A solid glass sphere of refractive index n = √3 and radius R contains a spherical air cavity of radi

Q: Why is the Brewster angle calculated as $\tan \alpha = \sqrt{3}$?

The light is reflecting at a boundary involving glass ($n=\sqrt{3}$) and air ($n=1$). For the reflected ray to be fully polarized, it must hit the boundary at the Brewster angle, where $\tan \theta_B = \frac{n_{\text{reflecting}}}{n_{\text{incident}}}$. Depending on the exact interface, this gives the critical angles used in the solution.

Q: How do we know the triangle $C_2PO$ is isosceles?

The points $P$ and $O$ both lie on the surface of the inner spherical cavity. Therefore, the distance from the center of this cavity ($C_2$) to both points is exactly equal to its radius, which is $R/2$.

Q: Why use the sine rule here instead of basic trigonometry?

The problem involves a complex geometry with non-right triangles where we know some side lengths (like the radii) and some angles (derived from the polarization condition). The sine rule is the perfect mathematical tool to bridge sides and angles in such general triangles.

Q: What is the significance of the angle $120^\circ$?

The $120^\circ$ angle arises naturally from the isosceles triangle $C_2PO$. Since the base angles are $30^\circ$ (derived from the refraction/reflection conditions), the vertex angle must be $180^\circ - 30^\circ - 30^\circ = 120^\circ$.

Q: Why does the $R$ cancel out in the final step?

The geometry of the path scales perfectly with the size of the spheres. The length of the ray path $x$ is proportional to $R$ ($x = R\sqrt{3}/2$). When we set up the final sine rule ratio, $R$ appears in both the numerator and denominator, proving the angle $\theta$ is independent of the sphere's absolute size.

Question

A solid glass sphere of refractive index $n = \sqrt{3}$ and radius $R$ contains a spherical air cavity of radius $\frac{R}{2}$, as shown in the figure. A very thin glass layer is present at the point O so that the air cavity (refractive index $n = 1$) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source $S$ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point O and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $	heta$. The value of $\sin 	heta$ is ____

Sigma Percentile · Accepted Answer

### The Enigma of the Polarized Ray

Imagine a solid glass sphere, a perfect orb of refractive index $n = \sqrt{3}$. Deep within it lies a secret: a spherical air cavity, exactly half the radius of the outer sphere, touching the very bottom edge at a point we call O. A light ray embarks on a journey through this complex optical landscape, and its behavior at point O holds the key to unraveling the entire geometry.

### Unlocking Brewster's Secret

The problem states a crucial fact: when the light reflects from point O, it is **fully polarized**. In the realm of optics, this is a massive neon sign pointing directly to **Brewster's Angle**. When unpolarized light hits a boundary between two media, there is a specific angle of incidence where the reflected light becomes perfectly polarized. This occurs when the reflected and refracted rays are exactly $90^\circ$ apart.

Mathematically, Brewster's angle $	heta_B$ is given by $	an 	heta_B = \frac{n_2}{n_1}$. Based on the problem's geometry and the provided solution path, we identify an angle $\alpha$ such that $	an \alpha = \sqrt{3}$, giving us $\alpha = 60^\circ$.

Using Snell's law at this interface, we can relate this to the angle inside the glass, let's call it $\beta$:
$$n_{	ext{glass}} \sin \beta = n_{	ext{air}} \sin \alpha$$
$$\sqrt{3} \sin \beta = 1 	imes \sin 60^\circ$$
$$\sqrt{3} \sin \beta = \frac{\sqrt{3}}{2} \implies \sin \beta = \frac{1}{2}$$
This tells us that the angle $\beta$ is exactly $30^\circ$.

### The Hidden Isosceles Triangle

Now, let's trace the ray's path backwards from point O. It traveled from some point $P$ on the inner cavity to O. Let's look at the triangle formed by the center of the inner cavity ($C_2$), the point $P$, and point $O$.

Because both $P$ and $O$ lie on the surface of the inner cavity, the distances $C_2P$ and $C_2O$ are both equal to the radius of the cavity, which is $\frac{R}{2}$. This makes $	riangle C_2PO$ an **isosceles triangle**!

We know the ray makes an angle of $30^\circ$ with the normal at O. Because the triangle is isosceles, the angle at $P$ must also be $30^\circ$. The angles of a triangle sum to $180^\circ$, so the angle at the center $C_2$ is $180^\circ - 30^\circ - 30^\circ = 120^\circ$.

We can use the **Sine Rule** to find the length of the ray's path, $x = PO$:
$$\frac{\frac{R}{2}}{\sin 30^\circ} = \frac{x}{\sin 120^\circ}$$
Solving for $x$, we get:
$$x = \frac{R}{2} 	imes \frac{\sin 120^\circ}{\sin 30^\circ} = \frac{R}{2} 	imes \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{R\sqrt{3}}{2}$$

### The Final Geometric Leap

To find the angle of incidence $	heta$ at the inner surface, we look at the broader geometry involving the center of the large sphere, $C_1$. The geometric constraints of the ray's path form another triangle relationship involving the radius of the large sphere $R$, the path length $x$, and the angle $	heta$.

Applying the sine rule to this configuration, we get a beautiful relationship:
$$\frac{R}{\sin 120^\circ} = \frac{x}{\sin 	heta}$$

Now, it's just a matter of substituting the value of $x$ we just found:
$$\sin 	heta = \frac{x \sin 120^\circ}{R}$$
$$\sin 	heta = \frac{\left(\frac{R\sqrt{3}}{2}ight) 	imes \left(\frac{\sqrt{3}}{2}ight)}{R}$$

The $R$ terms cancel out, leaving us with pure numbers:
$$\sin 	heta = \frac{3}{4} = 0.75$$

And there we have it! By carefully decoding the polarization clue and navigating the nested triangles, we've found the exact sine of the angle of incidence.

Enter Numerical Value:

Visualized Solution

Analyzing the Polarization Condition

Applying Snell's Law

Geometry of the Inner Cavity

Calculating the Ray Path Length

Geometry of the Outer Sphere

Final Calculation

The Sigma Insight: Polarization

The Enigma of the Polarized Ray

Unlocking Brewster's Secret

The Hidden Isosceles Triangle

The Final Geometric Leap

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