{2nr_1r_2^2}-\frac{1}{2r_1^2r_2}+\frac{1}{2nr_1^3})r^3\\\\&=(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})r+\frac{1}{2nr_1}(n-1)(\frac{1}{nr_1}-\frac{1}{r_2})(\frac{1}{r_1}+\frac{1}{r_2})\\\\ \end{aligned}

最后一行的化简结果要鸣谢另一位攀登计划的同学,这种观察能力令我深表佩服.

cosβ\cos\beta要利用sin(ij+α)\sin(i-j+\alpha)的结果来计算,这时要保留到三阶项,意味着只要用到sin(ij+α)\sin(i-j+\alpha)的一阶成分.

cosβ=1n2sin2(ij+α)1n2(1r1+1r21nr1)2r21n22(1r1+1r21nr1)2r2\begin{aligned} \cos\beta&=\sqrt{1-n^2\sin^2(i-j+\alpha)}\\\\&\approx\sqrt{1-n^2(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})^2r^2}\\\\&\approx1-\frac{n^2}{2}(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})^2r^2\\\\ \end{aligned}

这时我们可以开始计算xx的分子与分母.

分子保留二阶项:

A1r2r22[1n22(1r1+1r21nr1)2r2]+(1r1+1r21nr1)nr2r2+(12r1r2212n2r13+1nr12r212n2r12r2+12nr1r2212r12r2+12nr13)nr4r21r22r22n22(1r1+1r21nr1)2r2+(1r1+1r21nr1)nr2r2=1(12r22+n22r12+n22r22+12r12+n2r1r2nr12nr1r2nr1r2nr22+1r1r2)r2\begin{aligned} A&\approx\sqrt{1-\frac{r^2}{r_2^2}}\cdot[1-\frac{n^2}{2}(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})^2r^2]+\\\\&\quad\,(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})\frac{nr^2}{r_2}+(-\frac{1}{2r_1r_2^2}-\frac{1}{2n^2r_1^3}+\\\\&\quad\,\frac{1}{nr_1^2r_2}-\frac{1}{2n^2r_1^2r_2}+\frac{1}{2nr_1r_2^2}-\frac{1}{2r_1^2r_2}+\frac{1}{2nr_1^3})\frac{nr^4}{r_2}\\\\&\approx1-\frac{r^2}{2r_2^2}-\frac{n^2}{2}(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})^2r^2+(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})\frac{nr^2}{r_2}\\\\&=1-(\frac{1}{2r_2^2}+\frac{n^2}{2r_1^2}+\frac{n^2}{2r_2^2}+\frac{1}{2r_1^2}+\\\\&\quad\,\frac{n^2}{r_1r_2}-\frac{n}{r_1^2}-\frac{n}{r_1r_2}-\frac{n}{r_1r_2}-\frac{n}{r_2^2}+\frac{1}{r_1r_2})r^2\\\\ \end{aligned}

注意观察,r2r^2的系数是可以分组提取公因子,进行因式分解的:

12(1r12+1r22+2r1r2)n(1r12+1r22+2r1r2)+n22(1r12+1r22+2r1r2)=12(n1)2(1r1+1r2)2\begin{aligned} &\frac{1}{2}(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{2}{r_1r_2})-n(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{2}{r_1r_2})+\frac{n^2}{2}(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{2}{r_1r_2})\\\\&=\frac{1}{2}(n-1)^2(\frac{1}{r_1}+\frac{1}{r_2})^2\\\\ \end{aligned}

故而分子保留到二阶小量化为

A112(n1)2(1r1+1r2)2r2\begin{aligned} A&\approx1-\frac{1}{2}(n-1)^2(\frac{1}{r_1}+\frac{1}{r_2})^2r^2\\\\ \end{aligned}

分母要除去rr,有:

Bnr1r2r22[(1r1+1r21nr1)r+12nr1(n1)(1nr11r2)(1r1+1r2)][112(n1)2(1r1+1r2)2]1r2n(1r1+1r21nr1)+n12r1(1r1+1r2)(1nr11r2)r2n2r22(1r1+1r21nr1)r21r2+n22r2(1r1+1r21nr1)2r2=(n1)(1r1+1r2)+n12r1(1r1+1r2)(1nr11r2)r2+nr22r2(1r1+1r21nr1)(n1)(1r1+1r2)=(n1)(1r1+1r2)+(n1)(1r1+1r2)[12r1(1nr11r2)+n2r2(1r1+1r21nr1)]r2\begin{aligned} B&\approx \frac{n}{r}\sqrt{1-\frac{r^2}{r_2^2}}[(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})r+\frac{1}{2nr_1}(n-1)(\frac{1}{nr_1}-\frac{1}{r_2})(\frac{1}{r_1}+\frac{1}{r_2})]\\\\&\quad\,-[1-\frac{1}{2}(n-1)^2(\frac{1}{r_1}+\frac{1}{r_2})^2]\cdot\frac{1}{r_2}\\\\&\approx n(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})+\frac{n-1}{2r_1}(\frac{1}{r_1}+\frac{1}{r_2})(\frac{1}{nr_1}-\frac{1}{r_2})r^2\\\\&\quad\,\underline{-\frac{n}{2r_2^2}(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})r^2}-\frac{1}{r_2}\underline{+\frac{n^2}{2r_2}(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})^2r^2}\\\\&=(n-1)(\frac{1}{r_1}+\frac{1}{r_2})+\frac{n-1}{2r_1}(\frac{1}{r_1}+\frac{1}{r_2})(\frac{1}{nr_1}-\frac{1}{r_2})r^2\\\\&\quad\,\underline{+\frac{nr^2}{2r_2}(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})(n-1)(\frac{1}{r_1}+\frac{1}{r_2})}\\\\&=(n-1)(\frac{1}{r_1}+\frac{1}{r_2})+(n-1)(\frac{1}{r_1}+\frac{1}{r_2})[\frac{1}{2r_1}(\frac{1}{nr_1}-\frac{1}{r_2})+\\\\&\quad\,\frac{n}{2r_2}(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})]r^2\\\\ \end{aligned}

进入下一个环节:

Δ1(n1)(1r1+1r2)1(n1)(1r1+1r2){112(n1)2(1r1+1r2)2r2[n2r2(1r1+1r2)+12nr121r1r2]r2}=12(n1)2(1r1+1r2)2+n2r2(1r1+1r2)+12nr121r1r2(n1)(1r1+1r2)r2\begin{aligned} \Delta&\approx\frac{1}{(n-1)(\frac{1}{r_1}+\frac{1}{r_2})}-\frac{1}{(n-1)(\frac{1}{r_1}+\frac{1}{r_2})}\{1-\frac{1}{2}(n-1)^2(\frac{1}{r_1}+\frac{1}{r_2})^2r^2\\\\&\quad\,-[\frac{n}{2r_2}(\frac{1}{r_1}+\frac{1}{r_2})+\frac{1}{2nr_1^2}-\frac{1}{r_1r_2}]r^2\}\\\\&=\frac{\frac{1}{2}(n-1)^2(\frac{1}{r_1}+\frac{1}{r_2})^2+\frac{n}{2r_2}(\frac{1}{r_1}+\frac{1}{r_2})+\frac{1}{2nr_1^2}-\frac{1}{r_1r_2}}{(n-1)(\frac{1}{r_1}+\frac{1}{r_2})}r^2\\\\ \end{aligned}

至此,最终答案就已经被算出来了.(虽然这已经是十几天之后的结果,在机构改试卷的工作确实非常忙碌,根本无法抽身).

之前一直认为算不对的问题终于解决了,但是我其实并不惊讶. 毕竟这个答案形式本身非常丑陋,而且关于r1r_1r2r_2也不是对称的,所以我有理由猜测之前的所谓“算不对”实际上是“算不出CPhO-S所给出的原答案”,而且没有这么多时间来仔细打磨自己的计算过程,也就没有办法将答案化简到能用CASIO数值验证的程度. CPhO-S的原答案如下:

Δ=(11n3)nr13+n1r23+n(n21)(1r1+1r21nr1)3+2(n1)3(1r1+1r2)36(n1)2(1r1+1r2)2r2\begin{aligned} \Delta=\frac{(1-\frac{1}{n^3})\frac{n}{r^3_1}+\frac{n-1}{r^3_2}+n(n^2-1)(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{nr_1})^3+2(n-1)^3(\frac{1}{r_1}+\frac{1}{r_2})^3}{6(n-1)^2(\frac{1}{r_1}+\frac{1}{r_2})^2}r^2\\\\ \end{aligned}

这两个答案显然不太可能用肉眼证明是相等的. 所以说其实这一次只是打破了之前一直有的一个执念吧.

总结与反思

这一次的计算专题比我想象中的要容易许多. 这并不是说我的能力没有提升,只是说明我比我想象中的自己要强一些而已. 可惜的是,我并没有在打工前写完这一篇计算专题,而是搁到了快8月份才写完,之前定下的目标确实是没有达成,应该反思.

了结了之前在我的笔记本上备注为**“计算高峰,延年益寿”**的这道题目之后,我还有什么想要挑战的题目呢?我计划在开学前至少再写一篇计算专题,至于开学之后要做什么其实我自己也不太清楚. 这次在机构打工的最大收获其实就是认识了很多未来的同学,我对大学生活也有了更深入的了解. 希望接下来的路能够让我有更大的提升吧,至少我会继续挑战下去.


计算专题3 凸透镜的二阶像散
https://physnya.top/2024/07/09/计算专题3 凸透镜的二阶像散/
作者
菲兹克斯喵
发布于
2024年7月9日
更新于
2024年7月31日
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