Step-by-step solution:

1. Understanding the Phenomenon: Light from the point source S at the bottom of the glass slab travels outwards in all directions. When a ray of light reaches the top surface (glass-air interface), it goes from a denser medium (glass, μ=5/3) to a rarer medium (air, $μ_air ≈ 1$). This can lead to Total Internal Reflection (TIR).

2. Condition for Light Emergence: Light will emerge from the top surface only if the angle of incidence i at the glass-air interface is less than the critical angle $i_c$. If $i > i_c$, the light will be totally internally reflected back into the slab. The limiting case, which defines the boundary of the circular area of emergence, is when the angle of incidence equals the critical angle, $i = i_c$.

3. Calculating the Critical Angle ($i_c$): The critical angle is defined by Snell's law when the angle of refraction is 90°. $μ \sin(i_c) = μ_{air} \sin(90°)$ Given μ = 5/3 and $μ_{air} = 1$: $(5/3) \sin(i_c) = 1 \times 1$ $\sin(i_c) = \frac{1}{5/3} = \frac{3}{5}$

4. Geometrical Setup: Let's visualize the setup. Let the point source be S. Let O be the point on the top surface directly above S. The distance SO is the thickness of the slab, t = 8 cm. Let P be a point on the edge of the circular area from which light emerges. The radius of this circle is R = OP. A light ray SP strikes the surface at P at the critical angle $i_c$.

   The normal to the surface at P is a line parallel to SO. Therefore, the angle of incidence $i_c$ is the angle between the ray SP and the normal. By alternate interior angles, the angle OSP is also equal to $i_c$.

   We now have a right-angled triangle SOP, with:

   • SO = thickness, t = 8 cm (adjacent side to angle $i_c$)
   • OP = radius, R (opposite side to angle $i_c$)

5. Relating Radius, Thickness, and Critical Angle: From the trigonometry of the right-angled triangle SOP: $\tan(i_c) = \frac{\text{opposite}}{\text{adjacent}} = \frac{OP}{SO} = \frac{R}{t}$ Therefore, $R = t \tan(i_c)$.

6. Calculating $tan(i_c)$: We know $\sin(i_c) = 3/5$. We can find $\cos(i_c)$ using the identity $\sin^2(θ) + \cos^2(θ) = 1$. $\cos(i_c) = \sqrt{1 - \sin^2(i_c)} = \sqrt{1 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = \frac{4}{5}$ (We take the positive root as $i_c$ is an acute angle).

   Now, we can find $\tan(i_c)$: $\tan(i_c) = \frac{\sin(i_c)}{\cos(i_c)} = \frac{3/5}{4/5} = \frac{3}{4}$

7. Calculating the Radius (R): Substitute the values of t and $\tan(i_c)$ into the equation for R: $R = t \tan(i_c) = 8 \text{ cm} \times \frac{3}{4}$ $R = 6 \text{ cm}$

   The value of R is 6.