This regards the following problem:

A ray of light is traveling in glass and strikes a glass/liquid interface. The angle of incidence is $58.0^\circ$ and the index of refraction of the glass is $n = 1.50$. What is the largest index of refraction that the liquid can have, such that none of the light is transmitted into the liquid and all of it is reflected back into the glass?

My solution: We have $n_1 \sin \theta_1 = n_2 \sin \theta_2$. We wish to maximize $n_2$ provided $90^\circ \leq \theta \leq 180^\circ$. Plugging in our givens, we have $(1.50)(\sin 58^\circ) = n_2 \sin \theta_2 \rightarrow n_2 = \frac{(1.50)(\sin 58^\circ)}{\sin \theta_2}$. Since we wsh to maximize $n_2$, we should minimize $\sin \theta_2$, which can be any value in $(0, 1]$ so there is no theoretical limit on the size of $n_2$ (according to this math)

However, the textbook says the correct maximum is $1.27$, which is taking $\theta_2 = 90^\circ$. I don't think this is right at all, since if anything that is the minimum. Consider $1 < n_2 < 1.27$. Then $1.27 > \sin \theta_2 > 1$, which is impossible given the range of the sine function over the reals.

Who is right?

  • $\begingroup$ $sin\theta_1\geq n_2/n_1$ for no transmission in the medium $n_2$ where $\theta_1$ is the incident angle $\endgroup$ Apr 16, 2014 at 4:36
  • $\begingroup$ This is the case where common sense is more important than maths. You know that if the second medium has higher refractive index than the first, there won't be total internal reflection. So your conclusion was wrong. $\endgroup$
    – jinawee
    May 26, 2014 at 11:55

2 Answers 2


The textbook is right, just plug in some numbers: (let's say $n_2=1.3>1.27$)

$$1.5*\sin 58^\circ = 1.3* \sin \theta_2 $$ $$ \sin \theta_2=\frac{1.5}{1.3}*\sin 58^\circ$$ $$\theta_2=78^\circ <90^\circ$$

And if you go by the usual definition of $\theta_2$ (https://en.wikipedia.org/wiki/File:Snells_law2.svg) then this would mean that the beam would go into the liquid. This is why it's the largest refractive index: because for any larger index, as calculated above, the beam is not reflected.

(You can see your problem this way: the critical angle ($\theta_c=58^\circ$) is already given to you (this is why $\theta_2=90^\circ$) and you're looking for the refractive index of the liquid.)

And considering your $1 < n_2 < 1.27$ and the impossibility of $sin(\theta_2)>1$ argument: this is exactly the point of T.I.R. Say you have n=1 for air. Then $sin(\theta_2)=1.27$ obviously. Now you say this is impossible, and it is. But you have to remember that Snell's law is formulated for the incident and refracted beam.


Since $sin(\theta_2)>1$ is not possible for any refracted $\theta$, there is no refraction and only reflection.

  • $\begingroup$ So in this situation, is $1.27$ also the only possible index of refraction? How might one get a beam that reflects at an angle of greater than $90^\circ$ back into the glass? $\endgroup$
    – MT_
    Mar 17, 2014 at 3:09
  • $\begingroup$ No, it's not the only possible index of refraction, but it's the largest one where you still have reflection. $\endgroup$
    – user42076
    Mar 17, 2014 at 9:14
  • $\begingroup$ And for your second question (I hope I understood your question right): because when $\theta_2>90^\circ$ then you are not speaking of refracted beams any more. This also means that you get no meaningful results from Snell's law - like my calculation above with the air - because it's formulated for refracted beams only. $\endgroup$
    – user42076
    Mar 17, 2014 at 9:27

for Total internal reflection, $(\theta_2)\ge90$,so we can calculate the minimum $n_2$ required for Total internal reflection by setting $\theta_2=90$.

$$n_1 sin(\theta_1)=n_2 sin(\theta_2)$$ $$1.50 \cdot sin(58^{o})=n_2 sin(90)$$ $$\dfrac{1.50 \cdot sin(58^{o})}{sin(90)}=n_2$$ $$\therefore n_2=1.27$$

So, your textbook is right

  • $\begingroup$ It asks for "largest." Also I believe your first line you did not mean to include $\sin$ $\endgroup$
    – MT_
    Mar 16, 2014 at 22:31
  • $\begingroup$ Then you have a typo, the largest would obviously be 1. Unless it's referring the the largest numeric value of $n_2$, which would then be 1.27 $\endgroup$ Mar 16, 2014 at 22:32
  • $\begingroup$ Uhm, not to be rude, but if you read my question at all it should be pretty clear what I want. The original question asks for the largest numeric value of $n_2$ for T.I.R. and states that it is $1.27$, and I'm saying that that is not $1.27$. $\endgroup$
    – MT_
    Mar 16, 2014 at 22:38
  • $\begingroup$ So,summing up,you're right $\endgroup$ Mar 16, 2014 at 22:45

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