# 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)...????

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.

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

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.

• @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. – Andrew Steane Jul 11 at 8:56