On Hyperphysics, it says that the threshold index of refraction $n_t$ for some incident angle $\theta_i$ is (in radians);$$ n_isin\theta_i=n_tsin\frac\pi2 $$ but isn't $sin\frac\pi2$ equal to one? so would this be the correct equation for the threshold index of refraction?$$ n_isin\theta_i=n_t? $$ or displaying it as an inequality, total internal reflection occurs if:$$ n_2\geq n_isin\theta_i $$ When $0\leq\theta_i\leq\pi$ Is this correc?
Short answer: yes.
Long answer: You start with Snell's Law $n_i\sin\theta_i = n_r\sin\theta_r$ and set the angle of refraction $\theta_r = \pi/2$ (as total internal reflection occurs when the refracted wave would not leave the medium). That is why you are left with $n_i\sin\theta_i = n_r$. The angle of refraction would be even larger if $n_r$ was larger. Thus $n_r$ above is actually the threshold idex of refraction $n_t$ and for every $n_r$ greater than the threshold the total internal reflection still occurs. $$n_r \ge n_t = n_i\sin\theta_i$$
The critical angle is precisely the one that separates total and partial reflection. Consequently, it is only one value and it must be a equality, and not an inequality.
You are right, you get that equation when $\sin \theta_r=1$. That's because it means that the reflected ray exits at $90º$ from the normal, that is, parallel to the frontier.
If the angle is minor, it will be no longer in the frontier, but closer to the normal, and consequently it will be just another case or ordinary refraction.
If the angle is greater than $90º$, the equation has no real solution and there is not ordinary refraction but total reflection.
That's why $\theta_r=\pi/2$ is precisely the limiting case, and that's why it's corresponding $\theta_i$ is called critical angle or "limiting angle".