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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)...????

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Neglect

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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Chris Jul 11 at 12:05
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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.

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  • $\begingroup$ @user687961 For internal reflection you need the light to approach the interface from within the denser (higher index) medium. I was following your notation and therefore using $r$ for that angle, and it can take any value up to $90^\circ$. If instead you use $i$ always for the angle of incidence then you need to swap over the refractive indices in your math and then you will get $\sin r > 1$ when $i$ excedes the critical angle. $\endgroup$ – Andrew Steane Jul 11 at 8:56

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