One side of an equilateral prism is painted by a transparent material of refractive index \(n_2\). The refractive index of prism is \(1.6\). The minimum value of \(n_2\) required for total internal reflection from painted face is _______.

Given that the prism is an equilateral triangle, all its refracting angles are \(60^\circ\).

The incident ray falls normally on one of the faces of the prism. Therefore, the angle of incidence is \(0^\circ\) and the ray enters the prism undeviated.

From the geometry of the figure, the ray strikes the bottom painted face. In the right-angled triangle formed by the ray and the two faces of the prism, the angle between the left face and the bottom face is \(60^\circ\).

The angle the refracted ray makes with the bottom face is \(90^\circ - 60^\circ = 30^\circ\).

The angle of incidence \(i\) at the bottom painted face is the angle made with the normal to that face:
\(i = 90^\circ - 30^\circ = 60^\circ\)

For total internal reflection to occur at the painted face, the angle of incidence must be greater than or equal to the critical angle \(\theta_c\) for the prism-paint interface:
\(i \ge \theta_c\)
\(\sin i \ge \sin \theta_c\)

The critical angle is given by \(\sin \theta_c = \dfrac{n_2}{n_1}\), where \(n_1 = 1.6\) is the refractive index of the prism and \(n_2\) is the refractive index of the paint.

Substituting the values:
\(\sin 60^\circ \ge \dfrac{n_2}{1.6}\)
\(\dfrac{\sqrt{3}}{2} \ge \dfrac{n_2}{1.6}\)
\(n_2 \le 1.6 \times \dfrac{\sqrt{3}}{2}\)
\(n_2 \le 0.8\sqrt{3} = \dfrac{4\sqrt{3}}{5}\)

The limiting value of \(n_2\) required for total internal reflection is \(\dfrac{4\sqrt{3}}{5}\).

Answer: \(4\sqrt{3}/5\)