If the critical angle for a glass–air boundary is 45º, what is the refractive index of the glass?

Answer Details

The critical angle \(C\) is the angle of incidence inside the denser medium at which the refracted ray just grazes along the boundary, so the angle of refraction in air is \(90^\circ\). Applying Snell's law at the glass-air boundary, \[n_{g}\sin C = n_{a}\sin 90^\circ.\] Taking \(n_a = 1\) for air and \(\sin 90^\circ = 1\), this rearranges to the standard result \[n = \frac{1}{\sin C}.\]

Substituting \(C = 45^\circ\), for which \(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\): \[n = \frac{1}{1/\sqrt{2}} = \sqrt{2} \approx 1.41.\] The refractive index of the glass is \(\sqrt{2}\).

The frequent error is to write \(n = \sin C\), giving \(0.71\), a value less than one that would describe a medium in which light travels faster than in air. A refractive index for a denser medium relative to air is always greater than \(1\), so \(n = 1/\sin C\) is the correct arrangement, and \(\sin C\) small means \(n\) large. Keep the physical consequence in mind too: at any angle of incidence greater than \(45^\circ\) inside this glass, no light escapes and total internal reflection occurs, which is the principle behind optical fibres, prism periscopes and the sparkle of cut gemstones.