# Total Internal Reflection?

## Could somebody please explain the entirety of Total Internal Reflection in regards to the CIE O level syllabus? I usually use miniphysics.com for help, but even their notes are too difficult for me to comprehend. Please help! Thanks!

Apr 22, 2016

Snell's law helps to understand this.

#### Explanation:

Snell's law tells us how light behaves when it goes from one medium into another.

In the above, Snell's law tells us that :
${n}_{1} \sin {\theta}_{1} = {n}_{2} \sin {\theta}_{2}$

If light is travelling from one medium to a medium with a lower refractive index (or put another way, from one medium to a less optically dense medium) then ${n}_{1} > {n}_{2}$

By Snell's law then ${\theta}_{2} > {\theta}_{1}$ (following the maths of Snell's law)

If you increase the angle of incidence, ${\theta}_{1}$ eventually ${\theta}_{2}$ will increase until it is reflected along the boundary. At this point ${\theta}_{2}$ is ${90}^{0}$ (the critical angle).

Beyond this, and the ray will be reflected - this is total internal reflection.

For example, if medium 1 is glass (${n}_{1} = 1.55$) and medium 2 is a vacuum (${n}_{2} = 1$)

We can see that ${\theta}_{2} = {90}^{0}$ occurs when:

$1.55 \cdot \sin \left({\theta}_{1}\right) = 1 \cdot \sin \left({90}^{0}\right)$
hence
$\sin \left({\theta}_{1}\right) = \frac{1}{1.55}$
${\theta}_{1} = {40}^{0}$
Hence when ${\theta}_{1} > {40}^{0}$ the ray is reflected (total internal reflection)

If you try ${\theta}_{1} = {41}^{0}$ then
$1.55 \cdot \sin \left(41\right) = 1 \cdot \sin \left({\theta}_{1}\right)$
$1.02 = \sin \left({\theta}_{1}\right)$
And ${\theta}_{1}$ cannot be solved. Does this help?!