This is a problem from my textbook that I've been struggling to solve:

A ray of light from a laser pointer is incident on the "upper" surface of a glass cube at the angle of 70°. The ray passes through the cube and is incident on the perpendicular side at the critical angle of total reflection. Find the index of refraction of the glass cube.

Here's my attempt at solving this problem:

  1. Denote the index of reflection of the cube as $n_{cube}$ and index of reflection of the surrounding air as $n_{air} = 1$, the incident angle as $\alpha = 70^{\circ}$ and critical angle as $\alpha_{crit}$.
  2. Use Snell's law to express the index of reflection of the cube as: $$\frac{n_{cube}}{n_{air}} = \frac{\sin \alpha}{\sin \beta}$$
  3. Substitute known values: $$n_{cube} = \frac{\sin 70°}{\sin \beta}$$
  4. Notice that $\alpha_{crit} = 90^{\circ} - \beta$ and use the critical angle formula as: $$\frac{n_{air}}{n_{cube}} = \sin \alpha_{crit}$$ $$\frac{1}{n_{cube}} = \sin(90^{\circ} - \beta)$$

I've also come up with this diagram:

This is where I got stuck. I cannot easily substitute $\sin(90^{\circ} - \beta) = \cos \beta$ into $\sin \beta$ and cannot find any other relationship between these values. The textbook, however, uses all the same formulas and somehow arrives at the result of $n_{cube} = 1.37$


1 Answer 1


Your substitution is the right way. Now use $\cos^2(x)+\sin^2(x)=1$ to find $n^2.$

  • $\begingroup$ That worked, thank you. Totally forgot about that identity. $\endgroup$
    – nufflee
    Jun 9 at 19:08

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