0
$\begingroup$

When the first medium is air, then $\sin(c) = 1/n$ where $c$ is the critical angle and $n$ is the refractive index of $n_2/n_1$ (where $n_2$ is second medium and $n_1$ is in air). So, $1/(n_2/n_1)$ would be $n_1/n_2$.

However, when calculating with the first medium being not air, I find that you calculate $\sin c$ by $\sin(c)= n_2/n_1$. This is the complete reciprocal!

$\endgroup$
0
$\begingroup$

It should be just a problem of definitions of the mediums and their refractive indices. In the general case, when you have two mediums separated by a boundary and with refractive indices $n_1$ and $n_2$, the sine of the critical angle is equal to the ratio between the smaller index and the greater, i.e $$ \sin c = \begin{cases} n_1/n_2 ~ \quad \textrm{if}~n_1<n_2 \\ n_2/n_1 ~ \quad \textrm{if}~n_2<n_1 \\ \end{cases} $$ This is because the critical angle exists only if light passes from the medium with greater refractive index to the other with smaller value, and not viceversa. In that case, there exists an incidence angle $c$ for which the refraction angle of the light turns out to be $\pi/2$ (actually this is the operative definition of the critical angle). From this value and above there will be total internal reflection, i.e. no light passes to the other medium.

At this point I think that your doubt is just a matter of definitions in the different problems you encountered. Maybe in a problem in which air is one of the two mediums, the author prefers to say that $n_1$ is the index of the air and $n_2>n_1$ the other (usually the air is the medium with smaller refractive index in the problems). Conversely, when there is no air, I would say that usually $n_1$ and $n_2$ are respectively just the first and the second medium in which the light passes, with $n_1>n_2$ to allow the possibility of a total internal reflection.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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