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Can we get the refractive index of a prism by using white light ? I guess since white light is non chromatic, it is not possible.

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Of course you can. The prism is likely to disperse the light - that is, different colors will be refracted by different amounts. That means you don't just get "a" refractive index, but with careful experimental setup you will get the entire refractive index curve: a different value for every wavelength / color.

enter image description here

Setup: a narrow beam of white light incident on the prism. Measure the position where a particular color hits a screen far away. Calculate the angles, use Snell's law to compute $n$. Repeat for every color. If you know the wavelength of the colors (you can look these up) you can create a graph of refractive index vs wavelength; if you don't, you can create a table.

The precision of a setup like this is really only limited by your ability to set it up carefully; and the size. Usually, dispersion is strongest when you are working close to the critical angle of a configuration - this is why a diamond is cut the way that it is. In the diagram above, you are most likely far from the critical angle. This means that the differences in refractive index will matter less.

To get an accurate measure of refractive index, you would like to find the critical angle - that is the angle at which you get total internal reflection. If you know the internal angle at which this happens, you can compute the refractive index directly from Snell's law:

$$\frac{n_1}{n_2}=\frac{\sin\theta_2}{\sin\theta_1}$$

At the critical angle, $\sin\theta_2=1$ and $n_2=1$, so $n_1=\frac{1}{\sin\theta_1}$.

By looking at the color that just gets out of the prism, you find the color of the one that just didn't make it... and that is the color for which you thus find $n$. Now the tricky thing is that in order to do this experiment, the light will also bend at the other interface (the input face of the prism) which makes the math a bit more interesting.

If that is the setup you want to explore and you can't figure out the diagram / equations, please post a comment.

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It depends on how accurately you want to measure the refractive index. The refractive index typically only changes by about 1% over the visible spectrum so if you're happy with that accuracy just go ahead and estimate the central point of the refracted ray.

As Floris suggests, you could try and judge the refraction angle separately for the different colours in the refracted ray, but I suspect this is going to be a bit hit and miss. If you want better than 1% accuracy you should probably use monochromatic light.

If you're curious to know more about this, glass suppliers typically report the dispersion using the Abbe number - see for example the list here. To get the Abbe number the refractive index is measured at three wavelengths:

  • $n_D$ - 589.3nm (yellow sodium light)
  • $n_F$ - 486.1nm (greeny-blue)
  • $n_C$ - 656.3nm (red)

And the Abbe number is given by:

$$ V_D = \frac{n_d - 1}{n_F - n_C} $$

For fused silica the values from the link I gave are $n_D = 1.458$ and $V_D = 67.70$. This means $n_F - n_C = 0.0068$ i.e. the refractive index changes by $0.0068$ between greeny-blue and red light.

To get the wavelength at any particular wavelength you can approximate that variation of $n$ with $\lambda$ as a straight line.

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  • $\begingroup$ Your numbers basically say you can get to 0.5% accuracy without worrying about the color. If you observe dispersion in the light (because you are sufficiently far away) you may get down another factor 5 or so. If the initial challenge given is "do this with white light", I think that anything below 1% is heroic. Frankly I usually content that if you need to measure (directly) a quantity to better than that, you are probably using the wrong experiment... There's usually a way to amplify the difference with a known quantity - and you measure that instead. $\endgroup$
    – Floris
    Nov 7, 2014 at 17:36
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    $\begingroup$ @Floris: this is obviously more of an exercise for the student that a real measurement. I can rarely resist the temptation to prod students into looking into the subject that bit more deeply :-) $\endgroup$ Nov 7, 2014 at 17:46
  • $\begingroup$ +1 - and note I did a similar thing with my "now see how to set up for near-internal reflection which will amplify the effect if you do it right..." $\endgroup$
    – Floris
    Nov 7, 2014 at 18:04

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