This problem involves finding the radius of the circular area from which light emerges from a glass slab due to total internal reflection. Let's break it down step by step.

Step 1: Understand the Setup

A point source of light is placed at the bottom of a glass slab (on a plane surface). The slab has a refractive index μ = 5/3 and thickness t = 8 cm. Light rays from the source travel through the slab. Some rays hit the top surface at angles greater than the critical angle and undergo total internal reflection. Only rays hitting the top surface at angles less than the critical angle will refract out. The boundary between these two cases forms a circle of radius R on the top surface.

Step 2: Recall the Critical Angle

The critical angle (θ_c) is the angle of incidence in the denser medium (glass) for which the angle of refraction in the rarer medium (air) is 90°. It is given by:

$\sin \left({\theta }_c\right)=\frac{1}{\mu }$

Substituting μ = 5/3:

$\sin \left({\theta }_c\right)=\frac{1}{5/3}=\frac{3}{5}$

Step 3: Relate Geometry to the Critical Angle

Consider a light ray that leaves the point source and hits the top surface of the slab exactly at the critical angle. This ray defines the boundary of the circular area. For this ray, the angle of incidence at the top surface is θ_c. Inside the slab, the ray travels at an angle θ_c with the normal.

From the geometry of the right triangle formed:

$\tan \left({\theta }_c\right)=\frac{R}{t}$

Where R is the radius of the circle on the top surface and t = 8 cm is the thickness of the slab.

Step 4: Find tan(θ_c)

We know sin(θ_c) = 3/5. To find tan(θ_c), we can use the identity:

$\tan \left({\theta }_c\right)=\frac{\sin \left({\theta }_c\right)}{\cos \left({\theta }_c\right)}=\frac{\sin \left({\theta }_c\right)}{\sqrt{1-{\sin }^2\left({\theta }_c\right)}}$

Substituting sin(θ_c) = 3/5:

$\cos \left({\theta }_c\right)=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Therefore,

$\tan \left({\theta }_c\right)=\frac{3/5}{4/5}=\frac{3}{4}$

Step 5: Solve for R

From the geometry equation:

$\tan \left({\theta }_c\right)=\frac{R}{t}$

$\frac{3}{4}=\frac{R}{8}$

Solving for R:

$R=8\times \frac{3}{4}=6\text{cm}$

Final Answer

The value of R is 6 cm.

Related Topics and Formulae

Total Internal Reflection: This phenomenon occurs when light travels from a denser medium to a rarer medium and the angle of incidence is greater than the critical angle. No refraction occurs; all light is reflected back into the denser medium.

Critical Angle (θ_c): Given by $\sin \left({\theta }_c\right)=\frac{{\mu }_{\text{rarer}}}{{\mu }_{\text{denser}}}$. For a glass-air interface, μ_rarer ≈ 1 (air) and μ_denser = μ (glass).

Geometry: The path of the critical ray forms a right triangle within the slab, allowing the use of trigonometric functions like tangent to relate the slab's thickness to the radius of the emergent light circle.