A solid glass sphere of refractive index n= 3 and radius R contains a spherical air cavity of radius R2, 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 . The value of is [figure]
Answer
0.75
Solution
Reflected light is fully polarised at Brewster incidence: iB= n2n1.
The cavity touches the sphere at O, so a ray reaching O from inside comes through the cavity air and meets the thin glass layer: iB= 3 iB=60 .
In the cavity the ray is a chord making 60 with the cavity radius, so it entered the cavity at B with 60 refraction angle: 3 i= 60 i=30 . Taking O as origin (sphere centre at R, cavity centre at R2 above O), B is at height R4 and the incident ray in glass at B comes out horizontal.
This horizontal ray (after reflection at the sphere surface) is at distance R- R4= 3R4 from the sphere centre, so at the point of reflection = 3R/4R=0.75 (greater than 1 3, so it is total internal reflection).
If the cavity is ignored (glass–air Brewster angle 30 at O and the ray a chord of the sphere), = 30 =0.5.
Answer: 0.75
Official key: 0.5 or 0.75 accepted.