Angles of Refraction,Internal reflection and incidence First ill ask my question in brief and then ill elaborate it below...
Theoretically is it that internal reflection occurs within a solid sphere(which is considered to be the denser medium)  if and only if the angle that is incident to the sphere at the air-sphere interface  $(i) $ is equall to $90°$ and internal reflection will not occur if $i>90°$ or $i <90°$. (Regardless to the materials the sphere is made of).....
What a want to know is that if $i > 90°$ is possible?
Is there anyway $C $ can be greater if $i$ cannot exceed 90°?
Does internal reflection occur in a solid sphere if and only if $i= 90°$??..
I've added an image just to make it clear
Elaborated:-
When it comes to the refraction of light we know that when light travels from dense to rare and by any chance if the angle of incidence $(i) $ is greater than critical angle $(C) $ internal reflection takes place along with refraction....
So this means when the $(i)$ is closser to $C $ then intensity of Refracted light ray is greater than the intensity of the internally Reflected ray and the more the $i $ gets further away from $C $ intensity of internal reflected light increases where as intensity of Refracted light decreases right?.
So  now if we consider a solid spherical ball with refractive index $n $ and $n1$ being the refractive index of the outer medium (say air) $(n> n_1 )$
.....
To find the $C $ at the sphere-air interface... we use 
\begin{eqnarray}
n_1&\sin90°=n&\sin C\\
\end{eqnarray}
So solving this we get $C=n_1/n$ which is the critical angle.
And since its a sphere we find the $i $ at the air-sphere interface taking the refracted angle to be $r $ and using basic geometry since the normals of both instances is the radius of sphere and im considering light to travel on the same plane we get $r=C $
And now we find angle $i $ using
\begin{eqnarray}
n_1 \sin i&=&n \sin r\\
n_1 \sin i&=&n(n_1/n) [r=C]\\
\sin i&=&1\\
i&=&90°
\end{eqnarray}
Therefore only a ray of $i= 90°$ will cause the ray to internally reflect within the sphere as the angles inside the sphere becomes $C $ and anything less than that wont (since the angle inside the sphere gets less than $C $)
...but the angle $i $ cannot be greater than $90°$ as it will result the ray to enter the sphere from a different location. ....
So theoretically is it that internal reflection occurs within a solid sphere (which is considered to be the denser medium)  if and only if the angle that is incident to the sphere at the air-sphere interface  $(i)$ is equall to $90°$ and internal reflection will not occur if $i>90°$ or $i <90°$. (Regardless to the materials the sphere is made of)...????
 A: You made a mistake near the end of your derivation. When you say "but the angle $i$ cannot be greater than $90^\circ$" this is a true statement, but misleading. What the maths says is that when angle $r$ is greater than the critical angle then the refraction formula gives 
$$\sin i > 1 .$$ 
There is no (real) angle satisfying this condition! It is telling us not that $i$ is greater than $90^\circ$ but that there is no way for the light to pass out of the denser medium while satisfying laws of physics at the boundary---so the light stays inside the denser medium for all angles $r$ greater than the critical angle.
A: 
I AM NOT ASSUMING SPHERE TO BE MADE OF WATER(it just comes with the image).Let us assume for a second that total internal reflection does occur. So we have 
$sin(r)>\frac{n_\perp}{n}$ :condition for Total internal reflection.
Now let us find out at which value of $i$ this is possible.
$n_\perp sin(i)=n sin(r)$ 
This gives us $sin(i)>1$ which is not possible.
The conclusion is that TOTAL INTERNAL REFLECTION IS NOT POSSIBLE FOR A SPHERE OF HIGHER REFRACTIVE INDEX.
The case of $i=90$ is also not a case of total internal reflection because the light is not reflected it is just emitted parallel to the surface after undergoing normal refraction. Hope this is clear.
