Objective

• Deriving the equation for the critical angle

The critical angle formula can be derived using Snell’s law;

$n_{1} \sin{\theta_{1}} = n_{2} \sin{\theta_{2}}$

The image to the right shows an example of where the critical angle can be observed (lets assume the semicircular block is made from glass and the medium surrounding this block is air).

Using this image we can write the following;

$\theta_{1} = C$
$\theta_{2} = 90^{o}$
$n_{1} = n_{glass}$
$n_{2} = n_{air}$

And so, substituting these into Snell’s law gives us;

$n_{glass} \sin{C} = n_{air} \sin{90}$

Since  $\sin{90} = 1$ ;

$n_{glass} \sin{C} = n_{air}$

Rearranging for $\sin{C}$ gives;

$\sin{C} = \frac{n_{air}}{n_{glass}}$

The part many books brushes over and can therefore be confusing is changing $\frac{n_{air}}{n_{glass}}$ into $\frac{1}{n}$ .

The way it is done, is by considering the value for the absolute refractive index of air, $n_{air} = 1.0003 \approx 1$

Now we can write

$\sin{C} = \frac{1}{n_{glass}}$

Since we only need to deal with one refractive index, $n_{glass}$ becomes simply $n$ , and so;

$\sin{C} = \frac{1}{n}$

where n is the refractive index of the more dense medium.