# Physics 203 at Portland State 2014

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rainbow_project [2014/05/29 06:14]
wikimanager [Questions about theory and coding]
rainbow_project [2014/06/03 19:18] (current)
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* $P_\text{sc}$ $\,​=\,​I_\text{sc}\!\cdot\!{\text{Area}}_\text{sc}$ $\,​\approx\,​I_\text{sc}\!\cdot\!2\pi\,​(D\sin\gamma)\,​D\Delta \gamma\;$, where $\,D$ is the distance from the droplet to the observer.         * $P_\text{sc}$ $\,​=\,​I_\text{sc}\!\cdot\!{\text{Area}}_\text{sc}$ $\,​\approx\,​I_\text{sc}\!\cdot\!2\pi\,​(D\sin\gamma)\,​D\Delta \gamma\;$, where $\,D$ is the distance from the droplet to the observer.
* $P_\text{sc}$ $=s\!\cdot\!P_\text{inp}$         * $P_\text{sc}$ $=s\!\cdot\!P_\text{inp}$
-        * $I_\text{sc}$ $=s\!\cdot\!I_\text{inp}\frac{d}{D^2\sin\gamma}\frac{\Delta d}{\Delta\gamma}$ $=s \cdot I_\text{inp}\frac{R^2\delta}{D^2\sin\gamma}\frac{\Delta \delta}{\Delta\gamma}$ $\sim I_\text{inp}\frac{\delta}{\sin\gamma}\frac{\Delta \delta}{\Delta\gamma}\;​$:​ here the factors //s//, //R//, and //D// are pretty constant (or too complicated to calculate) and do not contribute to the relative intensity at various angles and wavelengths+        * $I_\text{sc}$ $=s\!\cdot\!I_\text{inp}\frac{d}{D^2\sin\gamma}\left|\frac{\Delta d}{\Delta\gamma}\right|$ $=s \cdot I_\text{inp}\frac{R^2\delta}{D^2\sin\gamma}\left|\frac{\Delta \delta}{\Delta\gamma}\right|$ $\sim I_\text{inp}\frac{\delta}{\sin\gamma}\left|\frac{\Delta \delta}{\Delta\gamma}\right|\;$: here the factors //s//, //R//, and //D// are pretty constant (or too complicated to calculate) and do not contribute to the relative intensity at various angles and wavelengths

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- Then, on the same grid, calculate the intensity of the deflected light, assuming some (any, really) intensity of the sunlight     - Then, on the same grid, calculate the intensity of the deflected light, assuming some (any, really) intensity of the sunlight
- (//​optional,​ for maximum grade//​) ​ - convert this intensity as a function of the $d/R$ grid to the total deflection angle grid.  That would be the model of what you observe when you look at the actual rainbow, for each color.     - (//​optional,​ for maximum grade//​) ​ - convert this intensity as a function of the $d/R$ grid to the total deflection angle grid.  That would be the model of what you observe when you look at the actual rainbow, for each color.
-  ​+  ​- I am having difficulty substituting $\beta_1$ into Snell'​s law. Is it possible I could have a hint based on the geometry of the diagram for the rainbow project to lead me in the right direction?​
+    * **Answer**: ​ If you look at a point where the beam enters the droplet, and notice that $\theta_i$ is equal to $\theta_i$ immediately on the other side of the water surface, then you can just see that $\beta_1$ is the difference $\theta_i-\theta_r$. See that on the drawing (Fig. 4)? The next reflection point is symmetric with this one, but the meaning of angles is different. Also, if you spot a right-angled triangle with $R$ (radius of the sphere) and $d$ as its sides, and the angle $\theta_i$ as one of its angles, that will allow you to relate $\sin\theta_i$ to the $d/R$ ratio. The flow of math is:
+      - from $\;d/R\;$ find $\;​\sin\theta_i$
+      - from that find $\theta_i$
+      - from that, using Snell'​s law, find $\theta_r$
+      - from that, find $\beta_1$
+  - How would I start solving for beta $\beta_2$, $\beta_3$, and $\beta_4$?​
+    * **Answer**: ​  If you look at the next reflection (Fig. 4), you can see that $\beta_2+\theta_r+\theta_r$ add up to 180$^\circ$ (or, $\pi$ radians). There, from symmetry, $\theta_r$ is the same as $\theta_r$ in the first refraction. All the other points are also symmetric to the original, so same logic applies. $\beta_3'​$ is the difference between two thetas already found earlier. And so on.
+
=====Data===== =====Data=====
Give yourself 20 minutes to google or search the literature for the index of refraction of water as a function of wavelength. Paste references and data (formulas and raw numbers if compact) below. Give yourself 20 minutes to google or search the literature for the index of refraction of water as a function of wavelength. Paste references and data (formulas and raw numbers if compact) below.
- **Index of refraction for water** ​ = 1.33.  This is the average value. We need its dependence on the wavelength $\lambda$!   - **Index of refraction for water** ​ = 1.33.  This is the average value. We need its dependence on the wavelength $\lambda$!
- **Average radius of water droplet** = 3 mm ([[rainbow project#​references|Ref. [4]]])   - **Average radius of water droplet** = 3 mm ([[rainbow project#​references|Ref. [4]]])
+  - **Average [[wp>​Sunlight]] Intensity on Earth** = 1367 $\frac{W}{m^2}$ ​

{{visible_spectrum.jpg}} {{visible_spectrum.jpg}}