Last year, our teacher made us an exam to check our knowledge in light reflection and refraction. I don't perfectly remember it, but I know that one of the exercises included a ray with its angle of incidence, and let's say 7 media with their respective indexes of refraction. We had to find the last angle of refraction. Even though it's fairly easy to do, it took my classmates a lot of time. Meanwhile I didn't even have time to reach that question: we were very limited on time.

I was thinking if there was another way to find it, a quicker one. Then, while looking at a figure of refraction in my physics book, I came up with an idea: what if we calculate it using only $n_1$, $n_7$, $\Theta_1$ and $\Theta_7$? I'll explain.

Let's say the first medium is air, and I'm calling it $n_1$. The seventh medium is glass, and I'm calling it $n_7$. There are 5 other media between air and glass. The angle of incidence will be $\Theta_1$ and the last angle of refraction, the one inside $n_7$, will be $\Theta_7$. Is it possible to calculate $\Theta_7$ as

$$\Theta_7 = \frac{n_7 \cdot \Theta_1}{n_1},$$

or is it required to calculate each angle of refraction one by one until we reach $\Theta_7$?

  • $\begingroup$ That should work. You're going to lose a bunch of light to reflection along the way though... $\endgroup$
    – M. Enns
    Apr 21 '19 at 16:48

Yes, you are right, except for that it is not the division of $\theta$, but $\sin\theta$. However, the real reason what you said is true can be seen by a bit of observation.

$$n_i\sin(\theta_i) = constant$$


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

And for the ray going to the third medium from the second,


What do you see in common (try rearraging them)?

$$n_1\sin\theta_1=n_2\sin\theta_2=n_3\sin\theta_3 ...$$

  • 1
    $\begingroup$ Yesss I forgot the sinuses! That's why I'm bad at physics 😅. Thankss now it totally makes sense!! $\endgroup$
    – Arbri Ibra
    Apr 21 '19 at 16:19

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