Radius of Circular Area of Light Emergence from Glass Slab

To find the radius $R$ of the circular area through which light emerges from the top surface of the glass slab, we consider the phenomenon of total internal reflection.

Light rays traveling from the point source at the bottom of the slab toward the upper surface (glass-air interface) will emerge into the air only if the angle of incidence $i$ at the interface is less than or equal to the critical angle $\theta_c$.

The critical angle $\theta_c$ is determined by Snell's law:
$\sin \theta_c = \frac{1}{\mu}$

Given that the refractive index of the glass slab is $\mu = \frac{5}{3}$:
$\sin \theta_c = \frac{1}{5/3} = \frac{3}{5}$

From this, we can calculate $\tan \theta_c$:
$\tan \theta_c = \frac{\sin \theta_c}{\cos \theta_c} = \frac{\sin \theta_c}{\sqrt{1 - \sin^2 \theta_c}} = \frac{3/5}{\sqrt{1 - (3/5)^2}} = \frac{3/5}{4/5} = \frac{3}{4}$

Let the thickness of the slab be $t = 8\text{ cm}$. The circular area from which light emerges corresponds to a cone of light with semi-vertical angle $\theta_c$. Therefore, the radius $R$ of the base of this cone at the top surface is:
$R = t \tan \theta_c$

Substituting the values:
$R = 8 \times \frac{3}{4} = 6\text{ cm}$

Thus, the value of $R$ is $6$.