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This is just an example I created to help you understand how to solve such problems. Snell's law is applicable in such cases. Hope this helps. Don't mind my handwriting though. Sorry I missed the 2 in denominator at last.


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The eikonal equation should be used when the index of refraction has varies with position; it is derived directly from Fermat's Princaple of Least Time. OTOH, if you just have slabs of finite thickness, each with its own index of refraction, then just apply Snell's Law multiple times; matrix methods help. Multi-layer hin film coatings can be designed this ...


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Yes, you can still use Snell's law if u=u(z), where u in the index of refraction and z is the axis parallel to the normal of the surface. Depending upon the length scale of the variation you may have to treat it as a differential.


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So, per JQK's suggestion, I began from the equation $\text(a)$ $$\frac{\frac{n^2}{\mu}-1}{\frac{n^2}{\mu}+2}=\frac{1}{3\epsilon_0}\frac{N}{V}\alpha$$ where $N$ is the number of gas particles, and $\alpha$ is molecular polarizability. From the ideal gas law $$PV=zRT$$ it follows that $\frac{N}{V}=\frac{N_A\rho}{M}$. Thus ...


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You have to account for the local fields and how they change as the density changes. Look up the Clausius–Mossotti Equation and its derivation.


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The middle part of the lens will just act as a rectangular glass slab. We may verify it by cutting the lens horizontally. Now we know that rectangular glass slab refract light in such a way that emergent is parallel to incident. Actually the same happens when the ray passes through optical centre. This can be observed in a thick lens. In thin lenses the ...


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The conditions for total internal reflection from an air-common glass interface are: The light has to be travelling in the glass. The angle of incidence has to be larger than the critical angle.


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For a pane of glass, the two surfaces are parallel to each other. That means that the exit angle from the 2nd surface will be the same as the entrance angle of the first. Whatever light enters (and isn't absorbed) will exit: $$n_{air}\sin\theta_{1}=n_{glass}\sin\theta_{g1}$$ and $$n_g\sin\theta_{g2}= n_{air}\sin\theta_2.$$ A quick sketch of two parallel ...


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The setup is such that you get refraction from air to liquid as the light enters the semicircle (via the material of the wall of whatever container you have), but that the ray will not be refracted again when the light leaves the container (assuming things are properly aligned, the angle of incidence of the light internal to the container is normal to the ...


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Recall that from energy and momentum conservation when the Snell's law is applicable, then R=0. But when R=1, then i=r.


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No.Law of refraction is $$n_1\sin \theta_1=n_2\sin \theta_2$$ If $\theta_1=0$ as $n_1\neq0$ and $n_2\neq 0$ we have $\theta_2=0$.There is conclusion on $n_1$ and $n_2$.And if $n_\neq n_2$ then $v_1 \neq v_2$ and $v=\frac{c}{n}$ .Velocity of light is same in any direction.It changes only if $n$ changes.


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we obtain the angle of the refracted or reflected ray to be equal to the angle of incidence The angle of reflection is always the same as the angle of incidence. The angle of refraction follows Snell's law. If the incidence angle is larger than the critical angle, the angle of refraction would be greater that 90°. This is not possible so all the light ...



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