Question 11

**Setting up the geometry** - Glass sphere: centre C, radius R. Air cavity: radius R/2, internally tangent to the outer sphere at O, so its centre lies on CO at distance R/2 from C. At O both surfaces share the same normal, namely the line CO. - The ray from S hits the inner side of the outer surface at point P (angle of incidence θ), reflects there, and travels along chord PO to O. **Condition at O (Brewster's law)** The reflected light at O is fully polarized, so the ray strikes the glass–air interface at O at Brewster's angle: tan θ_B = n_air / n_glass = 1/√3 ⟹ θ_B = 30°. Hence the chord PO makes 30° with the normal CO at O. **Relating to θ** Triangle CPO is isosceles (CP = CO = R). The angle at O between PO and CO is 30°, so the angle at P between PO and CP (the normal at P) is also 30°. This is the angle of reflection at P, which by the law of reflection equals the angle of incidence: θ = 30° ⟹ sin θ = 1/2. (Check: sin θ = 0.5 < 1/√3, so the reflection at P is partial, not total — consistent with the ray being "reflected" and light also exiting.) **Answer: sin θ = 0.5**

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