Does the magnitude of the refractive index affect reflection? Will a ray be more reflected the higher the refractive index of the other medium is?
[Physics] Refractive Index and Reflection
reflectionrefraction
Related Solutions
Everything is about electromagnetic waves.
At the border of medium one of the components of electric field have to be continuus and other component of electric displacement field. Such calculations leads to Fresnel equations
Suppose we define the following $\zeta = \ln{\varepsilon}$ and $\xi = \ln{\mu}$, where $\varepsilon$ and $\mu$ are the permittivity and permeability, respectively. In a system with no sources (i.e., $\mathbf{j} = 0$ and $\rho_{c} = 0$), then we know that $\nabla \cdot \mathbf{D} = 0$, where $\mathbf{D} = \varepsilon \ \mathbf{E}$ and $\mathbf{B} = \mu \ \mathbf{H}$. After a little vector calculus we can show that: $$ \nabla \cdot \mathbf{E} = - \nabla \zeta \cdot \mathbf{E} \tag{0} $$ Using this and some manipulation of Faraday's law and Ampêre's law, we can show that the general differential equation in terms of electric fields only is given by: $$ \left( \mu \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \nabla^{2} \right) \mathbf{E} = \left( \mathbf{E} \cdot \nabla \right) \nabla \zeta + \left( \nabla \zeta \cdot \nabla \right) \mathbf{E} + \nabla \left( \zeta + \xi \right) \times \left( \nabla \times \mathbf{E} \right) \tag{1} $$
We can get a tiny amount of reprieve from this by assuming that the permeability is that of free space, i.e., $\nabla \xi = 0$. If we further argue that the only direction in which gradients matter is along $\hat{x}$ and that the incident wave vector, $\mathbf{k}$, is parallel to this, then we can further simplify Equation 1 to: $$ \left( \mu_{o} \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \frac{ \partial^{2} }{ \partial x^{2} } \right) \mathbf{E} = \left( \mathbf{E} \cdot \nabla \right) \zeta' \hat{x} + \left( \zeta' \frac{ \partial }{ \partial x } \right) \mathbf{E} + \zeta' \hat{x} \times \left( \nabla \times \mathbf{E} \right) \tag{2} $$ where $\zeta' = \tfrac{ \partial \zeta }{ \partial x }$.
After some more manipulation, we can break this up into components to show that: $$ \begin{align} \text{x : } \left( \mu_{o} \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \frac{ \partial^{2} }{ \partial x^{2} } \right) E_{x} & = E_{x} \zeta'' + \zeta' \frac{ \partial E_{x} }{ \partial x } \tag{2a} \\ \text{y : } \left( \mu_{o} \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \frac{ \partial^{2} }{ \partial x^{2} } \right) E_{y} & = 0 \tag{2b} \\ \text{z : } \left( \mu_{o} \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \frac{ \partial^{2} }{ \partial x^{2} } \right) E_{z} & = 0 \tag{2c} \end{align} $$
In the limit where the incident wave is entirely transverse, then $E_{x} = 0$ and the x-component (Equation 2a) is entirely zero.
Next you assume that $\mathbf{E} = \mathbf{E}_{o}\left( x \right) e^{i \omega t}$, where $\omega$ is the frequency of the incident wave. Then there will be incident, reflected, and transmitted contributions to the total field at any given point (well the transmitted is always zero in the first medium, of course). Any incident and transmitted contributions with have $\mathbf{k} \cdot \mathbf{x} > 0$ while reflected waves will satisfy $\mathbf{k} \cdot \mathbf{x} < 0$. You define the ratio of the reflected-to-incident fields (well impedances would be more appropriate) to get the coefficient of reflection.
Simpler Approach
A much simpler approach is to know where to look for the answer to these types of questions. As I mentioned in the comments, there has been a ton of work on this very topic (i.e., spatially dependent index of refraction) done for the ionosphere. If we look at, for instance, Roettger [1980] we find a nice, convenient equation for the reflection coefficient, $R$, as a function of the index of refraction, given by:
$$
R = \int \ dx \ \frac{ 1 }{ 2 \ n\left( x \right) } \frac{ \partial n\left( x \right) }{ \partial x } \ e^{-i \ k \ x} \tag{3}
$$
There is no analytical expression for $R$ for your specific index of refraction. However, numerical integration is not difficult if one knows the values for $d$ and $\epsilon$. Note that if we do a Taylor expansion for small $\epsilon$, then the integrand (not including the exponential) is proportional to cosine, to first order in $\epsilon$ (cosine times sine if we go to second order).
References
- Gossard, E.E. "Refractive index variance and its height distribution in different air masses," Radio Sci. 12(1), pp. 89-105, doi:10.1029/RS012i001p00089, 1977.
- Roettger, J. "Reflection and scattering of VHF radar signals from atmospheric refractivity structures," Radio Sci. 15(2), pp. 259-276, doi:10.1029/RS015i002p00259, 1980.
- Roettger, J. and C.H. Liu "Partial reflection and scattering of VHF radar signals from the clear atmosphere," Geophys. Res. Lett. 5(5), pp. 357-360, doi:10.1029/GL005i005p00357, 1978.
Best Answer
Absolutely. See Fresnel's equations.
These show that reflection depends on angle of incidence, polarization, and refractive index.
The simplest case (for reflection at normal incidence, where polarization doesn't matter) tells us that the reflectance is
$$R = \left|\frac{n_1-n_2}{n_1+n_2}\right|^2$$
You can see that when $n_1=n_2$, there is no reflection; and that the intensity of reflection is the same whether you go from $n_1$ to $n_2$ or the other way around. Finally, when $n_1\gg n_2$, the reflectance gets very close to 1.