I look at what TIR is and then what reflection is – when the angle of incidence and the angle of election are the same. And TIR is when the angle of incidence is greater than the critical angle. TIR reflects with the angle of incidence=angle of refraction which is the same as reflection. are they both the same thing?
[Physics] the difference between reflection and total internal reflection (TIR)
opticsreflectionrefraction
Related Solutions
Answer: you can ignore the coating (assuming monotonic index of refraction); light does not exit if incident ray is beyond critical angle
Reasoning: Snell's Law states $$n_1 \sin \theta_1 = n_2 \sin \theta_2,$$ where $n_1$ and $n_2$ are indexes of reflection within media $M_1$ and $M_2$, and $\theta_1$ and $\theta_2$ are the angle between the normal and the light ray within the respective media.
I presume by "prism" you mean an interface $M_1$ to $M_2$, with $n_1 > n_2$.
Total internal reflections (TIR) occur when $\sin \theta_2 = \frac{n_1}{n_2} \sin\theta_1 > 1$, i.e., where transmission would be absurd since there is no real solution for $\theta_2$.
Now consider the case with $3$ media $M_1$, $M_2$, $M_3$ with parallel interfaces. Snell's law becomes $$n_1 \sin \theta_1 = n_2 \sin \theta_2 = n_3 \sin \theta_3 = \text{constant}.$$ So it doesn't mater what $n_2$ is. As long as $\sin\theta_3 = \frac{\text{constant}}{n_3} > 1$, then $\theta_3$ has no real solution, and light will not enter $M_3$ (in fact, TIR may occur from the $M_1$-$M_2$ interface and never reach $M_3$ in the first place).
We can generalize the above to multiple layers. In the limit we would have a (assumed monotonic) gradient of index of fractions from $n_1$ to $n_2$, for each plane parallel to a common plane (from which the normal is defined).
Now consider a light ray coming from $M_1$ that would normally undergo TIR against $M_2$. For the gradient case, the light ray would bend and become increasing parallel to the plane, until it encounters a layer $M_i$ with $\frac{n_1}{n_i} \sin \theta_1 = 1$. Thereafter the ray bends back toward $M_1$, effectively undergoing TIR.
(Edit: updated "Answer" section part to match what the question is asking)
The conditions for total internal reflection from an air-common glass interface are:
The light has to be travelling in the glass.
The angle of incidence has to be larger than the critical angle.
Best Answer
Total Internal Reflection is an example of reflection. In TIR and other forms of reflection (e.g. reflection off of a mirror or other barrier) the angle of incidence will be equal to the angle or reflection. You wrote "TIR reflects with the angle of incidence=angle of refraction." I'm not sure if this is a typo or if this is what you intended but "refraction" in that sentence should read "reflection" - there is no refraction in the case of TIR.
All TIR is reflection but not all reflection is TIR so in that sense they are not the same thing.