A light ray strikes the face AB of a right-angled prism ABC ( μ =...

When the light ray strikes the face $AB$ normally, it enters the prism undeviated and travels parallel to the base $BC$.

At the surface $AC$, the angle between the incident horizontal ray and the surface $AC$ is $\phi$.

Therefore, the angle of incidence $i$ at the surface $AC$ is given by:

$i = 90^{\circ} - \phi$

For total internal reflection to occur at the surface $AC$, the angle of incidence must be greater than or equal to the critical angle $\theta_c$ of the prism:

$i \ge \theta_c$

$90^{\circ} - \phi \ge \theta_c$

$\phi \le 90^{\circ} - \theta_c$

The maximum value of $\phi$ is:

$\phi_{\text{max}} = 90^{\circ} - \theta_c$

Taking the cosine of both sides, we get:

$\cos \phi_{\text{max}} = \cos(90^{\circ} - \theta_c) = \sin \theta_c$

We know that the critical angle is given by $\sin \theta_c = \dfrac{1}{\mu}$.

Given $\mu = 1.50 = \dfrac{3}{2}$, we have:

$\sin \theta_c = \dfrac{1}{1.50} = \dfrac{2}{3}$

Substituting this value, we get:

$\cos \phi_{\text{max}} = \dfrac{2}{3}$

$\phi_{\text{max}} = \cos^{-1}\left(\dfrac{2}{3}\right)