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What does it mean to say, "the refractive index of medium $B$ relative to medium $A$"? Assume that in this case, medium $A$ is optically denser than medium $B$.

I have difficulties trying to understand the word 'relative' in aforementioned quote. Does it mean refractive index of $A$ divided by the refractive index of $B$ or the other way round?

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    $\begingroup$ Can you please edit your question? It is unclear. $\endgroup$ – Ofek Gillon Mar 11 '17 at 7:53
  • $\begingroup$ I want to know whether answer for this is mediumB\mediumA (or) mediumA\mediumB $\endgroup$ – Sam Saravanan Mar 11 '17 at 8:20
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The fundamental thing about a refractive index is that it measures the speed of electromagnetic propagation in the medium in question. It is the ratio of the speed of light in the vacuum to that in the medium. Its a scale factor that measures the speed of the light in the medium to that in a vacuum.

A "relative" refractive index simply means we broaden this idea so that the speed ratio is calculated with the "reference" medium different from a vacuum. Suppose we have two mediums $A,\,B$, with vacuum relative refractive indices $n_a,\,n_b$. Then the index of $A$ relative to $B$ is:

$$\frac{\frac{c}{n_b}}{\frac{c}{n_a}} = \frac{n_a}{n_b}$$

Notice how we divide the lightspeed in medium $B$ by that in $A$; the definition resorts to the wonted, vacuum relative refractive index when we put $n_b=1$.

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When somebody states that the refractive index of glass is $\frac 3 2$ then that is taken to mean the refractive index when light is going from a vacuum in glass and it is the ratio

$_{\rm vacuum}n_{\rm glass} = \dfrac{\text{speed of light in a vacuum}}{\text{speed of light in glass}}\approx \dfrac{3 \times 10^8}{2 \times 10^8}= \dfrac 3 2$

and this is often called the absolute refractive index and just written as $n_{\rm glass}$.

A vacuum is taken as the reference with a refractive index of one.

So the term refractive index of glass can be taken to mean refractive index of glass (medium $B$) relative to a vacuum (medium $A$).

This being the case then you can say that the optical density of medium $B$ is greater than medium $A$ noting that this is not the statement that you made in your question.

If the light is going in the opposite direction, ie from glass into a vacuum (optically more dense into optically less dense), then if you so choose you could define a relative refractive index for a vacuum relative to glass and its value will be

$_{\rm glass}n_{\rm vacuum} = \dfrac{\text{speed of light in glass}}{\text{speed of light in a vacuum}}\approx \dfrac{2 \times 10^8}{3 \times 10^8}= \dfrac 2 3$.

If you designate that the refractive index of going from medium $A$ to medium $B$ is $_{\rm A} n _{\rm B}$ and the absolute refractive index of medium $A$ is $_{\rm V}n_{\rm A}=n_{\rm A}$ and that of medium $B$ is $_{\rm V}n_{\rm B}= n_{\rm B}$ then it is relatively easy to show that

$_{\rm A}n_{\rm B} = \dfrac {n_{\rm B}}{n_{\rm A}}$


When dealing and interface between two mediums $A$ and $B$ where the angle in medium $A$ is $\theta_{\rm A}$ and the angle in medium $B$ is $\theta_{\rm B}$ it is perhaps easier to use absolute refractive indices (relative to a vacuum) as follows

$n_{\rm A} \sin \theta_{\rm A}=n_{\rm B} \sin \theta_{\rm B}$

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protected by Qmechanic Mar 11 '17 at 11:14

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