A solid glass sphere of refractive index n = √3 and radius R contains a spherical air cavity of radius 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 θ. The value of sinθ is ________

The correct answer is 0.75.

To find the value of $\sin \theta$, we break down the path of the light ray using wave optics (Brewster's Law) and the geometry of the two spherical boundaries.

Step 1: Applying Brewster's Law at point O
Let the light ray inside the air cavity ($n_{\text{air}}=1$) be incident on the glass boundary ($n_{\text{glass}}=\sqrt{3}$) at point O. For the reflected light to be fully polarized, the angle of incidence $\alpha$ at point O must satisfy Brewster's Law for light incident from air to glass:

$\tan \alpha =\frac{n_{\text{glass}}}{n_{\text{air}}}=\frac{\sqrt{3}}{1}=\sqrt{3}$

This gives the angle of incidence in air at point O as:

$\alpha =60°$

Step 2: Snell's Law at the Cavity Surface (Point M)
The ray enters the air cavity at point M on its surface, refracting from the glass into the air cavity. Let the angle of incidence in the glass be $\beta$ and the angle of refraction in the air cavity be $\alpha$. Applying Snell's Law at point M:

$n_{\text{glass}}\sin \beta =n_{\text{air}}\sin \alpha$

Substitute the known values:

$\sqrt{3}\sin \beta =1·\sin 60°=\frac{\sqrt{3}}{2}\Rightarrow \sin \beta =\frac{1}{2}$

Thus, the angle in the glass at the cavity boundary is:

$\mathrm{��}=30°$

Step 3: Geometry of the Spherical Cavity and Glass Sphere
Let C be the center of the outer glass sphere of radius R, and C' be the center of the air cavity of radius R/2.
Since the air cavity is tangent to the glass sphere at point O, the points C, C', and O lie along the same vertical line (the diameter).
In the triangle C'OM within the cavity:
- The sides $\mathrm{C\text{'}M}=\mathrm{C\text{'}O}=\frac{R}{2}$ (radii of the cavity).
- Since the angle of incidence at O is $\alpha =60°$, the angle $\angle \mathrm{C\text{'}OM}=60°$.
- As C'OM is an isosceles triangle with a $60°$ angle, it is equilateral. Thus, $\mathrm{OM}=\frac{R}{2}$ and the central angle $\angle \mathrm{MC\text{'}O}=60°$.

Now, consider the triangle CC'M:
- The distance between the centers is $\mathrm{CC\text{'}}=\frac{R}{2}$.
- The radius of the cavity is $\mathrm{C\text{'}M}=\frac{R}{2}$.
- The angle $\angle \mathrm{CC\text{'}M}=180°-\angle \mathrm{MC\text{'}O}=180°-60°=120°$.
- Since $\mathrm{CC\text{'}}=\mathrm{C\text{'}M}=\frac{R}{2}$, triangle CC'M is isosceles, meaning:

$\angle \mathrm{C\text{'}MC}=\angle \mathrm{C\text{'}CM}=\frac{180°-120°}{2}=30°$

Using the Law of Sines in triangle CC'M to find the length $\mathrm{CM}=x$:

$\frac{x}{\sin 120°}=\frac{R/2}{\sin 30°}\Rightarrow x=\frac{R}{2}·\frac{\sin 120°}{\sin 30°}=\frac{R}{2}·\frac{\sqrt{3}/2}{1/2}=R\frac{\sqrt{3}}{2}$

Step 4: Finding the angle of incidence θ at the outer periphery
Let P be the point on the periphery of the glass sphere where the light ray is incident at angle $\theta$. The ray travels along the line PM in the glass.
At point M, the angle of refraction in glass is $\beta =30°$ relative to the normal C'M, and the angle $\angle \mathrm{C\text{'}MC}=30°$.
Thus, the angle between the lines CM and PM is:

$\angle \mathrm{CMP}=180°-(\angle \mathrm{C\text{'}MC}+\beta )=180°-(30°+30°)=120°$

Now apply the Law of Sines in triangle CMP, where $\mathrm{CP}=R$ and $\mathrm{CM}=x=R\frac{\sqrt{3}}{2}$:

$\frac{R}{\sin 120°}=\frac{x}{\sin \theta }$

Solve for $\sin \theta$:

$\sin \theta =\frac{x}{R}$