[物理] 平行光進入拋面後, 所有反射光都會通過焦點的數學證明

前設是反射定律。
Consider a general parabola $y^2=4px$.
The focus of the parabola is $(p, 0)$.

Let a general horizontal line be $y=k$, where $k$ is constant.
It cuts the parabola at $( \frac{k^2}{4p}, k)$.

It can be shown that the general normal line at $( \frac{k^2}{4p}, k)$ is:
$y=-\frac{k}{2p}x+k+\frac{k^3}{8p^2}$

and the reflection of the horizontal line through the normal at $( \frac{k^2}{4p}, k)$ is:
$y = -\frac{4pk}{4p^2 - k^2}x + \frac{k^3}{4p^2 - k^2} + k$, where $k \neq 2p$.

Substitute $(p, 0)$ into $y = -\frac{4pk}{4p^2 - k^2}x + \frac{k^3}{4p^2 - k^2} + k$,

we have
$L.H.S=0$

$R.H.S$ $\\=-\frac{4pk}{4p^2 - k^2}p + \frac{k^3}{4p^2 - k^2} + k
\\=\frac{-4p^2k+k^3+4p^2k - k^3}{4p^2 - k^2}
\\=0$

$L.H.S=R.H.S$

for $k=2p$, the reflection line is a vertical line pass through $(p,k)$ and $(p,0)$.
Thus, the reflection line passes the focus.

筆者利用Geogebra將反射的情況畫出來了, 可參考
http://tube.geogebra.org/m/2188507

[物理] 感生電動勢問題 (yahoo知識+：大一普物 電磁學 求詳解?)

剛剛在知識+看到這個問題，原本沒打算答，不過看到解答好像錯了：

我覺得有問題的原因是 B 在 較近wire和較遠wire的兩側在移動時，其中一邊上的dB/dt應該在增加，另一邊在減少，預期應該會看到一些像$\frac{1}{r}-\frac{1}{r+w}$的東西，他用的dB/dt是moving rod in magnetic field的情況，和本題有點不一樣，所以我也算一次：

The B field produced by the long wire is $B= \frac{\mu_0I}{2 \pi s}$

so $\phi = \int B \cdot da=\frac{\mu_0I}{2 \pi} \int_{r}^{r+w} \frac{l}{s} ds$

$\frac{\mathrm{d \phi} }{\mathrm{d} t} = \frac{\mu_0I}{2 \pi} \frac{\mathrm{d} }{\mathrm{d} t} \int_{r}^{r+w} \frac{l}{s} ds$

$= \frac{\mu_0Il}{2 \pi} (\frac{1}{r+w}-\frac{1}{r}) \frac{\mathrm{d r} }{\mathrm{d} t}=\frac{\mu_0Il}{2 \pi} (\frac{-w}{r(r+w)})v$

Thus $\varepsilon=-\frac{\mathrm{d \phi} }{\mathrm{d} t}=\frac{\mu_0Ilwv}{2 \pi r(r+w)}$

Since $\varepsilon=iR$, where i is the current in the loop,

$\frac{\mu_0Ilwv}{2 \pi r(r+w)}=iR$

$i=\frac{\mu_0Ilwv}{2 \pi r(r+w)R}$

很久沒碰了，不過看起來應該這樣才對。