A light ray is falling on a face of an equilateral glass prism as...

When the light ray falls perpendicularly on the first face of the equilateral glass prism, its angle of incidence is $0^{\circ}$. Thus, it enters the prism undeviated.

In the triangle formed at the top vertex, the angle of the prism is $60^{\circ}$, and the angle at the first face is $90^{\circ}$. The angle between the ray inside the prism and the second face is $180^{\circ} - (90^{\circ} + 60^{\circ}) = 30^{\circ}$.

The normal to the second face makes an angle of $90^{\circ}$ with the surface. The angle of incidence at the second face is $90^{\circ} - 30^{\circ} = 60^{\circ}$.

The critical angle for a glass-air interface is approximately $42^{\circ}$. Since the angle of incidence ($60^{\circ}$) is greater than the critical angle, the ray undergoes total internal reflection.

By the law of reflection, the angle of reflection is $60^{\circ}$, so the reflected ray makes an angle of $30^{\circ}$ with the second face.

In the triangle formed by the reflected ray, the second face, and the base, the prism's rightmost angle is $60^{\circ}$. The angle the ray makes with the second face is $30^{\circ}$. Therefore, the angle at which the ray strikes the base is $180^{\circ} - (60^{\circ} + 30^{\circ}) = 90^{\circ}$.

Since the reflected ray strikes the base perpendicularly, its angle of incidence at the base is $0^{\circ}$, and it emerges from the prism undeviated.

This complete path is correctly represented in the diagram corresponding to option (4).