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1

Two things that might help you. Finding Sellmeier Co-efficients Firstly, although the refractive index formulas you cite are not in the Sellmeier form, they are in the form: $$n(\lambda) = 1+\delta(\lambda) = 1+\sum_j\frac{K_j\,\lambda^2}{\lambda^2 - L_j}$$ where $\delta(\lambda) = \sum_j\frac{K_j\,\lambda^2}{\lambda^2 - L_j}$ is VERY small. So, squaring ...


1

You can reach the end of a rainbow, if you make one using a hose pipe on a sunny day. It is possible to locate the ends to within a few cm


2

Why can't we reach the ends of rainbow? The WIkipedia article you linked to contains the explanation The rainbow is not located at a specific distance, but comes from an optical illusion caused by any water droplets viewed from a certain angle relative to a light source. Thus, a rainbow is not an object and cannot be physically approached. Indeed, ...


1

I have discovered the source of the discrepancy with help from the researcher (thanks Dr. Lipinski!). It seems that many sources (including Wikipedia, the undergraduate heat transfer texts by Incropera & DeWitt and Lienhard & Lienhard) mention only breifly, or not at all, the dependence of the speed of light in the medium on the refractive index, ...


1

One approach is to try to use the original model as a ground truth and then redo the match to the desired form. I tried to do this in MATLAB: xs = 360.0 : 1.0 : 830.0; ys = 1 + 0.05792105./(238.0185-1.0./(xs.*xs)) + 0.00167917./(57.362-1.0./(xs.*xs)); cftool But, quality software that it is, it reliably crashed. So, I wrote my own nonlinear least ...


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Here is a slightly different take on this using the boundary conditions for electromagnetic fields at an interface. A key boundary condition, that is derived from Faraday's law, is that the component of the E-field tangential to the boundary must be continuous. So take an EM wave travelling at normal incidence with the electric field solely in a direction ...


2

Imagine you have a monochromator that you can tune to give you whatever wavelength you like. Now send light at a wavelength of 568 nm towards the interface. What has been calculated is that destructive interference takes place, so that less of the incident radiation is reflected. Let's take an extreme case where the reflection from the first and second ...


2

There are two sign conventions for doing lens calculations. In the Gaussian sign convention the object distance, $u$, is positive to the left of the lens and the image distance, $v$, is positive to the right of the lens. So in the Gaussian sign convention $u = +3000$cm. With this convention the lens equation and the the magnification are given by: $$ ...


2

Let me go a little further than @mark-h's answer: The behavior of light at an interface is described by the electro-magnetic field solution to the Helmhotlz equation. It gives the reflected and transmitted electric and magnetic components as a function of the refractive indices of the incident and exiting media. From those solutions we can derive the ...


2

Yes. The boundary conditions for Maxwell's equations gets you this. This reasoning is much more simple than it sounds. High refractive index means phasefronts of a plane wave are nearer together than for a low refractive index. When the waves in both mediums line up at the interface, the spacing between the intersection of the phasefronts and the interface ...


5

In general, reflection and refraction happen when light passes from one medium to another. You can see this if you see your own reflection in a window. Now, as a light ray approaches the critical angle, not only does the refracted ray get closer to the surface, but the amount of light transmitted gets less and less. At the critical angle, the refracted ray ...


1

If the light ray is normal to the surface, the maximum amount of light is transmitted. As the light ray bends, as in your part (b), a percentage of the light will be transmitted (refracted) and the remaining will be reflected (at the incidence angle). Very near the critical angle $\theta_c - d\theta$, likewise, some of the light will be transmitted ...



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