# Confusion over Snell's law [duplicate]

This question already has an answer here:

Does the angle of refraction always have $$n_2$$ as its refractive index? As far as I know, $$\frac{\sin\theta_i}{\sin\theta_r} = \frac{n_2}{n_1}.$$ So I got this question here:

Calculate the critical angle for a flint glass-air ($$n=1.58$$) boundary.

Since the critical angle has an angle of refraction of 90° that is $$\theta_r= 90^\circ$$, but I am confused on what values should I substitute for $$n_2$$ and $$n_1$$. So does the refractive index of the angle of incidence always equal $$n_1$$?

## marked as duplicate by Aaron Stevens, Jon Custer, ZeroTheHero, ahemmetter, John Rennie homework-and-exercises StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 21 '18 at 7:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – Stephan Nov 20 '18 at 9:28

## 2 Answers

To avoid confusion, here is a different way of writing Snell’s law. $$n_1\times\sin(i) = n_2\times\sin(r)$$ Here, $$n_1$$ is the refractive index of the medium in which the incident ray exists $$n_2$$ is the refractive index of the medium in which the refracted ray exists.

So for this question as the light goes from denser to rarer medium, $$n_1$$ would be the refractive index of flint and $$n_2$$ would be that of air.

• 'n1 is the refractive index of the medium in which the incident ray exists n2 is the refractive index of the medium in which the refracted ray exists' this is always the case right? – Fred Weasley Nov 20 '18 at 10:01
• @Tfue Yes, if $i$ represents the incident angle. The formula is set up in a nice way to remember it. You just keep things in the same medium on the same side of the equation. It would be even clearer if you wrote it as $$n_i\sin\theta_i=n_r\sin\theta_r$$ – Aaron Stevens Nov 20 '18 at 10:43

In physics, where possible, don't jump into applying a formula until you have a pretty good picture of the physical effect you are studying. In this example, the picture is easy to form: it is light rays going from one medium to another. Draw some example diagrams, sticking to the rule that in each case the angle between the ray and the normal from the interface is larger in the medium with lower index. In this way you can answer your own question about which refractive index refers to which medium in the formula you quoted.

(I have deliberately not told you the answer because I think if you try the method I advocate, you will learn more. I hope you will find this helpful rather than frustrating.)

• Andrew, many of the students that I taught physics to did NOT like to make a drawing, and I'm not sure why they were so reluctant to do so. Having said that, I note that the subscript "i" in Snell's law represents the incident ray, and the subscript "r" represents the refracted ray. Subscripts "1" and "2" are not as helpful as they could be, so I would recommend that the OP use the equation $n_i sin(\theta_i)=n_rsin(\theta_r)$ – David White Nov 20 '18 at 16:07