Light is incident from glass ( μ = 1.5 ) into air. Determine the...

The critical angle for the glass-air interface is given by $\sin \theta_c = \dfrac{1}{\mu} = \dfrac{1}{1.5} = \dfrac{2}{3}$.

For refraction, the angle of deviation is $\delta = r - i$. If \delta = 90^{}^{{}^{\circ}}, then r = 90^{}^{{}^{\circ}} + i, which is not possible since the maximum angle of refraction is 90^{}^{{}^{\circ}}.

Therefore, the light must undergo total internal reflection. For total internal reflection, the angle of deviation is \delta = 180^{}^{{}^{\circ}} - 2i.

Given \delta = 90^{}^{{}^{\circ}}, we have:

180^{}^{{}^{\circ}} - 2i = 90^{}^{{}^{\circ}}

2i = 90^{}^{{}^{\circ}} \Rightarrow i = 45^{}^{{}^{\circ}}

We must verify if this angle is greater than the critical angle. Since \sin 45^{}^{{}^{\circ}} = \dfrac{1}{\sqrt{2}} \approx 0.707 and $\sin \theta_c = \dfrac{2}{3} \approx 0.667$, we have $i > \theta_c$.

Thus, total internal reflection occurs and the required angle of incidence is 45^{}^{{}^{\circ}}.