1. Key idea: full polarization on reflection at point $O$

When light is reflected from the glass-air interface at point $O$ and the reflected ray is fully polarized, the incidence at $O$ must be the Brewster angle.

For light going from glass $(n_1=\sqrt{3})$ to air $(n_2=1)$,

Hence,

So the ray inside glass strikes the cavity surface at $O$ with angle of incidence $30^\circ$.

2. Use spherical geometry of the cavity

The air cavity has radius $\dfrac{R}{2}$ and is tangent to the outer glass sphere at $O$ (with a very thin glass layer there). Therefore the center of the cavity lies on the same radius through $O$.

Let:

• $C$ = center of the outer sphere,
• $C'$ = center of the cavity,
• $P$ = source point,
• $O$ = point where the ray reflects from the cavity.

Since the cavity radius is $R/2$ and it is internally tangent to the outer sphere at $O$,

Also,

Thus,

At point $O$, the normal to the cavity surface is along $C'O$, which is also along $CO$. So the incident ray $PO$ makes angle $30^\circ$ with $CO$.

3. Where can the source point be so that the reflected ray returns along the shown geometry?

Because the source is inside the glass sphere and the ray reflects at $O$, the relevant line from the source to $O$ makes $30^\circ$ with the radius $CO$.

Now consider the line through $O$ making angle $30^\circ$ with $CO$. Its perpendicular distance from the center $C$ is

That means this line is tangent to the concentric circle of radius $R/2$ centered at $C$. So the source point can correspond to the tangency point on that circle, giving

Then triangle $COP$ has:

• $CO=R$,
• $CP=\dfrac{R}{2}$,
• angle between $OP$ and $OC$ at $O$ equal to $30^\circ$.

This makes triangle $COP$ a right triangle at $P$, and

So,

4. Comparison with stored answer

Stored correct answer: 0.5 OR 0.75

Our derived answer is

So it matches one of the accepted stored answers.