A large glass slab (μ = 5/3) of thickness 8 cm is placed over a point source of light on a plane sur

Q: Why does the light form a circular area and not a square or another shape?

Because the point source emits light uniformly in all directions. The critical angle condition depends only on the angle of incidence, which forms a cone of light. The intersection of this cone with the flat top surface is a perfect circle.

Q: What happens to the light rays that hit the surface outside this circular area?

Rays hitting outside the circle have an angle of incidence greater than the critical angle. They undergo Total Internal Reflection (TIR) and bounce back into the glass slab, never escaping into the air.

Q: Why did we calculate $\tan \theta_C$ instead of just using $\sin \theta_C$?

The geometry of the setup forms a right triangle where the radius $R$ is the opposite side and the thickness $t$ is the adjacent side to the angle $\theta_C$. The tangent function directly relates the opposite and adjacent sides ($R/t$).

Q: Would the radius change if the glass slab was submerged in water instead of air?

Yes! The critical angle depends on the relative refractive index between the two media ($\sin \theta_C = \mu_{\text{outside}} / \mu_{\text{inside}}$). If the outside medium is water, the critical angle would increase, making the circular area larger.

Q: Does the color (wavelength) of the light source affect the size of the circle?

Technically, yes. The refractive index of glass slightly varies with the wavelength of light (dispersion). Therefore, different colors will have slightly different critical angles, causing the edges of the circle to have a faint rainbow fringe.

Question

A large glass slab $\left(\mu = \frac{5}{3}\right)$ of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius $R$ cm. What is the value of $R$?

Sigma Percentile · Accepted Answer

The phenomenon of light escaping from a denser medium into a rarer medium is governed by the principles of refraction and Total Internal Reflection (TIR). When a point source of light is placed at the bottom of a glass slab, it emits rays in all directions.

### The Phenomenon of the Trapped Light

As these rays travel upwards and hit the glass-air interface, they refract away from the normal. However, there is a strict limit to this escape. According to Snell's Law, as the angle of incidence increases, the angle of refraction also increases until it reaches exactly $90^\circ$. The angle of incidence at which this happens is called the **critical angle** ($	heta_C$).

Any light ray hitting the surface at an angle greater than $	heta_C$ will undergo Total Internal Reflection and bounce back into the glass. Therefore, the light can only emerge into the air within a cone defined by the critical angle. When this cone intersects the flat top surface of the slab, it forms a perfectly circular illuminated area.

### The Geometry of the Escape Cone

To find the radius $R$ of this circular area, we can visualize the geometry of the setup. The thickness of the slab $t$, the radius of the circle $R$, and the path of the critical ray form a right-angled triangle.

From this triangle, we can establish a simple trigonometric relationship:
$$ 	an 	heta_C = \frac{R}{t} $$
$$ R = t 	an 	heta_C $$

### The Mathematical Execution

We are given the refractive index of the glass slab as $\mu = \frac{5}{3}$. The sine of the critical angle is the reciprocal of the refractive index:
$$ \sin 	heta_C = \frac{1}{\mu} = \frac{1}{5/3} = \frac{3}{5} $$

To use our geometric formula, we need $	an 	heta_C$. If we imagine a right triangle where the perpendicular is $3$ and the hypotenuse is $5$, the base must be $\sqrt{5^2 - 3^2} = 4$. Therefore, the tangent of the critical angle is:
$$ 	an 	heta_C = \frac{3}{4} $$

Now, we simply substitute this value and the given thickness $t = 8 	ext{ cm}$ into our radius equation:
$$ R = 8 	imes \frac{3}{4} = 6 	ext{ cm} $$

The radius of the circular area from which light emerges is exactly **$6 	ext{ cm}$**.

Animated Solution for Physics - Optics: A large glass slab $(μ = \frac{5}{3})$ of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius $R$ cm. What is the value of $R$ ?

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{The Critical Ray}

\text{Geometry of the Illuminated Area}

\text{Finding } \sin \theta_C

\text{Finding } \tan \theta_C

\text{Calculating the Radius } R

\text{The Way Forward}

The Sigma Insight: Refraction and Total Internal Reflection

The Phenomenon of the Trapped Light

The Geometry of the Escape Cone

The Mathematical Execution

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