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The gradient of the graph is the same as the derivative of the graph and its importance is that it tells us how change in one variable (solution concentration) affects the other variable (refractive index). For the majority of situations where one is using such graphs, the slope is positive. A positive slope means that when the concentration of the ...


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Some minor points: First, the units of $\lambda^2$ will be $(\mu m)^2$. Second it seems you are asking for $a, b$ so that you can determine $n$? In fact, what is more often done is to plot $n$ versus $1/\lambda^2$ to determine $a, b$. At the moment you can't do anything (unless you follow Frobenius and look up $a, b$, however this means you know the ...


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If the question is about why red and blue light refract at different angles, this is because the refractive index, $n$ for many materials depend upon the frequency of light and they are referred to as dispersive. More often, the variation is given in terms of the wavelength ($\lambda$) and in the optics industry people quote the refractive index for the ...


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You have it backward, faster speed of light in a material corresponds to less refraction, not more. In the limit that the speed of light in a material was the same as the speed of light in vacuum, there would be no refraction at all. It can be shown that in a material the index of refraction is the speed of light in vacuum, $c$ divided by the speed of light ...


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(the following answer is included essentially in "The Feynman LECTURES ON PHYSICS-Mechanics, Radiation & Heat ,Vol. 1, 26-3 Fermat's principle of least time.) Suppose you are at point A in the land and a screaming girl is at point B in the sea. You can run with a speed $\:v_{1}\:$ on the land greater than the speed $\:v_{2}\:$ you can swim in the ...


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The lensmaker's formula states $$P_{lens}=\frac{n_{\rm{lens}}-n_{0}}{n_{0}}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$ This shows that the power of the lens does indeed depend on the refractive index of the medium ($n_0$) in which the lens is placed. In the extreme case where $n_{\rm{lens}}=n_0$, the lens will have no power at all.


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If you put unpolarized light in your Calcite crystal, then the beam will be divided into two beams, one for each polarization, with equal intensity. It you put polarized light (say, linearly), then the two beams will have an intensity that depends on the angle between the axis of polarization of the light and the crystal, following Malus' law. If the ...


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Hint : A key to the solution is what is meant by the complex wave 3-vector $\:\mathbf{k}\:$. This vector is not any complex 3-vector in $\: \mathbb{C}^{3}\:$ $$ \mathbf{k} \ne \left(k_{1}, k_{2}, k_{3} \right) \in \mathbb{C}^{3}, \:\:\text{that is with} \:\: k_{\rho} \in \mathbb{C} \tag{a-01} $$ but $$ \mathbf{k}=\left(k_{1}, k_{2}, k_{3} \right) ...


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The graph of angle of deviation vs angle of incidence is a $U$ shape. The fact that the angle of deviation is the same for these two rays means that a ray which is incident at $40$ degrees to the normal will emerge at $60$ degrees to the normal. This allows us to find the refractive index. $$\sin i_1 = \sin 40 = n\sin r_1$$ $$n\sin i_2 = \sin r_2 = \sin ...


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The refraction angle will be different for different materials provided these materials have a different index of refraction. Reversely, by measuring the angle you can determine the index of refraction of the material. You can look up Snell's law.


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One way to look at this is to go to the definition of the index of refraction $$n = \frac{c}{v}$$ v is related to the wavelength and frequency as $$v = \lambda f$$ So the index of refraction becomes $$n = \frac{c}{\lambda f}$$ Frequency remains the same when light goes into a new medium, so the only difference between light of different colors is the ...


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Forget for a while about Snell's Law, Maxwell's equations and take a look in the Figure. As Feynman said : "In Figure, our problem is again to go from A to B in the shortest time. To illustrate that the best thing to do is not just to go in a straight line, let us imagine that a beautiful girl has fallen out of a boat, and she is screaming for help in the ...


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Refraction occurs because of a change of speed of propagation of the wave. When light passes from air to water it slows down, whereas when sound travels from air to water it speeds up. Therefore sound is refracted away from the normal, whereas light is refracted towards the normal.


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The speed of sound is greater in water than in air, so the wavelength in water is greater than in air. In effect the refractive index of the water is less than the refractive index of the air. For light it is generally true that the refractive index is higher for denser materials, and this is because light interactions mainly with the electrons in a medium ...


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See Snell's Law. $$n_1sin(\theta_i)=n_2sin(\theta_o)$$ where $\theta_i$ is the angle the incoming ray makes with the perpendicular to the surface between the two substances, and $\theta_o$ is the angle the outgoing ray makes with the perpendicular to the surface between the two substances. This can actually be derived from Maxwell's equations, but that's ...


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The concept of "image" of a point means: the reflected/refracted light rays from it seem to come from the image point. So if you follow them backwards where they came from and prolongate them beyond the spot where they were reflected/refracted, they meet at the image point. Note: tt is not obviuos, that such a point must exist, the rays don't have to meet, ...


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It depends on from which material you are viewing. If you are viewing with your eye placed inside the water then you have to consider the "object" distance because now, only the phenomena reflection takes place. However if your eye is placed above the water, then "optical" distance is taken into consideration since both reflection(from the mirror) as well as ...



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