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quarto/precalc/trig_functions.qmd
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@@ -269,7 +269,7 @@ plot(sin, 0, 4pi)
The graph shows two periods. The wavy aspect of the graph is why this function is used to model periodic motions, such as the amount of sunlight in a day, or the alternating current powering a computer.
From this graph - or considering when the $y$ coordinate is $0$ - we see that the sine function has zeros at any integer multiple of $\pi$, or $k\pi$, $k$ in $\dots,-2,-1, 0, 1, 2, \dots$.
From this graph---or considering when the $y$ coordinate is $0$---we see that the sine function has zeros at any integer multiple of $\pi$, or $k\pi$, $k$ in $\dots,-2,-1, 0, 1, 2, \dots$.
The cosine function is similar, in that it has the same domain and range, but is "out of phase" with the sine curve. A graph of both shows the two are related:
@@ -693,14 +693,144 @@ atan(y, x)
##### Example
A (white) light shining through a [prism](http://tinyurl.com/y8sczg4t) will be deflected depending on the material of the prism and the angles involved (refer to the link for a figure). The relationship can be analyzed by tracing a ray through the figure and utilizing Snell's law. If the prism has index of refraction $n$ then the ray will deflect by an amount $\delta$ that depends on the angle, $\alpha$ of the prism and the initial angle ($\theta_0$) according to:
A (white) light shining through a [dispersive prism](https://en.wikipedia.org/wiki/Dispersive_prism) will be deflected depending on the material of the prism and the angles involved. The relationship can be analyzed by tracing a ray through the figure and utilizing Snell's law which relates the angle of incidence with the angle of refraction through two different media through:
$$
\delta = \theta_0 - \alpha + \arcsin(n \sin(\alpha - \arcsin(\frac{1}{n}\sin(\theta_0)))).
n_0 \sin(\theta_0) = n_1 \sin(\theta_1)
$$
If $n=1.5$ (glass), $\alpha = \pi/3$ and $\theta_0=\pi/6$, find the deflection (in radians).
:::{#fig-snells-law-prism}
```{julia}
#| echo: false
p1 = let
gr()
plot(; empty_style..., aspect_ratio=:equal)
n₀,n₁,n₂ = 1,3, 1
θ₀ = pi/7
α = pi/8
θ₁ = asin(n₀/n₁* θ₀)
θ₁′ = α - θ₁
θ₂′ = asin(n₁/n₂ * θ₁′)
θ₂ = θ₂′ - α
plot!([(-1,0), (1,0)]; line=(:black, 1))
plot!([(0,-1),(0,1)]; line=(:black, 1))
plot!([(0,1), (2tan(α),-1)]; line=(:black, 1))
S = Shape([(0,-1),(2tan(α),-1),(0,1)])
plot!(S, fill=(:gray80, 0.25), line=nothing)
xx = tan(α)/ (1 + tan(α)*tan(θ₁))
sl(x) = 1 - x/tan(α)
yy = sl(xx)
plot!(sl, xx-1/8, xx+1/9; line=(:red,3))
plot!(x -> yy + tan(α)*(x-xx), xx-1/4, xx+1/4; line=(:red, 3))
plot!([(-1,-sin(θ₀)), (0,0)]; line=(:black, 2))
plot!([(0,0), (xx, xx*tan(θ₁))]; line=(:black, 2))
plot!([(xx,yy), (xx + 5/8, yy - 5/8*tan(θ₂))]; line=(:black, 2))
annotate!([
(-1/2,1/2*sin(pi + θ₀/2), text(L"\theta_0")),
(1/5, 1/5*sin(θ₁/2), text(L"\theta_1")),
(2tan(α), -0.075, text(L"\theta_2")),
(-1/2, -3/4, text(L"n_0")),
(2tan(α/2), -3/4, text(L"n_1")),
(1 - (1-2tan(α))/2, -3/4, text(L"n_2"))
])
current()
end
p2 = let
plot(; empty_style..., aspect_ratio=:equal)
n₀,n₁,n₂ = 1,3, 1
θ₀ = pi/7
α = pi/8
θ₁ = asin(n₀/n₁* θ₀)
θ₁′ = α - θ₁
θ₂′ = asin(n₁/n₂ * θ₁′)
θ₂ = θ₂′ - α
xx = tan(α)/ (1 + tan(α)*tan(θ₁))
sl(x) = 1 - x/tan(α)
yy = sl(xx)
plot!(sl, xx-1/8, xx+1/9; line=(:red,3))
plot!(x -> yy + tan(α)*(x-xx), xx-1/4, xx+1/4; line=(:red, 3))
S = Shape([(0, sl(xx-1/8)), (xx-1/8,sl(xx-1/8)),
(xx+1/9, sl(xx+1/9)), (0, sl(xx+1/9))])
plot!(S, fill=(:gray80, 0.25), line=nothing)
#plot!([(-1,-sin(θ₀)), (0,0)]; line=(:black, 2))
plot!([(0,0), (xx, xx*tan(θ₁))]; line=(:black, 2))
plot!([(xx,yy), (xx + 2/8, yy - 2/8*tan(θ₂))]; line=(:black, 2))
annotate!([
(1/5, .1*sin(θ₁/2), text(L"\theta_1\prime")),
(xx + .1, 0.06, text(L"\theta_2\prime")),
(2tan(α/2), -1/8, text(L"n_1")),
(17/32, -1/8, text(L"n_2"))
])
current()
end
plot(p1, p2; layout=(1,2))
```
Light bending through a prism. The right graphic shows the second bending.
:::
```{julia}
#| echo: false
plotly()
nothing
```
Following Wikipedia, we have
$$
\theta_1 = \sin^{-1}\left( \frac{n_0}{n_1} \sin(\theta_0) \right)
$$
Both $\theta_0$ and $\theta_1$ are measured with respect to the coordinate system that looks like the $x-y$ plane. The red coordinate system is used to identify the angle of incidence for the second bending. Some right-triangle geometry relates the new angle $\theta'_1$ with $\theta_1$ through $\theta'_1 = \alpha - \theta_1$. With this new angle of incidence, the angle of refraction, $\theta'_2$, satisfies:
$$
n1 \sin(\theta'_1) = n2 \sin(\theta'_2)
$$
Or
$$
\theta'_2 = \sin^{-1}\left(\frac{n_1}{n_2}\sin(\theta'_1) \right)
$$
Finally, using right-triangle geometry, the angle $\theta_2 = \theta'_2 - \alpha$ can be identified.
For a prism, in air, we would have $n_0 = n_2 = 1$. Letting $n_1 = n$, and combining we get
$$
\begin{align*}
\delta &= \theta_0 + \theta_2\\
&=\theta_0 + \sin^{-1}\left(\frac{n_1}{n_2}\sin(\theta'_1) \right)- \alpha\\
&= \theta_0 - \alpha + \sin^{-1}\left(\frac{n}{1}\sin(\alpha -\theta_1) \right)\\
&= \theta_0 - \alpha + \sin^{-1}\left(n\sin\left(\alpha - \sin^{-1}\left( \frac{n_0}{n_1} \sin(\theta_0) \right)\right)\right)\\
&= \theta_0 - \alpha + \sin^{-1}\left(n\sin\left(\alpha -\sin^{-1}\left( \frac{1}{n} \sin(\theta_0) \right)\right) \right)
\end{align*}
$$
If the prism has index of refraction $n$ then the ray will deviate by this amount $\delta$ that depends on the initial incidence angle, $\alpha$ of the prism and $n$.
When $n=1.5$ (glass), $\alpha = \pi/3$ and $\theta_0=\pi/6$, find the deflection (in radians).
We have:
@@ -795,6 +925,8 @@ plot(abs ∘ T4, -1,1, label="|T₄|")
plot!(abs ∘ q, -1,1, label="|q|")
```
We will return to this family of polynomials in the section on Orthogonal Polynomials.
## Hyperbolic trigonometric functions