0
$\begingroup$

When light passes from an optically denser medium to an optically rarer medium, it bends away from the normal in accordance with the Snell's Law of refraction $\mu_1\sin i =\mu_2\sin r$ where $\mu_1$ and $\mu_2$ are the refractive indices of the two media, $\mu_1>\mu_2$, and $i$ and $r$ are the angles of incidence and refraction respectively.

When we increase $i$ gradually, $r$ also increases. Since, $r>i$ as $\mu_1>\mu_2$, $r$ becomes $90^\circ$ before $i$. The angle of incidence for which the angle of refraction is $90^\circ$ is called the critical angle $i_c$. From Snell's law we get, $\sin i_c =\mu_2/\mu_1$. This is how critical angle is explained in almost all sources. Beyond this angle it's said the Snell's law of refraction is no longer valid and refraction is not possible, only reflection or more precisely total internal reflection takes place.

I do not understand why we still consider the refractive index of the medium at grazing refracted ray ($r=90^\circ$) as $\mu_2$. How can we ascertain the speed of light (and hence the refractive index) in the interface separating two media? According to the derivations, the speed of light in the interface is same as that of the medium other than the one from which the ray emerged (here it's the rarer medium). But why do we choose it this way? Why can't it be the other way round?

$\endgroup$

1 Answer 1

1
$\begingroup$

The case of the refracted angle being $90^\circ$ is ill defined, because you can calculate and see that you get total internal reflection for all angles which are equal or larger than the critical angle. Meaning no energy is actually moving parallel to the interface, and you can't talk about its speed.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer. If possible, could you please support your answer with reliable resources? As far as I've learnt so far, a light ray incident at critical angle grazes the surface after refraction. The occurrence of TIR for rays incident at critical angle is new to me. This is how I thought internal reflection would take place as one increases the angle of incidence gradually. $\endgroup$
    – Vishnu
    Commented Dec 29, 2019 at 4:07
  • $\begingroup$ It's not like that 100% of the energy is transmitted and in the critical angle BAM, no transmission at all.The transmission gets weaker and weaker until at critical angle it approaches actual $0$ and after that it stays that way. See fresnel equations to have a quantitative idea of the numbers: en.wikipedia.org/wiki/Fresnel_equations,and you can see it in the following simulation as well (change the first medium refractive index and watch how as you increase the angle more is reglected and less transmitted: phet.colorado.edu/sims/html/bending-light/latest/… $\endgroup$ Commented Dec 29, 2019 at 19:29
  • $\begingroup$ @Vishnu What does graze the surface even mean? Stay within an atoms diameter of the surface for at least 1cm? Is the surface even defineable to that precision? Can a light ray even be so thin? $\endgroup$ Commented Aug 14, 2021 at 9:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.