1

Consider passing red and yellow lights separately through the same interfaces and with the same angles of incidence.

Light of which colour will be refracted more?

This question was given to me. All I am taught is Snell’s law of refraction which states for a given pair of media you have:

$$n_1\cdot sin(i)=n_2\cdot sin(r)$$ Since $n_1$, $n_2$, and $i$ remain constant it should be that the angle of refraction in both cases should remain constant. However, my instructor says that red light is “refracted more” than yellow light. My coursebook also mentions that the refractive index $n$ depends on the wavelength of the light, but does not elaborate.

How does refraction depend on the colour of light? Does the refractive index vary with wavelengths of light? If so, how? If I say that the refractive index of some medium is $n$, what wavelength of light does that correspond to?

Any help is appreciated.

Qmechanic
  • 201,751
Sahaj
  • 119

2 Answers2

3

The wikipedia article on the dispersive prism has a picture showing how violet light is refracted more than red light by the prism. This phenomenon is called dispersion. It is caused by the refractive index of the glass being greater for the violet light than for the red light. All light has the same speed in a vacuum, but the violet light will be slower than the red light as they pass through the glass. Since violet light has a shorter wavelength than red light, we see that the speed of light in the glass is less for light of shorter wavelength.

The differences in speed/refractive index between red and violet are different in different types of glass, and lenses made from two different glasses are sometimes used together so that the different dispersions of the two lenses approximately cancel each other. See the Wikipedia article on the Achromatic lens.

In the statement of Snell's Law $$\frac{\sin\theta_1}{\sin\theta_2}=\frac{n_2}{n_1}=\frac{v_1}{v_2}$$ the speeds and refractive indices $n_1, n_2, v_1, v_2$ are functions of the colour. There is no simple rule relating colour to refractive index; the values must be measured or looked up as required.

Peter
  • 1,859
  • Thanks. How can I calculate exactly the refractive index of a medium for a particular colour? Suppose I know refractive index $n$ for the medium, how would I calculate the refractive index of the medium in some coloured light? – Sahaj Aug 16 '23 at 13:30
  • You don't really calculate the wavelength-dependant refractive index for a particular medium or a particular colour, but you can measure and tabulate it. You can look up the values, or rather curves, in the literature; for example here https://refractiveindex.info/

    You can, for some mediums, say "the refractive index for visible (coloured) light as about x", but that value will not be exactly true for all colours of light, it will just be an approximation.

     – John Doe Aug 16 '23 at 14:08
  • Sorry to bother again. I understand dispersion now and how white light splits into component colours. I also know now that refractive index of a medium decreases with increasing wavelengths. But how can I use to justify that red light is refracted more at an interface than yellow light for constant $i$? For coloured light passing from air into glass, I can say that $$sin(r)= \frac{u_{air}(colour)*sin(i)}{u_{glass}(colour)}$$ – Sahaj Sep 21 '23 at 09:30
  • The picture in the dispersive prism article shows violet light being refracted more than red. This shows the speed of the violet light changes more than the red light, so the violet light is slowed more than the red. Yellow would be in between. – Peter Sep 21 '23 at 10:52
  • 1
    @Sahaj, I have edited my answer. I would not use $u$; standard notation is $n$ for refractive index or $v$ for speed. – Peter Sep 22 '23 at 05:43
3

No, $n_1$ and $n_2$ won't remain constant. They vary with the wavelength of light, which varies with color. Red has the longest wavelength. Wavelength is inversely proportional to the refractive index (given by Cauchy's equation).

Your instructor was right (at least when the light travels from rarer to denser). Red is indeed refracted more. This is because the refractive index for red will be less than that for yellow. You can then verify by using Snell's law. Take the first medium as air ($n_1=1$). $$\frac{\sin i}{\sin r} = \frac{n_2}{1}$$

$\sin i$ is constant for both colors. So... $$\sin r = \frac{1}{n_2}$$

$n_2$ will be greater for yellow; $\sin r$ will be smaller for yellow.

In the first quadrant, $\sin \theta$ increases with $\theta$. Thus, angle r for red has to be greater.

NOTE: In vacuum all colors travel with the same speed. In air the difference in refractive indices based on color is almost negligible, so I've taken it as 1.

Otherwise: Think of it like this. Take the interface between air and glass for instance. As the ray moves from rarer to denser it deviates towards the normal. Red deviates the least and thus would be at a greater angle from the normal than yellow. Try drawing out a ray diagram.

Here, too, angle of refraction is greater for red.

PS: The cases I've shown are both rarer to denser. This doesn't stand true the other way round. From denser to rarer, yellow deviates more and has a greater angle of refraction.

PPS: Most refracted in a prism is violet as it's deviated the most. I'm assuming you do not mean in that manner.

Hope it helps!

  • I hope this answers your query. I would say that the question, in the way it's framed, is quite ambiguous. It doesn't specify much. I've tried to explain through many scenarios. In case of any error, please do correct me. –  Feb 15 '24 at 15:48