The refractive index of haematite is 3.2 - Leaving Cert Physics - Question (c) - 2015

The critical angle can be calculated using Snell's law, which states that:

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

Where:

• n is the refractive index of the second medium (air in this case, approximately 1)
• C is the critical angle

Rearranging the equation gives:

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

Substituting the value of the refractive index (n = 3.2): $\sin C = \frac{1}{3.2}$

Now, calculating ( C ):

$C = \sin^{-1}\left(\frac{1}{3.2}\right)$

Calculating this using a calculator gives: $C \approx 18.21^\circ$

Therefore, the critical angle for haematite is approximately 18.21°.