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====== Differences ====== This shows you the differences between two versions of the page.
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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) delakins [Coding tasks] |
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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 |
| | {{ :projects:rainbow:rainbowdrawing2.png?nolink |}} | | | {{ :projects:rainbow:rainbowdrawing2.png?nolink |}} | | ||
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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}} | ||
rainbow_project.1401344051.txt.gz · Last modified: 2014/05/29 06:14 by wikimanager
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